# NCERT Solutions Class 9 Mathematics

Mathematics is considered to be a difficult subject. Students struggle in CBSE Class 9 Mathematics due to the increased complexity of themes and difficulty level of problems. It’s important to take the right guidance and optimise your potential. No wonder, NCERT Solutions come handy and it’s worth giving a shot because NCERT Solutions for Class 9 Mathematics will provide practical problem-solving solutions for CBSE 9th grade students while tackling the practice and in-text questions in the textbook.

For students  preparing for CBSE examinations and state or competitive tests such as Olympiads, NCERT solutions Class 9 are undoubtedly useful. The NCERT textbooks follow the guidelines for the CBSE examinations. Furthermore, these solutions assist students in preparing for class 10th and the equivalent board exams.

Benefits of NCERT Mathematics Solutions for Class 9

• Students can  improvise their preparation strategy by using the NCERT 9th Class Mathematics Books Solutions for Class 9 Mathematics syllabus.
• Experts at Extramarks have especially designed it to provide you with the finest in one spot. All solutions are written by subject specialists in  simple language for students to help themselves without any further assistance and become independent learners.
• Students are provided diagrams to assist them visualise the problems and solutions.
• Detailed and explanatory solutions will help you enhance your skills and gain a solid understanding of the subject.
• Students with patience and diligence can effortlessly grasp all subjects and achieve the highest possible grades in their tests.

Weightage For the First Term

Students should complete the higher-weightage subjects first before moving on to the lower-weightage units during their exam preparation. This allows students to focus on the most important topics while also building confidence in their ability to perform well in the annual exam.

It is recommended that students assess all of the concepts covered in the Class 9 Mathematics Syllabus and devise an appropriate study plan.

 Unit Name Marks Number Systems 8 Algebra 5 Coordinate Geometry 4 Geometry 13 Mensuration 4 Statistics and Probability 6

Weightage For the Second Term

 Unit Name Marks Algebra 12 Geometry 15 Mensuration 9 Statistics and Probability 4

NCERT Solutions for Class 9 Mathematics | Chapter-wise Solutions

(Add chapters 1 to 15 in a sequence)

CBSE Class 9 Mathematics NCERT Solutions enable you to properly planning your exam study strategy and ensure that you grasp all Mathematical concepts prior to the final exam. Practice these solutions on a daily basis to strengthen the fundamental ideas of class 9 Mathematics. The links to these solutions are provided below.

Chapter 1 – Number Systems

Chapter 2 – Polynomials

Chapter 3 – Coordinate Geometry

Chapter 4 – Linear Equations in Two Variables

Chapter 5 – Introduction to Euclid’s Geometry

Chapter 6 – Lines and Angles

Chapter 7 – Triangles

Chapter 9 – Areas of Parallelograms and Triangles

Chapter 10 – Circles

Chapter 11 – Constructions

Chapter 12 – Heron’s Formula

Chapter 13 – Surface Areas and Volumes

Chapter 14 – Statistics

Chapter 15 – Probability

Students can improve their theoretical skills and practical understanding by practising these NCERT Mathematics book class 9 solutions.   These Solutions will teach you how to approach problems and solve them efficiently using step-by-step methods. . As a student, you must constantly revise and practise all of the solutions. Begin by answering the questions in the NCERT book, and then cross check with  NCERT Solutions.

### NCERT Solutions for CBSE Class 9 Mathematics Chapters and Exercises

If you are confused and anxious about any of the problems, you can use our subject expert guidance for  CBSE class 9 Mathematics NCERT solutions. They have meticulously prepared these solutions, concentrating on the types of errors students commonly make during exams. The Solutions for Class 9 Mathematics contain 15 chapters and are open to all students.

### NCERT Solutions for Class 9 Mathematics Chapter 1 Number System

Students will learn about the notion of an irrational number, real numbers, and their decimal expansions. They will also learn about the procedures of real numbers on the number line, various types of operations on real numbers, and the laws of exponents for real numbers. There are 7 exercises in all; complete the activities and consult our book if you have any questions. The links to these exercises are given below.

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

Exercise 1.5

Exercise 1.6

### NCERT Solutions for Class 9 Mathematics Chapter 2 Polynomials

Students will study about algebraic identities, one-variable polynomials, polynomial zeros, the remainder theorem, and polynomial factorization. You learned factorization of algebraic expressions in earlier classes; now apply those concepts and learn some new ones when solving polynomials. There are seven exercises in this chapter. Download links to various exercises have been shared for further practice.

Exercise 2.1

Exercise 2.2

Exercise 2.3

Exercise 2.4

Exercise 2.5

Exercise 2.6

Exercise 2.7

Class 9 Mathematics NCERT Solutions Chapter 3 Coordinate Geometry

In earlier classes, you examined the position of you can practice through the links given below.

point on a given line. Through this chapter, you will learn how to find a point that is not on that line. Students will learn about the Cartesian system and how to use coordinates to map a point in the plane. Make a graph with a point on it using the X and Y axes as a reference. There are three exercises and a summary in this chapter.

Exercise 3.1

Exercise 3.2

Exercise 3.3

### NCERT Solutions for Class 9 Mathematics Chapter 4 Linear Equations in Two Variables

You learned linear equations in one variable in earlier classes, and now it’s time to add another variable. In this chapter, students will learn what a linear equation is, how to solve linear equations, how to draw a graphical representation of a linear equation in two variables, and how to solve equations of lines parallel to the x- and y-axes.  There are four exercises in this chapter.  The links to these exercises are given below.

Exercise 4.1

Exercise 4.2

Exercise 4.3

Exercise 4.4

### NCERT Solutions for Class 9 Mathematics Chapter 5 Introduction to Euclid’s Geometry

Students will study about Euclid’s definition, axioms, and postulates, as well as equivalent versions of Euclid’s fifth postulate. The theorem claims that two separate lines cannot have more than one common point. To better understand the chapter, complete the exercises and learn about Euclid’s postulates. Download the links below for further practice.

Exercise 5.1

Exercise 5.2

### NCERT Solutions for Class 9 Mathematics Chapter 6 Lines and Angles

This chapter explains to students the fundamental terms and definitions of lines and angles. Intersecting and non-intersecting lines, transversal and parallel lines, lines parallel to the same line, pairs of angles, and the angle sum property of a triangle will all be covered.  This chapter has three exercises.

Exercise 6.1

Exercise 6.2

Exercise 6.3

### NCERT Solutions for Class 9 Mathematics Chapter 7 Triangles

Students will learn about the SSS, SAS, ASA, and RHS congruence criteria in this chapter. Students can get a thorough understanding of the properties of triangles such as sides opposite to equal angles being equivalent, the total of any two sides of a triangle being more than the third side, and so on. As a result of studying this chapter, students will be able to master the fundamental ideas of a triangle. Download the links below for further practice.

Exercise 7.1

Exercise 7.2

Exercise 7.3

Exercise 7.4

Exercise 7.5

### NCERT Solutions for Class 9 Mathematics Chapter 8 Quadrilaterals

Questions in Chapter 8 quadrilaterals teach students about the  various types of quadrilaterals, such as squares, rhombuses, rectangles, and parallelograms, as well as their properties. . Angle sum property of quadrilaterals, qualities of a parallelogram, square, and rhombus that depend on multiple factors such as diagonal length, side measure, and so on are all examined. The mid-point theorem is also included. Download the links below for further practice.

Exercise 8.1

Exercise 8.2

### NCERT Solutions for Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles

In this chapter, students will study figures with the same base and parallels, parallelograms with the same base and parallels, and triangles with the same base and parallels. Learn about trapeziums as well. This chapter has four exercises.

Exercise 9.1

Exercise 9.2

Exercise 9.3

Exercise 9.4

### NCERT Solutions for Class 9 Mathematics Chapter 10 Circles

Students will learn about chord characteristics and how to determine the distance of equal chords from the centre. Other topics covered include angles subtended by an arc of the circle, cyclic quadrilaterals, and related properties. In Chapter 10, students will study about circles and the many terms related with it. You can practice through the links given below.

Exercise 10.1

Exercise 10.2

Exercise 10.3

Exercise 10.4

Exercise 10.5

Exercise 10.6

### NCERT Solutions for Class 9 Mathematics Chapter 11 Constructions

Triangles have already been addressed in previous chapters. However, those were simply for you . Students will learn about the foundations of construction and experiment with triangle construction in this chapter. Learn how to use a compass, draw perpendicular lines, and measure angles with a protector. This chapter may appeal to you if you have a solid understanding of construction. Practice with the links given below.

Exercise 11.1

Exercise 11.2

### NCERT Solutions for Class 9 Mathematics Chapter 12 Heron’s Formula

The Hero of Alexandria’s formula for calculating the area of a triangle when the lengths of all three sides are given.. Students will also learn how to compute the area of triangles and quadrilaterals in this chapter. Download the links below for further practice.

Exercise 12.1

Exercise 12.2

### NCERT Solutions for Class 9 Mathematics Chapter 13 Surface Areas and Volumes

Surface areas and volumes in Chapter 13 introduce students to the ideas necessary to calculate total surface areas, lateral or curved surface areas, and volumes of various shapes.  The concepts and formulas of this chapter  are  crucial because they  serve as the foundation for topics  covered in higher classes. . Download the links below for further practice.

Exercise 13.1

Exercise 13.2

Exercise 13.3

Exercise 13.4

Exercise 13.5

Exercise 13.6

Exercise 13.7

Exercise 13.8

Exercise 13.9

### NCERT Solutions for Class 9 Mathematics Chapter 14 Statistics

Statistics in Chapter 14 gives an in-depth analysis of problems based on the gathering and presentation of data for analytical purposes. Data representation in the form of frequency tables and drawing inferences from them are discussed. This chapter helps students improve their    Mathematical problem-solving skills and develops their spatial visualising abilities. Download the links below for further practice

Exercise 14.1

Exercise 14.2

Exercise 14.3

Exercise 14.4

### NCERT Solutions Class 9 Mathematics Chapter 15 Probability

This chapter supports students in dealing with real-world challenges that necessitate the use of probability concepts. This chapter introduces students to empirical probability, the definition of an event, and how to compute the probability of various events. It explains how to calculate the probability of a heads or tails coin flip, by tossing a die and drawing inferences based on the outcome. Check the given link below.

Exercise 15.1

### CBSE Class 9 Mathematics Unit-wise Marks Weightage

#### Weightage For the First Term

 Unit Name Marks Number Systems 8 Algebra 5 Coordinate Geometry 4 Geometry 13 Mensuration 4 Statistics and Probability 6 Total 40 Internal Assessment 10 Term I Total 50

Internal Assessment for Term I

 Internal Assessment Marks Periodic Tests 3 Multiple Assessments 2 Portfolio 2 Student Enrichment Activities- Practical Work 3 Total 10

Weightage For the Second Term

 Unit Name Marks Algebra 12 Geometry 15 Mensuration 9 Statistics and Probability 4 Total 40 Internal Assessment 10 Term II Total 50

Internal Assessment for Term II

 Internal Assessment Marks Periodic Tests 3 Multiple Assessments 2 Portfolio 2 Student Enrichment Activities- Practical Work 3 Total 10

Q.1 In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Ans

$\begin{array}{l}\mathrm{Since},\text{y:z}=\text{3:7}\\ \mathrm{Let}\text{y}=3\mathrm{a}\text{and z}=\text{7a}\\ \text{Since,}\mathrm{AB}\mathrm{CD}\text{and CD}\mathrm{EF}\\ \mathrm{so},\text{}\mathrm{AB}\mathrm{EF}\\ \mathrm{x}=\mathrm{z}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}\mathrm{CD}\\ \mathrm{x}+\mathrm{y}=180\mathrm{°}\left[\text{Cointerior angles}\right]\\ \mathrm{z}+\mathrm{y}=180\mathrm{°}\left[âˆµ\mathrm{x}=\mathrm{z}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7a}+3\mathrm{a}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{a}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{180\mathrm{°}}{10}=18\mathrm{°}\\ \mathrm{So},\text{x}=\text{7a}\\ =\text{7}\left(18\mathrm{°}\right)\\ =126\mathrm{°}\end{array}$

Q.2 In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{EF}\perp \mathrm{CD}\text{and}\angle \text{GED=126°.}\\ \text{To find:}\angle \mathrm{A}\text{GE,}\angle \text{GEF and}\angle \mathrm{F}\text{GE}\\ \text{AB}\text{CD}\\ \angle \text{AGE}=\angle \text{GE}\mathrm{D}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=126\mathrm{°}\\ \angle \text{GED}=\angle \text{GEF}+\angle \text{FED}\\ \text{126°}=\angle \text{GEF}+90\mathrm{°}\\ \angle \text{GEF}=126\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36\mathrm{°}\\ \angle \mathrm{A}\text{GE}+\angle \mathrm{F}\text{GE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}126\mathrm{°}+\angle \mathrm{F}\text{GE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{F}\text{GE}=180\mathrm{°}-126\mathrm{°}\\ =54\mathrm{°}\end{array}$

Q.3 In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\mathrm{ST},\text{}\angle \text{PQR}=\text{11}0\mathrm{°},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{RST}=\text{13}0\mathrm{°}\\ \mathrm{To}\text{prove:}\angle \text{QRS}\\ \text{Construction:\hspace{0.17em}Draw line}\mathrm{l}\text{paralllel to ST through R.}\\ \text{Since, ST}\stackrel{↔}{\mathrm{l}}\text{, then}\\ \angle \text{\hspace{0.17em}}\mathrm{TSR}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Co}-\mathrm{interior}\text{â€‹ angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}130\mathrm{°}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}-130\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50\mathrm{°}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\mathrm{ST}\text{and ST}\stackrel{↔}{\mathrm{l}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\stackrel{↔}{\mathrm{l}}\\ \angle \mathrm{PQR}=\angle \mathrm{SRW}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}110\mathrm{°}=\angle \mathrm{QRS}+\angle \mathrm{SRW}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\angle \mathrm{QRS}+50\mathrm{°}\\ \angle \mathrm{QRS}=110\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}\end{array}$

Q.4 In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\text{5}0\mathrm{°}\text{\hspace{0.17em}and}\mathrm{}\text{\hspace{0.17em}}\angle \text{PRD}=\text{127°}\\ \text{To find: x and y}\\ \text{Since, AB}\text{CD}\\ \text{So,}\angle \text{APR}=\angle \text{PRD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}50°}+\text{y}=127\mathrm{°}\\ ⇒\mathrm{y}=127\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=77\mathrm{°}\\ \text{And, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\angle \text{PQR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}50\mathrm{°}=\mathrm{x}\\ \mathrm{Thus},\text{x}=\text{50° and y}=\text{77°.}\end{array}$

Q.5 In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\mathrm{RS},\text{\hspace{0.17em}AB is incident ray to PQ and BC is incident ray}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}to RS.}\\ \text{To Prove: AB}\text{CD}\\ \text{Construction: Draw MB}\perp \text{PQ and NC}\perp \text{RS.}\\ \text{Proof}:\text{}\angle \text{ABM}=\angle \mathrm{MBC}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCN}=\angle \mathrm{NCD}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\end{array}$

$\begin{array}{l}\text{Since, perpendiculars to parallel lines are parallel.So,}\\ \text{\hspace{0.17em}\hspace{0.17em}MB}\text{NC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{M}\text{BN}=\angle \mathrm{BCN}\\ \text{\hspace{0.17em}}\frac{1}{2}\angle \mathrm{A}\text{BC}=\frac{1}{2}\angle \mathrm{BCD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}\text{BC}=\angle \mathrm{BCD}\\ \mathrm{Since},\text{alternate interior angles are equal, so}\\ \text{\hspace{0.17em}\hspace{0.17em}AB}\text{CD.}\end{array}$

Q.6 In Fig. 6.39, sides QP and RQ of Δ PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{PQR,\hspace{0.17em}}\mathrm{Ext}.\angle \text{SPR=135° and}\mathrm{Ext}.\angle \text{PQT=110°}\\ \text{To Find:}\angle \text{PRQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}+\angle \text{PQT}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}+110\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=180\mathrm{°}-110\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=70\mathrm{°}\\ \mathrm{Ext}.\text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{S}\text{PR}=\angle \text{PQR}+\angle \text{PRQ}\\ \text{\hspace{0.17em}135°}=70\mathrm{°}+\angle \text{PRQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PRQ}=135\mathrm{°}-70\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=65\mathrm{°}\end{array}$

Q.7 In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠ YOZ.

Ans

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{XYZ,\hspace{0.17em}\hspace{0.17em}}\angle \text{X}=\text{62}\mathrm{°},\angle \text{\hspace{0.17em}XYZ}=\text{54}\mathrm{°}\text{, YO}\mathrm{}\text{\hspace{0.17em}and ZO are the}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}bisectors of\hspace{0.17em}}\angle \text{XYZ and}\angle \text{XZY}\mathrm{}\text{\hspace{0.17em}respectively.}\\ \mathrm{To}\text{find:}\angle \text{OZY and}\angle \text{YOZ}\\ \text{In}\mathrm{\Delta }\text{XYZ,}\\ \text{}\angle \text{\hspace{0.17em}XYZ}+\angle \text{\hspace{0.17em}}\mathrm{Y}\text{XZ}+\angle \text{\hspace{0.17em}YZX}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}54\mathrm{°}+62\mathrm{°}+\angle \text{\hspace{0.17em}YZX}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}YZX}=180\mathrm{°}-116\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=64\mathrm{°}\\ \text{\hspace{0.17em}}\angle \text{\hspace{0.17em}OZY}=\frac{1}{2}\angle \text{\hspace{0.17em}YZX}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×64\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=32\mathrm{°}\\ \text{\hspace{0.17em}}\angle \text{\hspace{0.17em}OYZ}=\frac{1}{2}\angle \text{\hspace{0.17em}XYZ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×54\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=27\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{XYZ,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}OYZ}+\angle \text{\hspace{0.17em}OZY}+\angle \text{\hspace{0.17em}}\mathrm{Y}\text{OZ}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}27\mathrm{°}+32\mathrm{°}+\angle \text{\hspace{0.17em}}\mathrm{Y}\text{OZ}=180\mathrm{°}\\ \angle \text{\hspace{0.17em}}\mathrm{Y}\text{OZ}=180\mathrm{°}-59\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=121\mathrm{°}\end{array}$

Q.8 In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{AB}\mathrm{DE},\text{\hspace{0.17em}}\angle \text{BAC}=\text{35}\mathrm{°}\text{and}\mathrm{}\text{\hspace{0.17em}}\angle \text{CDE}=\text{53}\mathrm{°}\\ \mathrm{To}\text{find:\hspace{0.17em}}\angle \text{DCE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sol:\hspace{0.17em}Since, AB}\text{CD, so}\\ \angle \text{DE}\mathrm{A}=\angle \text{BAE}\left[\mathrm{Alternate}\text{â€‹â€‹ interior angles}\right]\\ \angle \text{DE}\mathrm{A}=35\mathrm{°}\\ ⇒\angle \text{CED}=35\mathrm{°}\\ \mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{CDE,}\\ \angle \text{CDE}+\angle \mathrm{D}\text{CE}+\angle \text{CED}=\text{180°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{angle sum property.}\right]\\ \text{\hspace{0.17em}}53\mathrm{°}+\angle \mathrm{D}\text{CE}+35\mathrm{°}=\text{180°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}\text{CE}=180\mathrm{°}-88\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}\text{CE}=92\mathrm{°}\end{array}$

Q.9 In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\angle \text{PRT}=\text{4}0\mathrm{°},\angle \text{RPT}=\text{95}°\mathrm{}\text{and\hspace{0.17em}\hspace{0.17em}}\angle \text{TSQ}=\text{75}\mathrm{°}\\ \mathrm{To}\text{find:}\angle \text{SQT}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{In}\text{}\mathrm{\Delta }\text{PRT,}\\ \angle \text{PRT}+\angle \text{PTR}+\angle \text{RPT}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}40\mathrm{°}+\angle \text{PTR}+95\mathrm{°}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PTR}=180\mathrm{°}-135\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=45\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QTS}=\angle \text{PTR}\left[\mathrm{Vertical}\text{opposite angels}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QTS}=45\mathrm{°}\\ \mathrm{In}\text{}\mathrm{\Delta }\text{QST,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QST}+\angle \text{SQT}+\angle \text{QTS}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}75\mathrm{°}+\angle \text{PTR}+45\mathrm{°}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PTR}=180\mathrm{°}-120\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}\end{array}$

Q.10 In Fig. 6.43, if PQ ⊥ PS, PQ||SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\text{PQ}\perp \text{PS},\text{PQ}||\text{SR},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{SQR}=\text{28}°\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\angle \text{QRT}=\text{65}\mathrm{°}\\ \mathrm{To}\text{find: x and y}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Sol}:\text{\hspace{0.17em}}\mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\mathrm{ST}\\ \angle \mathrm{PQR}=\angle \mathrm{PRT}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+28\mathrm{°}=65\mathrm{°}\\ \text{\hspace{0.17em}}\mathrm{x}=65\mathrm{°}-28\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=37\mathrm{°}\\ \mathrm{In}\text{}\mathrm{\Delta }\text{PQS,}\\ \text{\hspace{0.17em}}\angle \mathrm{QPS}+\angle \mathrm{PSQ}\text{\hspace{0.17em}}+\angle \mathrm{PQR}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{angle sum property.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}90\mathrm{°}+\mathrm{y}+37\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=180\mathrm{°}-127\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=53\mathrm{°}\end{array}$

Q.11 In Fig. 6.44, the side QR of Δ PQR is produced to a point S.
If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = (1/2) ∠QPR.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\text{The side QR of\hspace{0.17em}}\mathrm{\Delta }\text{PQR is produced}\text{to a point S}.\text{The}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}bisectors of}\angle \text{PQR and}\mathrm{}\text{\hspace{0.17em}}\angle \text{PRS meet at point T.}\\ \text{To Prove:}\angle \text{QTR}=\frac{1}{2}\angle \text{QPR}\\ \text{Proof}:\text{\hspace{0.17em}}\mathrm{Since},\angle \mathrm{PQT}=\angle \mathrm{TQR}=\mathrm{x}\left(\mathrm{let}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\angle \mathrm{PRT}=\angle \mathrm{TRS}=\mathrm{y}\left(\mathrm{let}\right)\\ \mathrm{In}\text{â€‹\hspace{0.17em}}\mathrm{\Delta PQR},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ext}.\text{\hspace{0.17em}}\angle \mathrm{PRS}=\angle \mathrm{PQR}\text{\hspace{0.17em}}+\angle \mathrm{QPR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{y}=2\mathrm{x}\text{\hspace{0.17em}}+\angle \mathrm{QPR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{y}-2\mathrm{x}=\angle \mathrm{QPR}\\ \mathrm{y}-\mathrm{x}=\frac{1}{2}\angle \mathrm{QPR}...\left(\mathrm{i}\right)\\ \mathrm{In}\text{â€‹\hspace{0.17em}}\mathrm{\Delta QTR},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Ext}.\text{\hspace{0.17em}}\angle \mathrm{TRS}=\angle \mathrm{TQR}\text{\hspace{0.17em}}+\angle \mathrm{QTR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\mathrm{x}\text{\hspace{0.17em}}+\angle \mathrm{QTR}\\ \text{\hspace{0.17em}}\mathrm{y}-\mathrm{x}=\angle \mathrm{QTR}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{QTR}=\frac{1}{2}\angle \mathrm{QPR}\\ \mathrm{Hence}\text{Proved.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{In}\mathrm{quadrilateral}\mathrm{ACBD},\mathrm{AC}=\mathrm{AD}\mathrm{and}\mathrm{AB}\mathrm{bisects}\mathrm{}\angle \mathrm{A}\\ \left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\mathrm{\Delta ABC}\cong \mathrm{\Delta ABD}.\mathrm{}\mathrm{What}\mathrm{can}\\ \mathrm{you}\mathrm{say}\mathrm{about}\mathrm{BC}\mathrm{and}\mathrm{BD}.\end{array}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{ABCD is a quadrilateral, AC}=\text{BD, AB bisects}\angle \text{A.}\\ \text{To prove:\hspace{0.17em}}\mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{ABD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:In}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{ABD}\\ \text{AC}=\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{ABD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \mathrm{AB}=\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \therefore \mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{ABD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ \mathrm{BC}=\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{CPCT}\right]\end{array}$

Q.13

$\begin{array}{l}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{quadrilateral}\mathrm{in}\mathrm{which}\mathrm{AD}=\mathrm{BC}\mathrm{and}\angle \mathrm{ }\mathrm{DAB}=\angle \mathrm{ }\mathrm{CBA}\\ \left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Prove}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{\Delta ABD}\cong \mathrm{\Delta BAC}\\ \left(\mathrm{ii}\right)\mathrm{BD}=\mathrm{AC}\\ \left(\mathrm{iii}\right)\angle \mathrm{ABD}=\angle \mathrm{ }\mathrm{BAC}.\end{array}$

Ans

$\begin{array}{l}\text{Given}:\text{ABCD is a quadrilateral, AD}=\text{BC,}\angle D\text{AB}=\angle C\text{BA}\text{.}\\ \text{To prove:}\text{\hspace{0.17em}}\\ \left(i\right)\text{\hspace{0.17em}}\Delta \text{ABD}\cong \Delta B\text{AC}\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}BD=AC\\ \left(iii\right)\text{\hspace{0.17em}}\angle \text{ABD}=\angle \text{BAC}\\ \text{Proof:}\left(i\right)\text{In}\Delta \text{ABD and}\Delta B\text{AC}\\ \text{AB}=B\text{A}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle BAD=\angle ABC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Given}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AD=BC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Given}\right]\\ \therefore \Delta \text{ABD}\cong \Delta \text{BAC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{SAS}\right]\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}BD=AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{CPCT}\right]\\ \left(iii\right)\text{\hspace{0.17em}}\angle \text{ABD}=\angle \text{BAC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{CPCT}\right]\\ \end{array}$

Q.14 AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Ans

$\begin{array}{l}\text{Given}:\text{AD and BC are equal perpendiculars to a line segment AB.}\\ \text{To Prove: CD bisects AB i.e., CO}=\text{OD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\text{\hspace{0.17em}}\mathrm{In}\text{â€‹}\mathrm{\Delta BOC}\cong \mathrm{\Delta AOD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBO}=\angle \mathrm{DAO}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOC}=\angle \mathrm{AOD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Vertical}\text{opposite angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{AD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \therefore \mathrm{\Delta BOC}\cong \mathrm{\Delta AOD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{AAS}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BO}=\text{OA\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\mathrm{CD}\text{bisects line segment AB.}\end{array}$

Q.15

$\begin{array}{l}\mathrm{l}\mathrm{and}\mathrm{m}\mathrm{are}\mathrm{two}\mathrm{parallell}\mathrm{ines}\mathrm{intersected}\mathrm{by}\mathrm{another}\mathrm{pair}\mathrm{of}\\ \mathrm{parallel}\mathrm{lines}\mathrm{p}\mathrm{and}\mathrm{q}\left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}\cong \mathrm{\Delta CDA}.\end{array}$

Ans

$\begin{array}{l}\text{Given}:\text{line}\mathrm{l}\text{and m are parallel and}\stackrel{â‡€}{\mathrm{p}}\stackrel{â‡€}{\mathrm{q}}.\\ \text{To prove:\hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}\cong \mathrm{\Delta CDA}\\ \text{Proof: In\hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABCand}\text{\hspace{0.17em}}\mathrm{\Delta CDA}\\ \angle \mathrm{BAC}=\angle \mathrm{DCA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternate}\text{\hspace{0.17em}interior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AC}=\mathrm{CA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \angle \mathrm{BCA}=\angle \mathrm{DAC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Alternate\hspace{0.17em}interior angles]}\\ \mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}\cong \text{\hspace{0.17em}}\mathrm{\Delta CDA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{â€‹ ASA}\right]\end{array}$

Q.16

$\begin{array}{l}\text{Line \hspace{0.17em}l \hspace{0.17em}is the bisector of an angle \hspace{0.17em}∠A and ∠B is any point on \hspace{0.17em}l.}\\ \text{BP and BQ are perpendiculars \hspace{0.17em}from B to the arms of\hspace{0.17em}∠A}\\ \left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{}\mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\mathrm{\Delta APB}\cong \mathrm{\Delta AQB}\mathrm{}\\ \left(\mathrm{ii}\right)\mathrm{BP}=\mathrm{BQ}\mathrm{or}\mathrm{B}\mathrm{is}\mathrm{equidistant}\mathrm{from}\mathrm{the}\mathrm{arms} \mathrm{of}\angle \mathrm{A}.\end{array}$

Ans

$\begin{array}{l}\text{Given}:\text{Line}\mathrm{l}\text{is bisector of}\angle \text{A, BP and BQ are perpendiculars to}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}the arms of angle.}\\ \mathrm{To}\text{prove:}\left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{APB}\cong \mathrm{\Delta AQB}\\ \left(\mathrm{ii}\right)\mathrm{BP}=\mathrm{BQ}\mathrm{or}\mathrm{B}\text{is equidistant from the arms \hspace{0.17em}of∠A.}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In\hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABQ}\mathrm{and}\text{\hspace{0.17em}}\mathrm{\Delta APB}\\ \angle \mathrm{BAQ}=\angle \mathrm{BAP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \angle \mathrm{AQB}=\angle \mathrm{APB}\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{each}\text{90°}\right]\\ \mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABQ}\cong \text{\hspace{0.17em}}\mathrm{\Delta APB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{â€‹ AAS}\right]\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BQ}=\mathrm{BP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BP}=\mathrm{BQ}\end{array}$

Q.17 In figure below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{}\mathrm{In}\text{}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{ADE, AC}=\text{AE},\text{AB}=\text{AD and}\angle \text{BAD}=\angle \text{EAC}\\ \text{To prove:BC=DE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: In\hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}\mathrm{and}\text{\hspace{0.17em}}\mathrm{\Delta ADE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \angle \mathrm{BAC}=\angle \mathrm{DAE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\angle \mathrm{BAD}+\angle \mathrm{DAC}=\angle \mathrm{DAC}+\angle \mathrm{EAC}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AC}=\mathrm{AE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}\cong \text{\hspace{0.17em}}\mathrm{\Delta ADE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{â€‹ SAS}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{DE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\end{array}$

Q.18

$\begin{array}{l}\mathrm{AB}\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{segment}\mathrm{and}\mathrm{P}\mathrm{is}\mathrm{its}\mathrm{mid}–\mathrm{point}.\mathrm{D}\mathrm{and}\mathrm{E}\mathrm{are}\\ \mathrm{points}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{AB}\mathrm{such}\mathrm{that}\angle \mathrm{BAD}=\angle \mathrm{ABE}\\ \mathrm{and}\angle \mathrm{EPA}=\angle \mathrm{DPB}\left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\text{}\mathrm{\Delta DAP}\cong \mathrm{\Delta EBP}\\ \left(\mathrm{ii}\right)\text{}\mathrm{AD}=\mathrm{BE}\end{array}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{}\mathrm{P}\text{is mid-point of AB,\hspace{0.17em}}\angle \text{BAD}=\angle \mathrm{ABE}\text{â€‹ and}\angle \mathrm{EPA}=\angle \mathrm{DPB}\\ \text{To prove:}\left(\mathrm{i}\right)\mathrm{\Delta DAP}\cong \mathrm{\Delta EBP}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\mathrm{BE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{In\hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{DAPand}\text{\hspace{0.17em}}\mathrm{\Delta EBP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DPA}=\angle \mathrm{EPB}\left[\angle \mathrm{APE}+\angle \mathrm{EPD}=\angle \mathrm{BPD}+\angle \mathrm{DPE}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\mathrm{P}=\mathrm{PB}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DAP}=\angle \mathrm{EBP}\left[\mathrm{Given}\right]\\ \mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{DAP}\cong \text{\hspace{0.17em}}\mathrm{\Delta EPB}\left[\mathrm{By}\text{â€‹ SAS}\right]\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\mathrm{BE}\left[\mathrm{By}\text{C.P.C.T.}\right]\end{array}$

Q.19

$\begin{array}{l}\mathrm{In}\mathrm{right}\mathrm{triangle}\mathrm{ABC},\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C},\mathrm{M}\mathrm{is}\mathrm{the}\mathrm{mid}–\mathrm{point}\\ \mathrm{of}\mathrm{hypotenuse}\mathrm{AB}.\mathrm{C}\mathrm{is}\mathrm{joined}\mathrm{to}\mathrm{M}\mathrm{and}\mathrm{produced}\mathrm{to}\mathrm{a}\mathrm{point}\\ \mathrm{D}\mathrm{such}\mathrm{that}\mathrm{DM}=\mathrm{CM}.\mathrm{Point}\mathrm{D}\mathrm{is}\mathrm{joined}\mathrm{to}\mathrm{point}\mathrm{B}\\ \left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\text{}\mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}\\ \left(\mathrm{ii}\right)\text{}\mathrm{\Delta DBC}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}.\\ \left(\mathrm{iii}\right)\text{}\mathrm{\Delta DBC}\cong \mathrm{\Delta ACB}\\ \left(\mathrm{iv}\right)\text{}\mathrm{CM}=\frac{1}{2}\mathrm{AB}\end{array}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{In right}\mathrm{\Delta }\text{ABC,}\angle \text{C}=90\mathrm{°}\text{and M is mid-point on}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}hypotenuse AB,\hspace{0.17em}CM}=\text{DM.}\\ \text{To Prove:}\left(\mathrm{i}\right)\mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\angle \mathrm{DBC}\text{is a right angle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{\Delta DBC}\cong \mathrm{\Delta ACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\mathrm{CM}=\frac{1}{2}\mathrm{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\left(\mathrm{i}\right)\text{In}\mathrm{\Delta AMC}\text{}\mathrm{and}\text{\hspace{0.17em}}\mathrm{\Delta BMD}\\ \mathrm{AM}=\mathrm{BM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AMC}=\angle \mathrm{BMD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Vertica}\mathrm{l}\text{opposite angles}\right]\\ \mathrm{CM}=\mathrm{DM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Given]}\end{array}$ $\begin{array}{l}\therefore \mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{S}.\mathrm{A}.\mathrm{S}.\right]\\ \mathrm{AC}=\mathrm{DB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACM}=\angle \mathrm{BDM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACD}=\angle \mathrm{BDC}\\ ⇒\mathrm{AC}\mathrm{DB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternate}\text{angles are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}+\angle \mathrm{DBC}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}}\left[\text{Co-interior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}90\mathrm{°}+\angle \mathrm{DBC}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DBC}=180\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}\\ \mathrm{So},\text{\hspace{0.17em}}\angle \mathrm{DBC}\text{is a right angle.}\\ \left(\mathrm{iii}\right)\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{DCB and}\mathrm{\Delta ACB}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Proved}\right]\\ \angle \mathrm{ACB}=\angle \mathrm{DBC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCB}\cong \mathrm{\Delta ACB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{S}.\mathrm{A}.\mathrm{S}.\right]\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DC}=\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{CM}=\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{CM}=\mathrm{DM}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CM}=\frac{1}{2}\mathrm{AB}\end{array}$

Q.20 In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠A.

Ans

$\begin{array}{l}\text{Given: In}\mathrm{\Delta }\text{ABC, AB}=\text{AC and bisectors of}\angle \text{B and}\angle \text{C intersect}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}at O.}\\ \text{To Prove:}\left(\text{i}\right)\text{OB}=\text{OC}\left(\text{ii}\right)\text{AO bisects}\angle \text{A}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In}\mathrm{\Delta }\text{ABC, AB}=\text{AC}\\ \text{So, \hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\angle \mathrm{ABC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Opposite angles of equal}\\ \text{sides are equal.}\end{array}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{ABC}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OCB}=\angle \mathrm{OBC}\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Given},\text{as OB and OC are angle}\\ \text{bisectors.}\end{array}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}\left[\begin{array}{l}\text{Opposite angles of equal}\\ \text{sides are equal.}\end{array}\right]\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\text{In}\mathrm{\Delta O}\text{AB and}\mathrm{\Delta O}\text{AC,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{OA}=\mathrm{OA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta O}\text{AB}\cong \text{}\mathrm{\Delta O}\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \mathrm{Hence},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OAB}=\angle \mathrm{OAC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{Thus},\text{OA bisects}\angle \text{A.}\end{array}$

Q.21 In Δ ABC, AD is the perpendicular bisector of BC (see figure below). Show that Δ ABC is an isosceles triangle in which AB = AC.

Ans

$\begin{array}{l}\text{Given}:\text{In}\mathrm{\Delta }\text{ABC, BD}=\text{DC and AD}\perp \text{BC.}\\ \text{To Prove: AB}=\text{AC}\\ \text{Proof: In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{ACD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\mathrm{AD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADB}=\angle \mathrm{ADC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{B}\mathrm{D}=\mathrm{DC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given]}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{ABD}\cong \text{}\mathrm{\Delta }\text{ACD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Therefore},\text{}\mathrm{\Delta }\text{ABC is isosceles triangle.}\end{array}$

Q.22 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure below). Show that these altitudes are equal.

Ans

$\begin{array}{l}\text{Given:In\hspace{0.17em}}\mathrm{\Delta ABC},\text{\hspace{0.17em}}\mathrm{AC}=\mathrm{AB}\text{, BE}\perp \text{AC and CF}\perp \text{AB.}\\ \text{To prove: BE}=\text{CF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta ABE}\text{and\hspace{0.17em}}\mathrm{\Delta ACF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAE}=\angle \mathrm{CAF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Common angle]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BEA}=\angle \mathrm{CFA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given]}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABE}\cong \text{\hspace{0.17em}}\mathrm{\Delta ACF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{A.A.S.}\right]\\ \therefore \mathrm{BE}=\mathrm{CF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.23

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{in}\mathrm{which}\mathrm{altitudes}\mathrm{BE}\mathrm{and}\mathrm{CF}\mathrm{to}\mathrm{sides}\mathrm{AC}\\ \mathrm{and}\mathrm{AB}\mathrm{are}\mathrm{equal}\left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\text{}\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}\\ \left(\mathrm{ii}\right)\text{}\mathrm{AB}=\mathrm{AC},\mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\end{array}$

Ans

$\begin{array}{l}\mathrm{ }\mathrm{Given}:\mathrm{In}\mathrm{ }\mathrm{\Delta ABC},\mathrm{ }\mathrm{BE}=\mathrm{CF},\mathrm{BE}\perp \mathrm{AC}\mathrm{and}\mathrm{CF}\perp \mathrm{AB}\mathrm{.}\\ \mathrm{To}\mathrm{prove}:\left(\mathrm{i}\right)\text{}\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}\\ \left(\mathrm{ii}\right)\text{}\mathrm{AB}=\mathrm{AC},\mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\\ \mathrm{Proof}:\left(\mathrm{i}\right)\text{}\mathrm{In}\mathrm{ }\mathrm{\Delta ABE}\mathrm{and}\mathrm{ }\mathrm{\Delta ACF}\\ \angle \mathrm{BAE}=\angle \mathrm{CAF} \mathrm{ }\left[\mathrm{Common}\mathrm{angle}\right]\\ \angle \mathrm{BEA}=\angle \mathrm{CFA} \mathrm{ }\left[\mathrm{Each}90°\right]\\ \mathrm{BE}=\mathrm{CF} \mathrm{ } \mathrm{ }\left[\mathrm{Given}\right]\\ \therefore \mathrm{\Delta ABE}\cong \mathrm{ }\mathrm{\Delta ACF} \left[\mathrm{By}\mathrm{A}.\mathrm{A}.\mathrm{S}\mathrm{.}\right]\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{AB}=\mathrm{AC} \left[\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\mathrm{.}\right]\\ \mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\\ \mathrm{Hence}\mathrm{proved}\mathrm{.}\end{array}$

Q.24 ABC and DBC are two isosceles triangles on the same base BC (see figure below). Show that ∠ABD = ∠ACD.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta ABC},\text{\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{and}\mathrm{in}\text{\hspace{0.17em}}\mathrm{\Delta DBC},\text{\hspace{0.17em}}\mathrm{DB}=\mathrm{DC}\text{.}\\ \text{To prove:}\angle \mathrm{ABD}=\angle \mathrm{ACD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta ABC}\\ \mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\text{\hspace{0.17em}}\angle \mathrm{ABC}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta DBC}\\ \mathrm{DB}=\mathrm{DC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DCB}=\text{\hspace{0.17em}}\angle \mathrm{DBC}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\mathrm{we}\text{}\mathrm{get}\\ \angle \mathrm{ACB}\text{\hspace{0.17em}}+\angle \mathrm{DCB}=\text{\hspace{0.17em}}\angle \mathrm{ABC}+\angle \mathrm{DBC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACD}=\text{\hspace{0.17em}}\angle \mathrm{ABD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABD}=\angle \mathrm{ACD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Hence}\text{proved.}\end{array}$

Q.25 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure below). Show that ∠BCD is a right angle.

Ans

$\begin{array}{l}\text{Given:Inâ€‹}\mathrm{\Delta }\text{ABC, AB}=\text{AC, side BAâ€‹ is produced to D such that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}=\text{AB.}\\ \text{To prove:}\angle \text{BCD is a right angle}.\\ \text{Proof:In}\mathrm{\Delta }\text{ABC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given]}\\ \angle \text{ACB}=\angle \text{ABC \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of equal sides are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{let}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta }\text{ACD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}=\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Given]}\\ \angle \text{ADC}=\angle \text{ACD \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of equal sides are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{let}\right)\\ \text{Now},\text{\hspace{0.17em}In}\mathrm{\Delta }\text{BCD}\\ \angle \text{BCD}+\angle \mathrm{C}\text{BD}+\angle \text{BDC}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{x}+\mathrm{y}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\left(\mathrm{x}+\mathrm{y}\right)=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}+\mathrm{y}\right)=\frac{180\mathrm{°}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCD}=90\mathrm{°}\\ ⇒âˆ†\text{BCD is right triangle.}\end{array}$

Q.26 ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}Given :In}\mathrm{\Delta }\text{ABC,}\angle \text{A}=\text{90° and AB}=\text{AC.}\\ \text{To find :\hspace{0.17em}}\angle \text{\hspace{0.17em}B and}\angle \text{\hspace{0.17em}C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sol : In}\mathrm{\Delta }\text{ABC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given]}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\angle \mathrm{ABC}=\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{let}\right)\\ \text{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}90\mathrm{°}+\mathrm{x}+\mathrm{x}=180\mathrm{°}\\ 2\mathrm{x}=180\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{90\mathrm{°}}{2}=45\mathrm{°}\\ \mathrm{So},\text{\hspace{0.17em}}\angle \mathrm{B}=45\mathrm{°}\text{and}\angle \mathrm{C}=45\mathrm{°}.\end{array}$

Q.27 Show that the angles of an equilateral triangle are 60° each.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{\Delta ABC}\text{is an equilateral triangle.}\\ \text{To prove :}\angle \text{A}=\angle \mathrm{B}=\angle \text{C}=60\mathrm{°}\\ \text{Proof}:\text{In}\mathrm{\Delta ABC},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}=\angle \mathrm{B}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Angles opposite to equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{\hspace{0.17em}}\mathrm{AB}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}=\angle \mathrm{A}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Angles}\text{opposite to equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{By angle sum property in a triangle,}\\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{C}+\angle \mathrm{C}+\angle \mathrm{C}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right)\right]\\ 3\angle \mathrm{C}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}=\frac{180\mathrm{°}}{3}=60\mathrm{°}\\ \mathrm{So},\text{}\angle \mathrm{A}\text{\hspace{0.17em}}=60\mathrm{°}\text{and}\angle \mathrm{B}\text{\hspace{0.17em}}=60\mathrm{°}\\ \text{Therefore},\text{each angle of an equilateral triangle is 60° each.}\end{array}$

Q.28

$\begin{array}{l}\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta DBC}\mathrm{are}\mathrm{two}\mathrm{isosceles}\mathrm{triangle}\mathrm{son}\mathrm{the}\mathrm{same}\\ \mathrm{base}\mathrm{BC}\mathrm{and}\mathrm{vertices}\mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{BC}\\ \left(\mathrm{see}\mathrm{the}\mathrm{figure}\mathrm{below}\right).\mathrm{If}\mathrm{AD}\mathrm{is}\mathrm{extended}\mathrm{to}\mathrm{intersect}\mathrm{BC}\mathrm{at}\mathrm{P},\mathrm{show}\\ \text{that}\end{array}$ $\begin{array}{l}\left(\mathrm{i}\right)\text{}\mathrm{\Delta ABD}\cong \mathrm{\Delta ACD}\\ \left(\mathrm{ii}\right)\text{}\mathrm{\Delta ABP}\cong \mathrm{\Delta ACP}\\ \left(\mathrm{iii}\right)\text{}\mathrm{AP}\mathrm{bisects}\angle \mathrm{A}\mathrm{as}\mathrm{well}\mathrm{as}\angle \mathrm{D}.\\ \left(\mathrm{iv}\right)\text{}\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{BC}.\end{array}$

Ans

$\begin{array}{l} \mathrm{Given}:\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta DBC}\mathrm{are}\mathrm{two}\mathrm{isosceles}\mathrm{triangles}\mathrm{on}\mathrm{the}\mathrm{same}\\ \mathrm{base}\mathrm{BC}\mathrm{}\mathrm{and}\mathrm{vertices}\mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{BC}.\\ \mathrm{To}\mathrm{prove}:\mathrm{ }\left(\mathrm{i}\right)\mathrm{\Delta ABD}\cong \mathrm{\Delta ACD}\\ \mathrm{ }\left(\mathrm{ii}\right)\mathrm{\Delta ABP}\cong \mathrm{\Delta ACP}\\ \mathrm{ }\left(\mathrm{iii}\right)\mathrm{AP}\mathrm{bisects}\angle \mathrm{A}\mathrm{as}\mathrm{well}\mathrm{as}\angle \mathrm{D}.\\ \\ \mathrm{ }\left(\mathrm{iv}\right)\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{BC}.\\ \mathrm{Proof}:\left(\mathrm{i}\right)\mathrm{In}\mathrm{}\mathrm{\Delta ABD} \mathrm{and}\mathrm{\Delta ACD}\\ \mathrm{AB}=\mathrm{AC} \mathrm{ }\left[\mathrm{Given}\right]\\ \mathrm{DB}=\mathrm{DC} \mathrm{ }\left[\mathrm{Given}\right]\\ \mathrm{ }\mathrm{AD}=\mathrm{AD} \mathrm{ }\left[\mathrm{Common}\right]\\ \therefore \mathrm{\Delta ABD}\mathrm{ }\cong \mathrm{\Delta ACD} \mathrm{ }\left[\mathrm{By}\mathrm{S}.\mathrm{S}.\mathrm{S}\mathrm{.}\right]\\ \angle \mathrm{BAD}=\angle \mathrm{CAD} \mathrm{ }\left[\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\mathrm{.}\right]\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\text{}\mathrm{In}\text{}\mathrm{\Delta ABP}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and\Delta ACP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAP}=\angle \mathrm{CAP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AP}=\mathrm{AP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \therefore \mathrm{\Delta ABP}\text{\hspace{0.17em}}\cong \mathrm{\Delta ACP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAP}=\angle \mathrm{CAP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BP}=\mathrm{CP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{iii}\right)âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAP}=\angle \mathrm{CAP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\text{\hspace{0.17em}}\mathrm{AP}\text{bisects}\angle \mathrm{A}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{In}\text{}\mathrm{\Delta BDP}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and\Delta }\text{\hspace{0.17em}}\mathrm{CDP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BD}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DP}=\mathrm{DP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BP}=\mathrm{CP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta BDP}\text{\hspace{0.17em}}\cong \mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{CDP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BDP}=\angle \mathrm{CDP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\mathrm{DP}\text{bisects\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BDC}\text{i.e.,}\angle \mathrm{D}.\\ \mathrm{Thus},\text{AP bisects}\angle \mathrm{A}\text{as well as}\angle \mathrm{D}.\\ \left(\mathrm{iv}\right)\text{}\angle \text{APB}=\angle \text{APC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{APB}+\angle \text{APC}=\text{180°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APB}+\angle \text{APB}=\text{180°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}2\angle \text{APB}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APB}=\frac{180\mathrm{°}}{2}=90\mathrm{°}\\ \text{And BP}=\mathrm{C}\text{P}\\ \text{So},\text{â€‹ AP is perpendicular bisector of BC.}\end{array}$

Q.29 AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{AD}\text{is altitude in isosceles}\mathrm{\Delta }\text{ABC in which AB}=\text{AC.}\\ \mathrm{To}\text{â€‹\hspace{0.17em}prove}:\left(\text{i}\right)\text{AD bisects BC}\left(\text{ii}\right)\text{AD bisects\hspace{0.17em}\hspace{0.17em}}\angle \text{A}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\left(\mathrm{i}\right)\mathrm{In}\text{}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{ACD}\\ \text{AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADB}=\angle \mathrm{ADC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Eachâ€‹ 90°]}\\ \text{AD}=\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{ABD}\cong \mathrm{\Delta }\text{ACD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \text{So},\text{â€‹ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BD}=\text{CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Therefore},\text{AD bisects BC.}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAD}=\angle \mathrm{CAD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Therefore},\text{AD bisects}\angle \mathrm{A}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.30

$\begin{array}{l}\mathrm{Two}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{BC}\mathrm{and}\mathrm{median}\mathrm{AM}\mathrm{of}\mathrm{one}\mathrm{triangle}\mathrm{ABC}\mathrm{are}\\ \mathrm{respectively}\mathrm{equal}\mathrm{to}\mathrm{sides}\mathrm{PQ}\mathrm{and}\mathrm{QR}\mathrm{and}\mathrm{median}\mathrm{PN}\mathrm{of}\\ \mathrm{\Delta PQR}\left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\mathrm{\Delta ABM}\cong \mathrm{\Delta PQN}\\ \left(\mathrm{ii}\right)\mathrm{\Delta ABC}\cong \mathrm{\Delta PQR}\end{array}$

Ans

$\begin{array}{l}\text{Given:In}\Delta \text{ABC and}\Delta \text{PQR,}\text{\hspace{0.17em}}\text{AB}=\text{PQ,}\text{\hspace{0.17em}}\text{BC}=\text{QR and AM}=PN.\\ \mathrm{To}\text{prove:}\text{\hspace{0.17em}}\left(i\right)\Delta ABM\cong \Delta PQN\\ \left(ii\right)\Delta ABC\cong \Delta PQR\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Proof}:\text{(i)In}\mathrm{\Delta }\text{ABM and}\Delta \text{PQN}\text{\hspace{0.17em}}\text{}\\ \text{AB}=\text{PQ}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\mathrm{Given}\right]\text{\hspace{0.17em}}\\ \text{BM}=\text{QN}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\frac{1}{2}BC=\frac{1}{2}QR\right]\\ \text{AM}=PN\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Given\right]\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta \text{ABC}\cong \Delta \text{PQR}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{S}\text{.S}\text{.S}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle ABM=\angle PQN\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ ⇒\angle B=\angle Q\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}In\text{}\Delta \text{ABC and}\Delta \text{PQR}\\ \text{AB}=\text{PQ}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Given}\right]\text{\hspace{0.17em}}\\ \angle B=\angle Q\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\mathrm{Proved}\text{above}\right]\\ \text{BC}=\text{QR}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Given}\right]\text{\hspace{0.17em}}\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta ABC\cong \Delta PQR\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{S}\text{.A}\text{.S}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Hence Proved}\text{.}\end{array}$

Q.31 BE and CF are two equal altitudes of a triangle ABC.
Show that:
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e ABC is an isosceles triangle.

Ans

$\begin{array}{l}\text{Given}:\mathrm{In}\text{â€‹}\mathrm{\Delta }\text{ABC, BE}\perp \text{AC and CF}\perp \text{AB and BE = CF}\\ \text{Toâ€‹ prove:}\mathrm{\Delta }\text{ABC is isosceles triangle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:(i)}\mathrm{In}\text{â€‹}\mathrm{\Delta }\text{BCE and}\mathrm{\Delta }\text{CBF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BEC}=\angle \mathrm{CFB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}}\mathrm{BC}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \mathrm{BE}=\mathrm{CF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given]}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{BCE}\cong \mathrm{\Delta }\text{CBF\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \left(\mathrm{ii}\right)\mathrm{AB}=\mathrm{AC}\left[\mathrm{By}\mathrm{CPCT}\right]\\ \text{i.e ABC is an isosceles triangle}\end{array}$

Q.32 ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Ans

$\begin{array}{l}\text{Given:Inâ€‹}\mathrm{\Delta }\text{ABC, AB}=\text{AC and AP}\perp \text{BC}\\ \text{Toâ€‹ prove:}\mathrm{\Delta }\text{ABC is isosceles triangle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:Inâ€‹}\mathrm{\Delta }\text{ABP and}\mathrm{\Delta A}\text{CP}\\ \angle \mathrm{APB}=\angle \mathrm{APC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \text{AP}=\text{AP\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{ABP}\cong \mathrm{\Delta A}\text{CP\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \therefore \angle \mathrm{ABP}=\angle \mathrm{ACP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}=\angle \mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence Proved.}\end{array}$

Q.33 Show that in a right angled triangle, the hypotenuse is the longest side.

Ans

$\begin{array}{l}\text{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{ABC,}\angle \text{B}=\text{90°}\\ \mathrm{To}\text{prove: AC is the longest side of}\mathrm{\Delta }\text{ABC}.\\ \text{Proof:In\hspace{0.17em}}\mathrm{\Delta }\text{ABC,}\\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{A}+90\mathrm{°}+\angle \mathrm{C}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{C}=180\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{C}=90\mathrm{°}\\ \text{Hence},\text{the other two angles have to be acute}\left(\text{i}.\text{e}.,\text{less than 9}0\mathrm{º}\right).\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}\text{is the largest angle of the triangle.}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}>\angle \mathrm{A}\text{and}\angle \mathrm{B}>\angle \mathrm{C}\\ \therefore \mathrm{AC}>\mathrm{BC}\text{and AC}>\mathrm{A}\text{B}\\ \left[\begin{array}{l}\text{In any triangle},\text{}\\ \text{the side opposite to the larger}\left(\text{greater}\right)\text{angle is longer}.\end{array}\right]\\ \text{Therefore,â€‹ AC is the largest side of the triangle.}\end{array}$

Q.34 In Fig. shown below, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Ans

$\begin{array}{l}\text{Given: In}\mathrm{\Delta }\text{ABC, AB and AC are produced upto P and Q respectively.}\angle \text{PBC}<\angle \text{QCB.}\\ \text{To prove: AC}>\text{AB}\\ \text{Proof: Since,}\angle \text{PBC}+\angle \text{ABC}=\text{180°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{Pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PBC}=\text{180°}-\angle \text{ABC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{QCB}+\angle \text{ACB}=\text{180°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{Pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QCB}=\text{180°}-\angle \text{ACB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{It}\text{â€‹â€‹ is given that, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PBC}<\angle \text{QCB}\\ \text{180°}-\angle \text{ABC}<\text{180°}-\angle \text{ACB}\\ ⇒\text{\hspace{0.17em}}-\angle \text{ABC}<-\angle \text{ACB}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ACB}<\angle \text{ABC}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}<\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Side opposite to smaller}\\ \text{angle is smaller.}\end{array}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}>\text{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Henceâ€‹ proved.}\end{array}$

Q.35 In Fig. shown, ∠B<∠A and ∠C<∠D. Show that AD
<bc< p=””></bc<>

Ans

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{In}\text{â€‹}\mathrm{\Delta }\text{OAB,}\angle \text{B}<\angle \text{A and in}\mathrm{\Delta }\text{OCD,}\angle \text{C}<\angle \text{D}.\\ \mathrm{To}\text{prove:\hspace{0.17em}AD}<\text{BC}\\ \text{Proof}:\text{\hspace{0.17em}Inâ€‹}\mathrm{\Delta }\text{OAB,}\\ \text{}\angle \text{B}<\angle \text{A\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ ⇒\mathrm{OA}<\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Side}\text{opposite to smaller angle is}\\ \text{smaller.}\end{array}\right]\\ \mathrm{In}\text{â€‹}\mathrm{\Delta }\text{OCD,}\\ \text{}\angle \text{C}<\angle \text{D\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ ⇒\mathrm{OD}<\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\text{Side opposite to smaller angle is}\\ \text{smaller.}\end{array}\right]\\ \text{Adding relation}\left(\mathrm{i}\right)\text{â€‹ and}\left(\mathrm{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OA}+\mathrm{OD}<\mathrm{OB}+\mathrm{OC}\\ ⇒\mathrm{AD}<\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence Proved.}\end{array}$

Q.36 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure below). Show that ∠A > ∠C and ∠B > ∠D.

Ans

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{In}\text{â€‹}\mathrm{quadrilateral}\text{ABCD,}\mathrm{A}\text{B is the smallest side and CD is the largest side.}\\ \mathrm{To}\text{prove:\hspace{0.17em}}\angle \text{A}>\angle \text{C and}\angle \mathrm{B}>\angle \mathrm{D}.\end{array}$

$\begin{array}{l}\text{Proof}:\text{\hspace{0.17em}In}\mathrm{\Delta }\text{ABC,}\\ \text{BC}>\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAC}>\angle \text{ACB\hspace{0.17em}}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\end{array}$ $\begin{array}{l}\text{In}\mathrm{\Delta }\text{ACD,}\\ \text{CD}>\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{CAD}>\angle \text{ACD\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \mathrm{Adding}\text{relation}\left(\mathrm{i}\right)\text{and relation}\left(\mathrm{ii}\right),\text{we get}\\ \angle \text{BAC}+\angle \text{CAD}>\angle \text{ACB}+\angle \text{ACD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAD}>\angle \text{BCD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{A}>\angle \text{C\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{\hspace{0.17em}In}\mathrm{\Delta }\text{ABD,}\\ \text{AD}>\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABD}>\angle \text{ADB\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iv}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \text{\hspace{0.17em}In}\mathrm{\Delta }\text{BCD,}\\ \text{CD}>\text{BC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{CBD}>\angle \text{BDC\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{v}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \mathrm{Adding}\text{relation}\left(\mathrm{iv}\right)\text{and relation}\left(\mathrm{v}\right),\text{we get}\\ \angle \text{ABD}+\angle \text{CBD}>\angle \text{ADB}+\angle \text{BDC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABC}>\angle \text{ADC}\\ ⇒\text{\hspace{0.17em}}\angle \text{B}>\angle \text{D\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{vi}\right)\\ \mathrm{From}\text{relation}\left(\mathrm{iii}\right)\text{â€‹ and relation}\left(\mathrm{vi}\right),\text{we have}\\ \angle \text{A}>\angle \text{C and}\angle \text{B}>\angle \text{D.\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.37 In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Ans

$\begin{array}{l}\text{\hspace{0.17em}Given}:\mathrm{In}\text{â€‹}\mathrm{\Delta }\text{PQR, PR > PQ and PS bisects}\angle \text{QPR.}\\ \text{To prove:\hspace{0.17em}}\angle \text{PSR}>\angle \text{PSQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: In}\mathrm{\Delta }\text{PQR},\text{\hspace{0.17em}PS bisects}\angle \text{QPR}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QPS}=\angle \text{RPS\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{In}\mathrm{\Delta }\text{PQS},\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSR}=\angle \mathrm{PQS}+\angle \mathrm{QPS}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSR}=\angle \mathrm{PQR}+\angle \mathrm{QPS}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PQR}=\angle \mathrm{PSR}-\angle \mathrm{QPS}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{In}\text{}\mathrm{\Delta }\text{PRS},\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSQ}=\angle \mathrm{PRS}+\angle \mathrm{SPR}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSQ}=\angle \mathrm{PRQ}+\angle \mathrm{RPS}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PRQ}=\angle \mathrm{PSQ}-\angle \mathrm{RPS}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PRQ}=\angle \mathrm{PSQ}-\angle \mathrm{QPS}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}PR}>\text{PQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PQR}>\angle \mathrm{PRQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angle of greater side}\\ \text{is greater.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PSR}-\angle \mathrm{QPS}>\angle \mathrm{PSQ}-\angle \mathrm{QPS}\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PSR}>\angle \mathrm{PSQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Hence}\text{proved.}\end{array}$

Q.38 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans

$\begin{array}{l}\text{Given}:\text{A line l and a point A which is away from line l.}\\ \text{To prove: Perpendicular line segment from A to line l is the shortest.}\\ \text{Contruction: Draw AB}\perp \stackrel{↔}{\mathrm{l}}\text{\hspace{0.17em}},\text{AC and AD.}\\ \text{Proof: In}\mathrm{\Delta ABC},\text{}\angle \text{B=90°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Then},\text{}\angle \text{A}+\angle \text{C}=\text{90°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}>\angle \mathrm{C}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AC}>\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Side}\text{opposite to greater angle is greater.}\right]\\ \text{Similarly,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}>\angle \mathrm{D}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}>\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Side}\text{opposite to greater angle is greater.}\right]\\ \text{Therefore},\text{it can be observed that of all line segments drawn}\\ \text{from a given point not on it},\text{the perpendicular line segment is}\\ \text{the shortest}.\mathrm{}\end{array}$

Q.39 ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of ΔABC.

Ans

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In âˆ†ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of âˆ†ABC.

Q.40 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Ans

A point in the interior of a triangle which is equidistant from all the sides is in centre. In centre is a point which is obtained by the intersection of angle bisectors.

Here, in âˆ†ABC, we can find the in centre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of âˆ†ABC.

Q.41 In a huge park, people are concentrated at three points (see figure below):

A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit. Where should an icecream parlour be set up so that maximum number of persons can approach it?

Ans

Icecream parlour should be set up at equidistant point from three points A, B and C so that maximum number of people will approach there.

When we join three points A, B and C, we get a triangle. A point which is equidistant from three vertex is called Circumcentre.

In figure, Circumcentre O is obtained by intersection of perpendicular bisectors of sides of triangle ABC.

At this point of icecream parlour, maximum number of people will approach.

Q.42 Complete the hexagonal and star shaped Rangolies [see the figure below (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\left(i\right)Area\left(\Delta AOB\right)=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(5\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\end{array}$

$\begin{array}{l}\text{Area of hexagonal – shaped}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{rangoli}=6×\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\left(\Delta AOB\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6×\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{75\sqrt{3}}{2}{\text{cm}}^{\text{2}}\\ Area\text{}of\text{equilateral triangle having its side as 1 cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(1\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ Number\text{of equilateral triangles of 1 cm side that can be filled}\\ \text{in hexagonal-shaped rangoli}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\left(\frac{75\sqrt{3}}{2}\right)}{\left(\frac{\sqrt{3}}{4}\right)}=150\end{array}$

$\begin{array}{l}\end{array}$ $\begin{array}{l}\text{Star}-\text{shaped rangoli has 12 equilateral triangles of side}\text{5 cm in it}.\\ \text{Area of equilateral triangle}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(5\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{Area of star-shaped rangoli}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12×\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=75\sqrt{5}{\text{cm}}^{\text{2}}\\ \text{Area}of\text{equilateral triangle having its side as 1 cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(1\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{Number of equilateral triangles of 1 cm side that can be filled}\\ \text{in hexagonal-shaped rangoli}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\left(75\sqrt{3}{\text{cm}}^{\text{2}}\right)}{\left(\frac{\sqrt{3}}{4}\right)}=300\\ \text{Therefore},\text{star-shaped rangoli has more equilateral}\\ \text{triangles in it}.\end{array}$

Q.43 The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Ans

Let four angles of quadrilateral be 3x, 5x, 9x and 13x. Then, by angle sum property in quadrilateral

3x + 5x + 9x + 13x = 360°

30x = 360°
x = 360°/30
= 12°

So, the first angle of quadrilateral
= 3(12°)
= 36°

= 5(12°)
= 60°

= 9(12°)
= 108°

= 13(12°)
= 156°

Q.44 If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Ans

$\begin{array}{l}\text{Given: Let ABCD be a parallelogram, in which AC}=\text{BD.}\\ \text{To prove: ABCD is a rectangle.}\\ \text{Proof: In ΔADC and ΔBCD},\\ \text{\hspace{0.17em}AD}=\text{BC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite sides of parallelogram are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{DC}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\mathrm{C}=\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ADC}\cong \mathrm{\Delta BCD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}=\angle \mathrm{BCD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{but\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}+\angle \mathrm{BCD}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Co}-\text{interior angles}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}=\angle \mathrm{BCD}=90\mathrm{°}\\ \text{Since},\text{one angle of parallelogram is 90°.}\\ \text{So, ABCD is a rectangle.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.45 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Ans

$\begin{array}{l}\text{Given}:\text{Let ABCD be a quadrilateral},\text{in which AO}=\mathrm{OC},\text{}\\ \text{}\mathrm{BO}=\mathrm{OD}.\text{}\mathrm{AC}\text{and BD bisect each other at 9}0\mathrm{°}.\\ \text{To prove}:\text{ABCD is a rhombus}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\text{â€‹\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AO}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}}\mathrm{\Delta BOC}\text{and}\mathrm{\Delta DOC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BO}=\mathrm{OD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOC}=\angle \mathrm{DOC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OC}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\text{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BOC}\cong \mathrm{\Delta DOC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{DC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}}\mathrm{\Delta COD}\text{and}\mathrm{\Delta DOA}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CO}=\mathrm{OA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{COD}=\angle \mathrm{AOD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OD}=\mathrm{OD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta COD}\cong \mathrm{\Delta DOA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DC}=\mathrm{DA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{From equation}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}\\ \text{Since},\text{all sides of quadrilateral ABCD are equal, so}\\ \text{ABCD is a rhombus.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.46 Show that the diagonals of a square are equal and bisect each other at right angles.

Ans

$\begin{array}{l}\text{Given: Let ABCD be a square.}\\ \mathrm{To}\text{prove:AC}=\text{BD, A}\mathrm{O}=\mathrm{OC},\text{}\mathrm{BO}=\text{OD and AC}\perp \text{BD.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: In}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta DCB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{DC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Sides of square}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABC}=\angle \mathrm{DCB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Each 90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABC}\cong \mathrm{\Delta DCB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AC}=\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{In}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Sides of square}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Vertical opposite angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAO}=\angle \mathrm{DCO}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Alternate interior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AO}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BO}=\mathrm{OD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{In}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Sides of square}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AO}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}+\angle \mathrm{COB}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=90\mathrm{°}\\ \text{Thus},\text{diagonals AC and BD are equal and bisect each}\\ \text{other at 90°.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.47 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Ans

$\begin{array}{l}\text{Given}:\mathrm{Let}\text{ABCD be a quadrilateral. AC}=\text{BD and AC}\perp \text{BD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}OA}=\text{OC and OB}=\text{OD.}\\ \text{To prove: ABCD is a square.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:In}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COB}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\mathrm{O}=\mathrm{OC}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COB}\text{}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OB}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COB}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta BOC}\text{and}\mathrm{\Delta COD}\\ \text{\hspace{0.17em} \hspace{0.17em}}\mathrm{BO}=\mathrm{OD}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOC}=\angle \mathrm{DOC}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OC}=\mathrm{OC}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta BOC}\cong \mathrm{\Delta COD}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{In}\mathrm{\Delta COD}\text{and}\mathrm{\Delta DOA}\\ \text{\hspace{0.17em} \hspace{0.17em}}\mathrm{CO}=\mathrm{OA}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DOC}=\angle \mathrm{DOA}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OD}=\mathrm{OD}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta COD}\cong \mathrm{\Delta DOA}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CD}=\mathrm{DA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{From equation}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right),\text{we have}\\ \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}\\ \text{Thus},\text{ABCD is a square.Hence proved.}\end{array}$

Q.48 Diagonal AC of a parallelogram ABCD bisects ∠A . Show that
(i) it bisects ∠C and
(ii) ABCD is a rhombus.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{ABCD is a parallelogram in which diagonal AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}bisects}\angle \text{A.}\\ \text{To prove}:\left(\mathrm{i}\right)\text{}\mathrm{it}\text{bisects\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}\text{also},\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{ABCD}\text{}\mathrm{is}\text{}\mathrm{a}\text{rhombus}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{Since, AD}\text{BC}\\ \text{so,}\angle \mathrm{DAC}=\angle \mathrm{BCA}\text{}...\left(\mathrm{i}\right)\left[\text{Alternate interior angles}\right]\\ \mathrm{and}\text{AB}\text{DC}\\ \text{so,\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{DCA}...\left(\mathrm{ii}\right)\left[\text{Alternate interior angles}\right]\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{DAC}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\left[\mathrm{Given}\right]\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DCA}=\angle \mathrm{BCA}\left[\text{From equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right)\right]\\ ⇒\mathrm{AC}\text{bisects}\angle \mathrm{C}.\\ \text{(ii) From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{BCA}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{BA}\left[\begin{array}{l}\text{Sides opposite to equal angles}\\ \text{are equal.}\end{array}\right]\\ \text{Since},\text{adjacent sides of parallelogram are equal, so ABCD}\\ \text{is a rhombus. Hence proved.}\end{array}$

Q.49 ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{ABCD}\text{is a rhombus.}\\ \text{To prove: AC bisects}\angle \text{A as well as}\angle \text{C.}\\ \text{\hspace{0.17em}\hspace{0.17em}BD bisects}\angle \text{B as well as}\angle \text{D.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:In}\mathrm{\Delta ABC},\text{AB}=\text{BC}\\ \text{So,}\angle \text{BAC}=\angle \text{BCA}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{Since,}\mathrm{AB}\mathrm{DC}\\ \text{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAC}=\angle \text{DCA}...\left(\mathrm{ii}\right)\left[\text{Alternate interior angles.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCA}=\angle \text{DCA}\\ ⇒\mathrm{diagonal}\text{AC bisects}\angle \text{C.}\\ \text{Since, ADBC}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{DAC}=\angle \text{BCA}\dots \left(\mathrm{iii}\right)\left[\text{Alternate interior angles.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAC}=\angle \text{DAC}\\ ⇒\text{diagonal AC bisects}\angle \text{A.}\\ \text{Thus, AC bisects}\angle \text{A as well as}\angle \text{C.}\\ \text{Similarly, BC bisects}\angle \text{B as well as}\angle \text{D.Hence proved.}\end{array}$

Q.50 ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square.
(ii) diagonal BD bisects ∠B as well as ∠D.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{ABCD}\text{is a rectangle. Diagonal}\mathrm{AC}\text{bisects}\angle \mathrm{A}\text{as well}\\ \text{\hspace{0.17em}\hspace{0.17em}as}\angle \mathrm{C}.\\ \text{To prove:}\left(\mathrm{i}\right)\text{ABCD is a square.}\\ \left(\mathrm{ii}\right)\text{Diagonal}\mathrm{BD}\text{bisects}\angle \mathrm{B}\text{as well as}\angle \mathrm{D}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{Since,}\angle 1=\angle 2\text{and}\angle 3=\angle 4\\ âˆµ\mathrm{AB}\mathrm{CD}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle 2=\angle 3\text{}\left[\text{Alternate interior angles}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle 1=\angle 3\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CD}=\mathrm{AD}\left[\begin{array}{l}\text{Opposite sides of equal angles}\\ \text{are equal.}\end{array}\right]\\ \end{array}$ $\begin{array}{l}\mathrm{Since},\mathrm{adjecent}\text{sides of a rectangle are equal, so ABCD is}\\ \text{a square.}\\ \left(\mathrm{ii}\right)\mathrm{As}\text{ABCD is a square.}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADB}=\angle \mathrm{ABD}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{Since},\mathrm{AB}\mathrm{CD}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABD}=\angle \mathrm{CDB}...\left(\mathrm{ii}\right)\left[\mathrm{Alternate}\text{interior angles}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADB}=\angle \mathrm{CDB}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}⇒\mathrm{BD}\text{bisects}\angle \mathrm{D}.\\ \text{Similarly},\text{we can prove that BD bisects}\angle \mathrm{B}.\\ \text{Therefore},\text{BD bisects}\angle \mathrm{B}\text{as well as}\angle \mathrm{D}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.51

$\begin{array}{l}\text{In parallelogram ABCD},\text{two points P and Q are taken on diagonal BD such that DP}=\text{BQ}.\text{Show that}:\\ \left(\mathrm{i}\right)\text{}\mathrm{\Delta APD}\cong \mathrm{\Delta CQB}\\ \left(\mathrm{ii}\right)\text{}\mathrm{AP}=\mathrm{CQ}\mathrm{}\\ \left(\mathrm{iii}\right)\text{}\mathrm{\Delta AQB}\cong \mathrm{\Delta CPD}\mathrm{}\\ \left(\mathrm{iv}\right)\text{}\mathrm{AQ}=\mathrm{CP}\mathrm{}\\ \left(\mathrm{v}\right)\text{}\mathrm{APCQ}\text{is a parallelogram}\end{array}$

Ans

$\begin{array}{l}\text{Given}:\mathrm{ABCD}\text{is a parallelogram and DP}=\text{BQ.}\\ \text{To prove:}\mathrm{}\left(\mathrm{i}\right)\mathrm{\Delta APD}\cong \mathrm{\Delta CQB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{AP}=\mathrm{CQ}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\text{}\mathrm{\Delta AQB}\cong \mathrm{\Delta CPD}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\text{}\mathrm{AQ}=\mathrm{CP}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{v}\right)\text{}\mathrm{APCQ}\text{is a parallelogram}\\ \text{\hspace{0.17em}Proof:\hspace{0.17em}In}\mathrm{\Delta APD}\text{}\mathrm{and}\text{}\mathrm{\Delta CQB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite sides of parallelogram.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADP}=\angle \mathrm{CBQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Alternate}\mathrm{r}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DP}=\mathrm{BQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta APD}\cong \mathrm{\Delta CQB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AP}=\mathrm{CQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{iii}\right)\text{In}\mathrm{\Delta AQB}\text{}\mathrm{and}\text{}\mathrm{\Delta CPD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Opposite}\text{sides of parallelogram.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABQ}=\angle \mathrm{CDP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternater}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BQ}=\mathrm{DP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AQB}\cong \mathrm{\Delta CPD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AQ}=\mathrm{PC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{v}\right)\text{}\mathrm{Since},\text{AP}=\text{CQ and AQ}=\text{PC}\\ \text{So, APCQ is a parallelogram.}\text{[Opposite sides are parallel.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Hence}\text{proved.}\end{array}$

Q.52

$\begin{array}{l}\text{ABCD isaparallelogramandAPandCQareperpendiculars}\\ \text{fromverticesAandCondiagonalBD}\left(\text{seethefigurebelow}\right)\text{.}\\ \text{Showthat}\\ \text{\hspace{0.17em}}\left(\text{i}\right)\text{ΔAPB}\cong \text{ΔCQD}\\ \left(\text{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AP=CQ}\end{array}$

Ans

$\begin{array}{l}\text{Given:ABCD is a parallelogram and AP}\perp \text{BD and CQ}\perp \text{BD.}\\ \text{To prove:}\left(\text{i}\right)\text{ΔAPB}\cong \text{ΔCQD}\\ \left(\text{ii}\right)\text{AP=CQ}\\ \text{Proof:}\left(\text{i}\right)\text{\hspace{0.17em}In ΔAPB\hspace{0.17em}and\hspace{0.17em}ΔCQD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABP=}\angle \text{CDQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Alternate interior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APB=}\angle \text{CQD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Each 90°]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB=CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite sides of parallelogram.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ΔAPB}\cong \text{ΔCQD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{By A.A.S.}\right]\\ \left(\text{ii}\right)\text{AP=CQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{By C.P.C.T.}\right]\text{Hence proved.}\end{array}$

Q.53

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{DEF},\text{AB}=\text{DE},\text{AB}||\text{DE},\text{BC}=\text{EF and BC}||\text{EF}.\text{}\\ \text{Vertices A},\text{B and C are joined to vertices D},\text{E and F respectively}\left(\text{see the figure below}\right).\text{Show that}\\ \left(\mathrm{i}\right)\mathrm{quadrilateral}\mathrm{ABED}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\\ \left(\mathrm{ii}\right)\mathrm{quadrilateral}\mathrm{BEFC}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\\ \left(\mathrm{iii}\right)\mathrm{AD}||\mathrm{CF}\mathrm{and}\mathrm{AD}=\mathrm{CF}\\ \left(\mathrm{iv}\right)\mathrm{quadrilateral}\mathrm{ACFD}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\\ \left(\mathrm{v}\right)\text{}\mathrm{AC}=\mathrm{DF}\\ \left(\mathrm{vi}\right)\text{}\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.\end{array}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{In}\text{}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta DEF},\text{AB}=\text{DE},\text{AB}||\text{DE},\mathrm{}\text{BC}=\text{EF and}\\ \text{BC}||\text{EF.}\\ \text{To prove:}\left(\mathrm{i}\text{)quadrilateral}\mathrm{ABED}\mathrm{is}\mathrm{a}\text{parallelogram}\\ \left(\mathrm{ii}\right)\text{quadrilateral}\mathrm{BEFC}\mathrm{is}\mathrm{a}\text{parallelogram}\\ \left(\mathrm{iii}\right)\text{}\mathrm{AD}||\mathrm{CF}\mathrm{and}\mathrm{AD}=\mathrm{CF}\\ \left(\mathrm{iv}\right)\text{quadrilateral A}\mathrm{CFD}\mathrm{is}\mathrm{a}\text{parallelogram}\\ \left(\mathrm{v}\right)\text{}\mathrm{AC}=\mathrm{DF}\\ \left(\mathrm{vi}\right)\text{}\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{Since, AB}=\text{DE and AB}\text{DE}\\ \text{So, ABED is a parallelogram.}\\ \left[\mathrm{One}\text{pair of opposite sides is parallel and equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii)Since},\text{BC}=\text{EF and BC}||\text{EF}\\ \text{So, BEFC is a parallelogram.}\\ \left[\text{One pair of opposite sides is parallel and equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}(iii)Since, ABED is a prallelogram.}\\ \text{So, AD}=\text{BE and AD}\text{BE\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\mathrm{One}\text{pair of opposite sides is parallel and equal.}\right]\\ \text{Since},\text{BEFC is a prallelogram.}\\ \text{So, BE}=\text{CF and BE}\text{CF\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\text{One pair of opposite sides is parallel and equal.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{AD}=\text{CF and AD}\text{CF}\\ \text{(iv)Since,\hspace{0.17em}\hspace{0.17em}AD}=\text{CF and AD}\text{CF}\\ \text{So},\text{ADFC is a prallelogram.}\\ \left[\text{One pair of opposite sides is parallel and equal.}\right]\\ \text{(v)Since},\text{ADFC is a prallelogram.}\\ \text{So, AC}=\text{DF}\left[\text{Opposite sides of parallelogram.}\right]\\ \left(\mathrm{vi}\right)\text{In}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta DEF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{DE}\left[\text{Given}\right]\\ \mathrm{BC}=\mathrm{EF}\left[\text{Given]}\\ \mathrm{AC}=\mathrm{DF}\left[\text{Proved above}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}\left[\mathrm{By}\text{\hspace{0.17em}}\mathrm{S}.\mathrm{S}.\mathrm{S}.\right]\mathrm{Hence}\text{proved.}\end{array}$

Q.54

$\begin{array}{l}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{trapezium}\mathrm{in}\mathrm{which}\mathrm{AB}||\mathrm{CD}\mathrm{and}\mathrm{AD}=\mathrm{BC}\\ \left(\mathrm{see}\mathrm{the}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\text{}\angle \mathrm{A}=\angle \mathrm{B}\\ \left(\mathrm{ii}\right)\text{}\angle \mathrm{C}=\angle \mathrm{D}\\ \left(\mathrm{iii}\right)\text{}\mathrm{\Delta ABC}\cong \mathrm{\Delta BAD}\\ \left(\mathrm{iv}\right)\text{}\mathrm{diagonal}\mathrm{AC}=\mathrm{diagonal}\mathrm{BD}\end{array}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\mathrm{In}\text{trapezium ABCD, AB}\text{CD and AD}=\text{BC.}\\ \text{To prove:}\left(\mathrm{i}\right)\angle \mathrm{A}=\angle \mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\angle \mathrm{C}=\angle \mathrm{D}\\ \left(\mathrm{iii}\right)\mathrm{\Delta ABC}\cong \mathrm{\Delta BAD}\\ \left(\mathrm{iv}\right)\text{diagona}\mathrm{l}\mathrm{AC}=\text{diagonal}\mathrm{BD}\\ \text{Construction}:\text{Draw CE}\text{DA.\hspace{0.17em}Join AC.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:Since, AB}\mathrm{CD}\text{}⇒\text{AE}\mathrm{CD}\\ \text{and CE}\mathrm{DA}\\ \text{So},\text{AECD is a prallelogram.}\\ \text{Then, AD}=\text{CE}\\ \text{But, AD}=\text{BC}\\ ⇒\mathrm{CE}=\mathrm{BC}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}=\angle \mathrm{CEB}\left[\begin{array}{l}\mathrm{Opposite}\text{sides of equal angles are}\\ \text{equal in}\mathrm{\Delta }\text{CEB.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}+\angle \mathrm{CBA}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\left[\mathrm{Linear}\text{pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CEA}+\angle \mathrm{DAE}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}}\left[\text{Cointerior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}+\angle \mathrm{DAE}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\left[âˆµ\angle \mathrm{CBE}=\angle \mathrm{CEA}\right]\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}+\angle \mathrm{CBA}=\angle \mathrm{CBE}+\angle \mathrm{DAE}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBA}=\angle \mathrm{DAE}\\ ⇒\angle \mathrm{B}=\angle \mathrm{A}\\ ⇒\angle \mathrm{A}=\angle \mathrm{B}\\ \left(\mathrm{ii}\right)\mathrm{S}\mathrm{ince},\text{AB}\text{DC}\\ \text{So,\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{D}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ ⇒\angle \mathrm{A}+\angle \mathrm{D}=\angle \mathrm{B}+\angle \mathrm{C}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}=\angle \mathrm{C}\left[âˆµ\angle \mathrm{A}=\angle \mathrm{B}\right]\\ \text{(iii) In}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta BAD}\\ \mathrm{AB}=\mathrm{BA}\left[\mathrm{Common}\right]\\ \angle \mathrm{A}=\angle \mathrm{B}\left[\mathrm{Proved}\text{above}\right]\\ \mathrm{AD}=\mathrm{BC}\left[\mathrm{Given}\right]\\ \therefore \mathrm{\Delta ABC}\cong \mathrm{\Delta BAD}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{iv}\right)\text{}\mathrm{AC}=\mathrm{BD}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Thus},\text{diagonal AC = diagonal BD.Hence proved.}\end{array}$

Q.55 ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the figure below). AC is a diagonal. Show that :
(i) SR||AC and SR = (1/2)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Ans

$\begin{array}{l}\text{Given}:\text{ABCD is a quadrilateral in which P, Q, R and S \hspace{0.17em}are}\\ \text{\hspace{0.17em}mid-points of the sides AB, BC, CD and DA.}\\ \text{To prove:}\left(\text{i}\right)\text{SR||AC and SR}=\frac{\text{1}}{2}\text{AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{ii}\right)\text{PQ}=\text{SR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{iii}\right)\text{PQRS is a parallelogram.}\\ \text{Proof: \hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{I}\mathrm{n}\text{}\mathrm{\Delta ADC},\text{\hspace{0.17em}}\mathrm{S}\text{and R are the mid-points of DA and DC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}SR}\mathrm{AC}\text{and SR=}\frac{1}{2}\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii) In}\mathrm{\Delta ABC},\text{\hspace{0.17em}}\mathrm{P}\text{and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}PQ}\mathrm{AC}\text{and PQ}=\frac{1}{2}\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}PQ}=\text{SR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equaton}\left(\mathrm{ii}\right),\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{PQ}=\mathrm{SR}\text{and PQ}\mathrm{SR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQSR}\text{is a prallelogram.}\left[\begin{array}{l}\text{One pair of opposite}\\ \text{sides is equal and parallel.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.56 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Ans

$\begin{array}{l}\text{Given}:\text{ABCD is a rhombus and P},\text{Q},\text{R and S are the}\mathrm{}\text{mid-points}\\ \text{of the sides AB},\text{BC},\text{CD and DA}\mathrm{}\text{respectively.}\\ \text{To prove:PQRS is a rectangle.}\\ \text{Construction}:\text{Draw AC and BD.}\\ \text{Proof: In}\mathrm{\Delta ABC},\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively.So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}=\frac{1}{2}\mathrm{AC}\text{and PQ}\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta ADC},\text{S and R are the mid-points of AD and DC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} SR}=\frac{1}{2}\mathrm{AC}\text{and SR}\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}From equation(i) and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}=\text{SR and PQ}\mathrm{SR}\\ \text{So, PQRS is a parallelogram.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{If one pair of opposite sides of a quadrilateral is parallel}\\ \text{and equal, then it is a parallelogram.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta ABD},\text{P and S are the mid-points of AB and AD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PS}=\frac{1}{2}\mathrm{BD}\text{and PS}\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{PN}\mathrm{MO}\text{and MP}\mathrm{ON}\\ \text{So, PMON is a parallelogram.}\\ \text{}\angle \mathrm{P}=\angle \mathrm{MON}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of parallelogram are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}\\ \text{Since,}\angle \mathrm{P}=90\mathrm{°}\text{in parallelogram PQRS, so PQRS is a rectangle.}\end{array}$

Q.57 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Ans

$\begin{array}{l}\text{Given}:\text{ABCD is a rectangle and P},\text{Q},\text{R and S are the}\\ \text{mid-points of the sides AB},\text{BC},\text{CD and DA}\text{respectively}\text{.}\\ \text{To prove:PQRS is a rhombus}\text{.}\\ \text{Construction}:\text{Draw AC and BD}\text{.They are the diagonals of the quadrilateral.}\\ \text{Proof: In}\Delta ABC,\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\frac{1}{2}AC\text{and PQ}AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In ΔADC},\text{S and R are the mid-points of AD and DC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{SR}=\frac{1}{2}AC\text{and SR}AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{From equation(i) and equation}\left(ii\right),\text{we have}\\ \text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\text{SR and PQ}SR\\ \text{So, PQRS is a parallelogram}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}\text{If one pair of opposite sides of a quadrilateral is parallel}\\ \text{and equal, then it is a parallelogram}\text{.}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In ΔABD},\text{P and S are the mid-points of AB and AD}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PS}=\frac{1}{2}BD\text{and PS}BD\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(iii\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Since},\text{AC}=\text{BD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Diagonals\text{of rectangle are equal}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}PQ=PS\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\frac{1}{2}AC=\frac{1}{2}BD\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Since},\text{adjecent sides of a parallelogram ABCD are equal}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{So, PQRS is rhombus}\text{.}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.58 ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC.

Ans

$\begin{array}{l}\text{Given}:\text{}\mathrm{ABCD}\text{is a trapezium in which}AB||DC,\text{BD is a}\text{\hspace{0.17em}}\text{diagonal}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and E is the mid}-\text{point of AD}\text{. A line through E is prallel}\\ \text{}\text{\hspace{0.17em}}\text{to AB}\text{.}\\ \text{To prove: F is the mid-point of BC i}\text{.e}\text{., BF}=\text{CF}\text{.}\\ \text{Proof: In ΔABD},\text{E is mid-point of AD and EO}AB\\ \text{So, by converse of mid-point theorem,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{OD}=\text{OB}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In ΔBCD},\text{O is mid-point of BD and EF}CD\\ \text{So, by converse of mid-point theorem,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{CF}=\text{FB}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Therefore, F is the mid-point of BC}\text{.}\text{\hspace{0.17em}}\text{\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.59 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31).
Show that the line segments AF and EC trisect the diagonal BD.

Ans

$\begin{array}{l}\text{Given}:\text{In parallelogram ABCD, E and F are the mid-points of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sides AB and CD respectively.}\\ \mathrm{To}\text{prove: DP}=\text{PQ}=\text{QB}\\ \text{Proof: In}\mathrm{\Delta DQC},\text{F is mid-point of DC and PF}\mathrm{QC},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}So},\text{by converse of mid-point theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DP}=\text{PQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In ΔABP},\text{E is mid-point of AB and EQ}\mathrm{AP},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{by converse of mid-point theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}=\text{QB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DP}=\text{PQ}=\mathrm{QB}\\ \text{This implies that line segments AF and EC trisect the diagonal BD}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.60 Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Ans

$\begin{array}{l}\text{Given}:\text{Let ABCD is a quadrilateral and P},\text{Q},\text{R and S are the}\\ \text{mid-points of the sides AB},\text{BC},\text{CD and DA}\text{respectively}\text{.}\\ \text{To prove:PR and QS bisect each other}\text{.}\\ \text{Construction}:\text{Draw AC}\text{.}\\ \text{Proof: In}\Delta ABC,\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\frac{1}{2}AC\text{and PQ}AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(i\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In}\Delta ADC,\text{S and R are the mid-points of AD and DC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{SR}=\frac{1}{2}AC\text{and SR}AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{From equation(i) and equation}\left(ii\right),\text{we have}\\ \text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\text{SR and PQ}SR\\ \text{So, PQRS is a parallelogram}\text{.} \text{\hspace{0.17em}}\text{[}\begin{array}{l}\text{If one pair of opposite sides of a quadrilateral is parallel}\\ \text{and equal, then it is a parallelogram.}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{Since, diagonals of a parallelogram bisect each other, so}\\ \text{PR and QS bisect each other}\text{.}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.61

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C}.\mathrm{A}\mathrm{line}\mathrm{through}\mathrm{the}\\ \mathrm{mid}–\mathrm{point}\mathrm{M}\mathrm{of}\mathrm{hypotenuse}\mathrm{AB}\mathrm{and}\mathrm{parallel}\mathrm{to}\mathrm{BC}\mathrm{intersects}\\ \mathrm{AC}\mathrm{at}D.\text{Show that}\\ \left(\mathrm{i}\right)\text{}\mathrm{D}\text{is the mid-point of}\mathrm{AC}\\ \left(\mathrm{ii}\right)\text{MD⊥AC}\\ \left(\mathrm{iii}\right)\text{}\mathrm{CM}=\mathrm{MA}=\frac{1}{2}\mathrm{AB}\end{array}$

Ans

$\begin{array}{l}\text{Given:In\hspace{0.17em}}\mathrm{\Delta ABC},\text{}\angle \mathrm{C}=90\mathrm{°}\text{and M is mid-point of AB, MD}\mathrm{BC}.\\ \mathrm{To}\text{prove:}\left(\mathrm{i}\right)\text{D is the mid-point of AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii)MD⊥AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(iii)CM=MA=}\frac{1}{2}\mathrm{AB}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In}\mathrm{\Delta ABC},\text{MD}\mathrm{BC}\text{and M is mid-point of AB, then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} by converse of mid-point theorem,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}=\text{DC i.e.,}\mathrm{D}\text{is the mid-point of}\mathrm{AC}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\text{Since, MD}\mathrm{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{}\angle \mathrm{MDA}=\angle \mathrm{BCA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Corresponding angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Therefore},\text{}\mathrm{MD}\perp \mathrm{AC}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}ΔMDC and ΔMDA}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DC}=\mathrm{DA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{MDC}=\angle \mathrm{MDA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°\hspace{0.17em}as}\mathrm{MD}\perp \mathrm{AC}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{MD}=\mathrm{MD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta MDC}\cong \mathrm{\Delta MDA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}CM=MA\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Since},\text{AM}=\frac{1}{2}\mathrm{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{\hspace{0.17em}}\mathrm{CM}=\mathrm{MA}=\frac{1}{2}\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.62 Which of the following figures lie on the same base and between the same parallels.

In such a case, write the common base and the two parallels.

Ans

(i)

Yes. It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB CD.

(ii)

No. It can be observed that parallelogram PQRSand trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line.

(iii)

Yes. It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR.

(iv)

No. It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base.

(v)

Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a common base AD and these are lying between the same parallel lines AD and BQ.

(vi)

No. It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines.

Q.63 In the given figure, ABCD is parallelogram, AE ⊥DC and CF ⊥AD.
If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Ans

$\begin{array}{l}\text{In parallelogram ABCD},\text{CD}=\text{AB}=\text{16 cm}\\ \mathrm{}\left[\text{Opposite sides of a parallelogram are equal}\right]\mathrm{}\\ \text{We know that}\mathrm{}\text{Area of a parallelogram}=\text{Base}×\text{Corresponding altitude}\\ \mathrm{}\text{Area of parallelogram ABCD}=\text{CD}×\text{AE}=\text{AD}×\text{CF}\\ \mathrm{}\text{16 cm}×\text{8 cm}=\text{AD}×\text{1}0\text{cm}\\ \mathrm{AD}=\frac{16×8}{10}=12.8\text{\hspace{0.17em}}\mathrm{m}\\ \text{Thus},\text{\hspace{0.17em}the\hspace{0.17em}length\hspace{0.17em}of\hspace{0.17em}}\mathrm{AD}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}2.8\text{\hspace{0.17em}}\mathrm{cm}.\end{array}$

Q.64

$\begin{array}{l}\mathrm{If}\mathrm{E},\mathrm{F},\mathrm{G}\mathrm{and}\mathrm{H}\mathrm{are}\mathrm{respectively}\mathrm{the}\mathrm{mid}–\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{sides}\\ \mathrm{ofa}\mathrm{parallelogram}\mathrm{ABCD}\mathrm{show}\mathrm{that}\mathrm{ar}\left(\mathrm{EFGH}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right).\end{array}$

Ans

$\begin{array}{l}\text{Given}:\text{ABCD is a parallelogram in which E, F, G and H are}\\ \text{mid-points of AB, BC, CD and DA respectively}\text{.}\\ \text{To Prove:}ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right)\\ \text{Contruction}:\text{Join HF}\text{.}\\ \text{Proof: Since, ABCD is a parallelogram}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{So, AD}BC\text{and AD}=\text{BC}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}AD=\frac{1}{2}BC\text{and AD}BC\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AH=BF\text{and AH}BF\\ \text{So, ABFH is a parallelogram}\text{.}\\ \text{Then, AB}HF.\\ \text{Since}\Delta \text{HEF and parallelogram ABFH are on the same base}\\ \text{HF and between the same}\text{parallel lines AB and HF},\\ \therefore \text{Area (}\Delta \text{HEF)}=\frac{1}{2}\text{Area (ABFH}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \text{\hspace{0.17em}}\left(i\right)\\ \text{Similarly, it can be proved that}\\ \therefore \text{Area (}\Delta \text{HGF)}=\frac{1}{2}\text{Area (DCFH}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \text{\hspace{0.17em}}\left(ii\right)\\ \text{On adding equations (1) and (2), we get}\\ \text{Area (}\Delta \text{HEF)}+\text{Area (}\Delta \text{HGF)}=\frac{1}{2}\text{Area (ABFH}\right)+\frac{1}{2}\text{Area (DCFH}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left\{\text{Area (ABFH}\right)+\text{Area (DCFH}\right)\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area (}\Delta \text{EFGH)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\text{Area (ABCD}\right)\end{array}$

Q.65 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

Ans

Given: ABCD is a parallelogram. Points P and Q lies on sides CD and AD respectively.

To prove: area(BQC) = area(APB)

Proof: It can be observed that âˆ†BQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.

So, area(BQC) = (1/2) area ABCD …(i)

Similarly, âˆ†APB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.
So, area(APB) = (1/2) area ABCD …(ii)

From equation (i) and equation(ii), we have area(BQC) = area(APB). Hence proved.

Q.66

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{given}\mathrm{figure},\mathrm{P}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{in}\mathrm{the}\mathrm{interior}\mathrm{of}\mathrm{a}\\ \mathrm{parallelogram}\mathrm{ABCD}.\mathrm{Show}\mathrm{that}\mathrm{}\\ \left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{APB}\right)+\mathrm{ar}\left(\mathrm{PCD}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)\mathrm{}\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{APD}\right)+\mathrm{ar}\left(\mathrm{PBC}\right)=\mathrm{ar}\left(\mathrm{APB}\right)+\mathrm{ar}\left(\mathrm{PCD}\right)\mathrm{}\end{array}$

Ans

$\begin{array}{l}\text{Given}:ABCD\text{is a parallelogram}\text{. P is any point inside of it}\text{.}\\ \mathrm{To}\text{prove:}\left(i\right)ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(ii\right)ar\left(APD\right)+ar\left(PBC\right)=ar\left(APB\right)+ar\left(PCD\right)\\ Construction:\text{â€‹}\text{}Draw\text{EF}\text{AB and QR}AD\text{through point P}\text{.}\\ \text{Proof: In parallelogram ABCD,}\\ \text{AD}BC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ ⇒\text{AE}BF\\ \text{and\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AB}EF\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{construction}\right]\\ \text{Thus},\text{ABFE is a prallelogram}\text{.}\\ \text{Since,}\Delta \text{APB and parallelogram ABFE are lying on the same base}\\ \text{AB and between the same parallel lines AB and EF}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}area}\left(\Delta \text{APB}\right)=\frac{1}{2}area\left(ABFE\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{Now EF}CD\text{because EF}AB\text{and AB}\text{CD}\text{.}\\ \text{So,}\Delta \text{DPC and parallelogram ABFE are lying on the same base}\\ \text{DC and between the same parallel lines AB and EF}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}area\left(\Delta \text{DPC}\right)=\frac{1}{2}area\left(ABFE\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{Adding equation}\left(i\right)\text{and equation}\left(ii\right),\text{we get}\\ area\left(\Delta \text{APB}\right)+area\left(\Delta \text{DPC}\right)=\frac{1}{2}area\left(ABFE\right)+\frac{1}{2}area\left(ABFE\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}area\left(ABCD\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(A\right)\\ \left(ii\right)\text{In parallelogram ABCD,}\\ \text{AB}DC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Opposite sides of parallelogram}\text{.}\right]\\ ⇒\text{AQ}DR\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AD}QR\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{construction}\right]\\ \text{Thus},\text{AQRD is a prallelogram}\text{.}\\ \text{Since,}\Delta \text{APD and parallelogram AQRD are lying on the same base}\\ \text{AD and between the same parallel lines AD and QR}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}area\left(\Delta \text{APD}\right)=\frac{1}{2}area\left(AQRD\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(iii\right)\\ \text{Now QR}BC\text{because QR}BC\text{and AD}B\text{C}\text{.}\\ \text{So,}\Delta \text{BPC and parallelogram QBCR are lying on the same base}\\ \text{BC and between the same parallel lines QR and BC}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}area\left(\Delta \text{BPC}\right)=\frac{1}{2}area\left(\text{QBCR}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(iv\right)\\ \text{Adding equation}\left(iii\right)\text{and equation}\left(iv\right),\text{we get}\\ \text{area}\left(\Delta \text{APD}\right)+area\left(\Delta \text{BPC}\right)=\frac{1}{2}area\left(AQRD\right)+\frac{1}{2}area\left(\text{QBCR}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}area\left(ABCD\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(B\right)\\ \text{From equatin}\left(A\right)\text{and}\left(B\right),\text{we have}\\ \text{area}\left(\Delta \text{APB}\right)+area\left(\Delta \text{DPC}\right)=area\left(\Delta \text{APD}\right)+area\left(\Delta \text{BPC}\right).\end{array}$

Q.67

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{given}\mathrm{figure},\mathrm{PQRS}\mathrm{and}\mathrm{ABRS}\mathrm{are}\mathrm{parallelograms}\mathrm{and}\\ \mathrm{X}\mathrm{is}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{side}\mathrm{}\mathrm{BR}.\mathrm{Show}\mathrm{that}\mathrm{}\\ \left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{PQRS}\right)=\mathrm{ar}\left(\mathrm{ABRS}\right)\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{\Delta PXS}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{PQRS}\right)\end{array}$

Ans

$\begin{array}{l}\text{Given:}\mathrm{PQRS}\mathrm{and}\mathrm{ABRS}\mathrm{are}\mathrm{parallelograms}\mathrm{and}\mathrm{X}\mathrm{is}\mathrm{any}\mathrm{point}\mathrm{on}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{side}\text{\hspace{0.17em}}\mathrm{}\mathrm{BR}.\\ \text{To prove:\hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{PQRS}\right)=\mathrm{ar}\left(\mathrm{ABRS}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta PXS}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{PQRS}\right)\\ \text{Proof:}\left(\mathrm{i}\right)\text{Since, the parallelogram PQRS and ABRS lie on the}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}same base SR}\mathrm{}\text{and also},\text{these lie in between the same}\\ \text{parallel lines SR and PB}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area}\left(\text{PQRS}\right)\text{}=\text{Area}\left(\text{ABRS}\right)\text{}...\text{}\left(\text{i}\right)\mathrm{}\\ \left(\mathrm{ii}\right)\text{Since,}\mathrm{\Delta }\text{AXS and parallelogram ABRS lie on the same base}\\ \text{and are between the same parallel lines AS and BR.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}Area}\left(\mathrm{\Delta AXS}\right)=\frac{1}{2}\mathrm{area}\left(\mathrm{ABRS}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{}\mathrm{we}\text{have}\\ \text{\hspace{0.17em}\hspace{0.17em}Area}\left(\mathrm{\Delta AXS}\right)=\frac{1}{2}\mathrm{area}\left(\text{PQRS}\right).\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.68 A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Ans

$\begin{array}{l}\text{In the figure},\text{point A divides the field into three parts}.\\ \text{These parts are triangular in shape:}\Delta \text{PSA},\text{}\Delta \text{PAQ},\text{and}\Delta S\text{RA}\\ \text{Area of}\Delta \text{PSA}+\text{Area of}\Delta \text{PAQ}+\text{Area of}\Delta S\text{RA}=\text{Area of PQRS}\dots \text{}\left(\text{i}\right)\\ \text{We know that if a parallelogram and a triangle are on}\text{the same}\\ \text{base and between the same parallels},\text{then the area of the}\\ \text{triangle is half the area of the}\text{parallelogram}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area}\left(\Delta \text{PAQ}\right)=\frac{\text{1}}{2}\text{Area}\left(\text{PQRS}\right)\text{}\text{\hspace{0.17em}}\dots \text{}\left(\text{ii}\right)\\ \text{From equation}\left(\text{i}\right)\text{and equation}\left(\text{ii}\right),\text{we get}\\ \text{Area of}\Delta \text{PSA}+\frac{1}{2}\text{Area of}\Delta \text{PAQ}+\text{Area of}\Delta S\text{RA}=\text{Area of PQRS}\\ \text{Area of}\Delta \text{PSA}+\text{Area of}\Delta S\text{RA}=\text{Area of PQRS}-\frac{1}{2}\text{Area of}\Delta \text{PAQ}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\text{Area of}\Delta \text{PAQ}\\ \text{Therefore},we\text{can say that the farmer must sow wheat in}\\ \text{triangular part PAQ and pulses in other two triangular parts PSA}\\ \text{and SRA or wheat in triangular parts PSA and SRA and pulses in}\\ \text{triangular parts PAQ}.\end{array}$

Q.69 In the given figure, E is any point on median AD of a âˆ†ABC. Show that ar(ABE) = ar(ACE).

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{ABC, AD is median and E is any point on AD.}\\ \text{To prove :}\mathrm{ar}\left(\mathrm{ABE}\right)=\mathrm{ar}\left(\mathrm{ACE}\right)\\ \text{Proof: In}\mathrm{\Delta ABC},\text{AD is median.}\\ \text{So, ar}\left(\mathrm{\Delta ABD}\right)=\text{ar}\left(\mathrm{\Delta ACD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{In}\mathrm{\Delta EBC},\text{ED is median.}\\ \text{So, ar}\left(\mathrm{\Delta EBD}\right)=\text{ar}\left(\mathrm{\Delta ECD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{Subtracting equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ \text{ar}\left(\mathrm{\Delta ABD}\right)-\text{ar}\left(\mathrm{\Delta EBD}\right)=\text{ar}\left(\mathrm{\Delta ACD}\right)-\text{ar}\left(\mathrm{\Delta ECD}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area}\left(\mathrm{\Delta }\text{ABE}\right)=\text{Area}\left(\mathrm{\Delta }\text{ACE}\right)\end{array}$

Q.70

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{triangle}\mathrm{ABC},\mathrm{E}\mathrm{is}\mathrm{the}\mathrm{mid}–\mathrm{point}\mathrm{of}\mathrm{median}\mathrm{AD}.\mathrm{Show}\\ \mathrm{that}\mathrm{ar}\left(\mathrm{BED}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right).\end{array}$

Ans

$\begin{array}{l}\text{Given:Inâ€‹}\mathrm{\Delta ABC},\text{AD is the median and E is the mid-point of AD.}\\ \mathrm{To}\text{prove :}\mathrm{ar}\left(\mathrm{BED}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{Proof : In}\mathrm{\Delta ABC},\text{AD is median.}\\ \text{So, ar}\left(\mathrm{\Delta ABD}\right)=\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}..\left(\mathrm{i}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{In}\mathrm{\Delta BDA},\text{BE is median.}\\ \text{So, ar}\left(\mathrm{\Delta BED}\right)=\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\text{Median divides a triangle into two triangles of equal areas.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BED}\right)=\frac{1}{2}\left\{\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABC}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BED}\right)=\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.71 Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Ans

Given: ABCD is a parallelogram; in which diagonals AC and BD bisect each other at O.

To prove: area(DAOB)= area(DBOC)= area(DCOD)= area(DDOA)

Q.72 In the figure shown below, ABC and ABD are two triangles on the same base AB.
If line- segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).

Ans

$\begin{array}{l}\text{Given:ΔABC and}\mathrm{\Delta ADB}\text{have same base AB and OC}=\text{OD.}\\ \text{To prove:}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta ABD}\right)\\ \text{Proof:In}\mathrm{\Delta ADC},\text{AO is median, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}area}\left(\mathrm{\Delta AOC}\right)=\text{area}\left(\mathrm{\Delta AOD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\text{Median divides a triangle into two triangles of equal areas.}\right]\\ \text{In}\mathrm{\Delta BDC},\text{BO is median, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}area}\left(\mathrm{\Delta BOC}\right)=\text{area}\left(\mathrm{\Delta BOD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{Adding equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{area}\left(\mathrm{\Delta AOC}\right)+\text{area}\left(\mathrm{\Delta BOC}\right)=\text{area}\left(\mathrm{\Delta AOD}\right)+\text{area}\left(\mathrm{\Delta BOD}\right)\\ \text{area}\left(\mathrm{\Delta ABC}\right)=\text{area}\left(\mathrm{\Delta DAB}\right)\text{â€‹\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.73

$\begin{array}{l}\mathrm{D},\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{respectively}\mathrm{the}\mathrm{mid}–\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{sides}\\ \mathrm{BC},\mathrm{CA}\mathrm{and}\mathrm{AB}\mathrm{of}\mathrm{a}\mathrm{\Delta ABC}.\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{BDEF}\mathrm{is}\mathrm{a}\mathrm{parallelogram}.\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{DEF}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{BDEF}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)\end{array}$

Ans

$\begin{array}{l}\text{Given}:\text{In}\mathrm{\Delta }\text{ABC, D, E and F are the mid-points of BC, CA and AB}\\ \text{respectively.}\\ \text{To prove:}\left(\mathrm{i}\right)\mathrm{BDEF}\text{is a parallelogram.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{DEF}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{BDEF}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{Proof}:\left(\mathrm{i}\right)\text{In}\mathrm{\Delta }\text{ABC, F and E are the mid-points of AB and AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}respectively, so by mid-point theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} FE}=\frac{1}{2}\mathrm{BC}\text{and FE}\text{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{FE}=\mathrm{BD}\text{and FE}\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}âˆµ\mathrm{D}\text{is the mid-point}\\ \text{of BC.}\end{array}\right]\\ \text{BDEF is a parallelogram because one pair of opposite sides}\\ \text{is parallel and equal.}\\ \text{(ii) Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}FE}=\frac{1}{2}\mathrm{BC}\text{and FE}\text{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{FE}=\mathrm{DC}\text{and FE}\text{DC}\\ \text{Thus, EFDC is a parallelogram.\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Since},\text{one pair of opposite sides}\\ \text{is equal and parallel.}\end{array}\right]\\ \text{Similarly},\text{\hspace{0.17em}DEAF is a parallelogram.}\\ \text{Since},\text{diagonal of a parallelogram divides it into two congruent}\\ \text{triangles. So,}\\ \text{ar}\left(\mathrm{\Delta AFE}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)\text{, ar}\left(\mathrm{\Delta BFD}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)\text{and}\\ \text{ar}\left(\mathrm{\Delta DFE}\right)=\text{ar}\left(\mathrm{\Delta DCE}\right)\\ ⇒\text{\hspace{0.17em}ar}\left(\mathrm{\Delta AFE}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)=\text{ar}\left(\mathrm{\Delta BFD}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)\\ ⇒\text{ar}\left(\mathrm{\Delta DFE}\right)=\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)\\ \left[âˆµ\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\text{ar}\left(\mathrm{\Delta AFE}\right)+\text{ar}\left(\mathrm{\Delta DFE}\right)+\text{ar}\left(\mathrm{\Delta BFD}\right)+\text{ar}\left(\mathrm{\Delta DFE}\right)\right]\\ \left(\mathrm{iii}\right)\mathrm{a}\mathrm{r}\left(\mathrm{BDEF}\right)=\text{ar}\left(\mathrm{\Delta BFD}\right)+\text{ar}\left(\mathrm{\Delta DFE}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)+\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABC}\right)\end{array}$

Q.74 In figure shown below, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC)= ar (AOB)
(ii) ar (DCB)= ar (ACB)
(iii) DA||CB or ABCD is a parallelogram.

Ans

$\begin{array}{l}\text{Given:In quadrilateral ABCD, AC and BD are diagonals which}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}intersect at O,}\mathrm{OB}\text{}=\text{}\mathrm{OD}\text{and}\mathrm{AB}\text{}=\text{}\mathrm{CD}.\\ \text{To prove:\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right)\text{}\mathrm{ar}\text{}\left(\mathrm{DOC}\right)=\text{}\mathrm{ar}\text{}\left(\mathrm{AOB}\right)\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{ar}\text{}\left(\mathrm{DCB}\right)=\text{}\mathrm{ar}\text{}\left(\mathrm{ACB}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\text{}\mathrm{DA}||\mathrm{CB}\text{}\mathrm{or}\text{}\mathrm{ABCD}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{parallelogram}.\\ \text{Construction}:\text{Draw BM}\perp \text{AC and DN}\perp \text{AC.}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In}\mathrm{\Delta BOM}\text{and}\mathrm{\Delta DON}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOM}=\angle \mathrm{DON}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Vertical}\text{opposite angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BMO}=\angle \mathrm{DNO}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{each}\text{\hspace{0.17em}90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{OB}=\mathrm{OD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \mathrm{\Delta BOM}\cong \mathrm{\Delta DON}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{â€‹ A.A.S}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BM}=\mathrm{DN}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BOM}\right)=\mathrm{ar}\left(\mathrm{\Delta DON}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{In}\text{}\mathrm{\Delta BMA}\text{and}\mathrm{\Delta DNC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{BM}=\mathrm{DN}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BMA}=\angle \mathrm{DNC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{each}\text{\hspace{0.17em}90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta BMA}\cong \mathrm{\Delta DNC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BMA}\right)=\mathrm{ar}\left(\mathrm{\Delta DNC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \mathrm{ar}\left(\mathrm{\Delta BOM}\right)+\mathrm{ar}\left(\mathrm{\Delta BMA}\right)=\mathrm{ar}\left(\mathrm{\Delta DON}\right)+\mathrm{ar}\left(\mathrm{\Delta DNC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta AOB}\right)=\mathrm{ar}\left(\mathrm{\Delta DOC}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta DOC}\right)=\mathrm{ar}\left(\mathrm{\Delta AOB}\right)\\ \left(\mathrm{ii}\right)\text{Since},\text{}\mathrm{ar}\left(\mathrm{\Delta DOC}\right)=\mathrm{ar}\left(\mathrm{\Delta AOB}\right)\\ \mathrm{Adding}\text{ar}\left(\mathrm{\Delta BOC}\right)\text{both sides, we get}\\ \mathrm{ar}\left(\mathrm{\Delta DOC}\right)+\text{ar}\left(\mathrm{\Delta BOC}\right)=\mathrm{ar}\left(\mathrm{\Delta AOB}\right)+\text{ar}\left(\mathrm{\Delta BOC}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}\left(\mathrm{\Delta DCB}\right)=\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \left(\mathrm{iii}\right)\text{ar}\left(\mathrm{\Delta DCB}\right)=\text{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{If two triangles having same base and equal area, then}\\ \text{they lie between same parallels.}\\ \text{So, BC}\text{AD}\\ \text{or DA}\mathrm{BC}\\ \mathrm{As}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COD}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OAB}=\angle \mathrm{OCD}\\ ⇒\angle \mathrm{CAB}=\angle \mathrm{ACD}\\ ⇒\mathrm{AB}\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternate}\text{angles are equal.}\right]\\ \text{So},\text{ABCD is a parallelogram because opposite side}\\ \text{are parallel.}\end{array}$

Q.75 D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC)=ar(EBC). Prove that DE || BC.

Ans

$\begin{array}{l}\text{Given : In\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABC},\text{D and E are the points on AB and AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively and}\mathrm{ar}\left(\mathrm{DBC}\right)=\mathrm{ar}\left(\mathrm{EBC}\right).\\ \text{To prove : DE}\text{BC}\\ \text{Proof : Since,}\mathrm{ar}\left(\mathrm{DBC}\right)=\mathrm{ar}\left(\mathrm{EBC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and both triangles have same base BC.}\\ \text{So, DE}\text{BC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}âˆµ\mathrm{If}\text{two triangles have equal area and same}\\ \text{base, then they lie between same parallels.}\end{array}\right]\end{array}$

Q.76 XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ABE)= ar(ACF).

Ans

$\begin{array}{l}\text{Given:In}\mathrm{\Delta ABC},\text{XY}\mathrm{BC},\text{BE}\mathrm{AC}\text{and CF}\mathrm{AB}.\\ \text{To prove:\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABE}\right)=\text{}\mathrm{ar}\left(\mathrm{ACF}\right)\\ \text{Proof:Since, parallelograms BCYE and BCFX have same base}\\ \text{BC and lie between same parallels.So,}\\ \text{ar}\left(\text{BCYE}\right)=\mathrm{ar}\left(\text{BCFX}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{ΔAEB and parallelogram BCYE have same base and lie between}\\ \text{same parallels. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta ABE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BCYE}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}..\left(\mathrm{ii}\right)\\ \text{Similarly,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta ACF}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BCFX}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right),\text{we have}\\ \text{}\frac{1}{2}\text{ar}\left(\text{BCYE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BCFX}\right)\\ \text{From equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta ABE}\right)=\text{ar}\left(\mathrm{\Delta ACF}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence Proved.}\end{array}$

Q.77 The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the figure below). Show that ar (ABCD) = ar (PBQR).

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{ABCD}\text{and PBQR are parallelogram. AQ}\mathrm{CP}.\\ \mathrm{To}\text{prove:}\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{PBQR}\right)\\ \mathrm{Construction}:\text{Join AC and PQ.}\\ \text{Proof: Since, AQ}\text{CP}\\ \text{}\mathrm{\Delta ACP}\text{and}\mathrm{\Delta QPC}\text{have same base CP and both triangles are}\\ \text{between same parallels. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ACP}\right)=\mathrm{ar}\left(\mathrm{\Delta QPC}\right)\\ \mathrm{Subtracting}\text{ar}\left(\mathrm{\Delta BCP}\right)\text{from both sides, we get}\\ \mathrm{ar}\left(\mathrm{\Delta ACP}\right)-\text{ar}\left(\mathrm{\Delta BCP}\right)=\mathrm{ar}\left(\mathrm{\Delta QPC}\right)-\text{ar}\left(\mathrm{\Delta BCP}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta QBP}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2×\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=2×\mathrm{ar}\left(\mathrm{\Delta QBP}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{PBQR}\right)\\ \left[âˆµ\mathrm{Diagonal}\text{divides a parallelogram in two triangles of equal area.}\right]\end{array}$

Q.78 Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.

Prove that ar(AOD) = ar(BOC).

Ans

$\begin{array}{l}\text{Given:In trapezium ABCD, AB}\text{CD.}\\ \text{To prove: ar (AOD) = ar (BOC)}\\ \text{Proof: Area of two triangles is equal if they have same base}\\ \text{and lie between same parallels.}\\ \text{Here, AB}\text{CD,}\mathrm{\Delta ADC}\text{and}\mathrm{\Delta BDC}\text{have same base.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ADC}\right)=\mathrm{ar}\left(\mathrm{\Delta BDC}\right)\\ \mathrm{⇒}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta AOD}\right)+\mathrm{ar}\left(\mathrm{\Delta DOC}\right)=\mathrm{ar}\left(\mathrm{\Delta BDC}\right)+\mathrm{ar}\left(\mathrm{\Delta DOC}\right)\\ \mathrm{⇒}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta AOD}\right)=\mathrm{ar}\left(\mathrm{\Delta BDC}\right)\end{array}$

Q.79 In figure shown below, ABCDE is a pentagon.
A line through B parallel to AC meets DC produced at F. Show that
(i) ar(ACB) = ar (ACF)
(ii) ar(AEDF) = ar (ABCDE)

Ans

$\begin{array}{l}\text{Given: ABCDE is a pentagon and BF}\mathrm{AC}.\\ \text{To prove :}\left(\mathrm{i}\right)\text{}\mathrm{ar}\left(\mathrm{ACB}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{a}\mathrm{r}\left(\mathrm{AEDF}\right)=\mathrm{ar}\left(\mathrm{ABCDE}\right)\\ \text{Proof:}\left(\mathrm{i}\right)\text{}\mathrm{Since},\text{\hspace{0.17em}}\mathrm{AC}\mathrm{BF},\text{}\mathrm{\Delta ACB}\text{and}\mathrm{\Delta ACF}\text{have same base AC.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ACB}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)\\ \left[\begin{array}{l}\text{Areas of two triangles are equal if they have same base}\\ \text{and lie between same parallels.}\end{array}\right]\\ \left(\mathrm{ii}\right)\mathrm{W}\mathrm{e}\text{proved above:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ACB}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)\\ \text{Adding ar}\left(\mathrm{AEDC}\right)\text{both sides, we get}\\ \mathrm{ar}\left(\mathrm{ACB}\right)+\text{ar}\left(\mathrm{AEDC}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)+\text{ar}\left(\mathrm{AEDC}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{ABCDE}\right)=\text{ar}\left(\mathrm{AEDF}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{AEDF}\right)=\text{ar}\left(\mathrm{ABCDE}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.80 A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Let}\text{ABCD be the shape of the plot of land of Itwaari and}\\ \text{AED is his adjacent plot. ED be the side of his adjoining land.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{To}\text{prove: ar}\left(\mathrm{\Delta AOB}\right)=\text{ar}\left(\mathrm{\Delta EOD}\right)\\ \mathrm{Construction}:\text{}\\ \text{The proposal may be implemented as follows}.\mathrm{}\\ \text{Join diagonal BD and draw a line parallel to BD through point A}.\text{}\\ \text{Let it meet}\mathrm{}\text{the extended side CD of ABCD at point E}.\text{Join BE and}\\ \text{AD}.\text{Let them intersect each other at O}.\text{Then},\text{portion}\mathrm{\Delta }\text{AOB can}\\ \text{be cut from the original field so that the new shape of the field}\\ \text{will be}\mathrm{\Delta }\text{BCE}.\end{array}$

$\begin{array}{l}\text{Proof: Since,}\mathrm{\Delta EDB}\text{}\mathrm{and}\text{}\mathrm{\Delta ADB}\text{have same base and lie}\\ \text{between same parallels. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}\left(\mathrm{\Delta EDB}\right)=\text{ar}\left(\mathrm{\Delta ADB}\right)\\ \mathrm{Subtracting}\text{ar}\left(\mathrm{\Delta BOD}\right)\text{from both sides, we get}\\ \text{ar}\left(\mathrm{\Delta EDB}\right)-\text{ar}\left(\mathrm{\Delta BOD}\right)=\text{ar}\left(\mathrm{\Delta ADB}\right)-\text{ar}\left(\mathrm{\Delta BOD}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}\left(\mathrm{\Delta EOD}\right)=\text{ar}\left(\mathrm{\Delta BOA}\right)\\ \mathrm{Adding}\text{ar}\left(\mathrm{BCDO}\right)\text{both sides, we get}\\ \text{ar}\left(\mathrm{\Delta EOD}\right)+\text{ar}\left(\mathrm{BCDO}\right)=\text{ar}\left(\mathrm{\Delta BOA}\right)+\text{ar}\left(\mathrm{BCDO}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BCE}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)\\ \mathrm{Thus},\text{we can replace land in the shape of}\mathrm{\Delta BOA}\text{by the land}\\ \text{in the shape of}\mathrm{\Delta EOD}\text{to implement the proposal of Itwaari.}\end{array}$

Q.81 ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar(ACY).

Ans

$\begin{array}{l}\text{Given}:\text{}\mathrm{ABCD}\text{is a trapezium in which AB}\text{CD and XY}\mathrm{AC}.\\ \mathrm{To}\text{prove:}\mathrm{ar}\left(\mathrm{ADX}\right)=\mathrm{ar}\left(\mathrm{ACY}\right)\\ \text{Construction}:\mathrm{Join}\text{CX.}\\ \text{Proof:Since, AB}\mathrm{CD},\text{then}\\ \text{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACX}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{\hspace{0.17em}}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Both}\text{}\mathrm{\Delta s}\text{have same base and}\\ \text{lie between same parallels.}\end{array}\right]\\ \text{and AC}\mathrm{XY},\text{then}\\ \text{ar}\left(\mathrm{\Delta ACX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{\hspace{0.17em}}\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equqation}\left(\mathrm{ii}\right),\text{we have}\\ \text{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.82 In Fig.9.28, AP||BQ||CR. Prove that ar(AQC) = ar(PBR).

Ans

$\mathrm{Given}:\mathrm{In}\text{given figure,}\mathrm{AP}||\mathrm{BQ}||\mathrm{CR}$

$\begin{array}{l}\mathrm{To}\text{prove:}\mathrm{ar}\left(\mathrm{AQC}\right)=\mathrm{ar}\left(\mathrm{PBR}\right)\\ \text{Proof:Since,\hspace{0.17em}\hspace{0.17em}}\mathrm{AP}||\mathrm{BQ}||\\ \text{So, ar}\left(\mathrm{\Delta BPQ}\right)=\text{ar}\left(\mathrm{\Delta ABQ}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\mathrm{Both}\text{triangle}\mathrm{s}\text{have same base and lie between same parallels.}\right]\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{BQ}\mathrm{CR}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BQR}\right)=\text{ar}\left(\mathrm{\Delta BQC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\mathrm{Both}\text{triangle}\mathrm{s}\text{have same base and lie between same parallels.}\right]\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{ar}\left(\mathrm{\Delta BPQ}\right)+\text{ar}\left(\mathrm{\Delta BQR}\right)=\text{ar}\left(\mathrm{\Delta ABQ}\right)+\text{ar}\left(\mathrm{\Delta BQC}\right)\\ ⇒\text{ar}\left(\mathrm{\Delta BPR}\right)=\text{ar}\left(\mathrm{\Delta AQC}\right)\\ ⇒\text{\hspace{0.17em}ar}\left(\mathrm{\Delta AQC}\right)=\text{ar}\left(\mathrm{\Delta BPR}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Hence}\text{proved.}\end{array}$

Q.83 Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD)=ar (BOC). Prove that ABCD is a trapezium.

Ans

$\begin{array}{l}\text{Given}:\text{ABCD is a quadrilateral and its diagonals AC and BD}\\ \text{intersect at O and}\mathrm{ar}\left(\mathrm{AOD}\right)=\mathrm{ar}\left(\mathrm{BOC}\right).\\ \mathrm{To}\text{prove:}\mathrm{ABCD}\text{is a trapezium.}\\ \text{Proof:}\mathrm{It}\text{â€‹ is given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{AOD}\right)=\mathrm{ar}\left(\mathrm{BOC}\right)\\ \text{Adding}\mathrm{ar}\left(\mathrm{AOB}\right)\text{both sides, we get}\\ \mathrm{ar}\left(\mathrm{AOD}\right)+\mathrm{ar}\left(\mathrm{AOB}\right)=\mathrm{ar}\left(\mathrm{BOC}\right)+\mathrm{ar}\left(\mathrm{AOB}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABD}\right)=\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{Since},\text{\hspace{0.17em}both triagles have same base AB and having equal area.}\\ \text{So they will lie between same parallels.}\\ \text{Thus, AB}\text{CD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}ABCD is a trapezium.}\end{array}$

Q.84 In the figure shown below, ar (DRC) = ar (DPC) and ar(BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Ans

$\begin{array}{l}\text{Given}:\text{In given figure,}\\ \text{ar}\left(\text{DRC}\right)=\text{ar}\left(\text{DPC}\right)\text{and}\text{ar}\left(\text{BDP}\right)=\text{ar}\left(\text{ARC}\right).\\ \text{To prove: The}\text{quadrilaterals ABCD and DCPR are trapeziums}\text{.}\\ \text{Proof: It is given that}\\ \text{ar}\left(\text{DRC}\right)=\text{ar}\left(\text{DPC}\right)\\ or\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ar}\left(\text{DPC}\right)=\text{ar}\left(\text{DRC}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{Since, both triangles have same base and equal area}\text{. Then,}\\ \text{DC}\text{RP}\\ ⇒\text{DCPR is a trapezium}\text{.}\\ \text{Given that}\\ \text{ar}\left(\text{BDP}\right)=\text{ar}\left(\text{ARC}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{Subtracting equation}\left(i\right)\text{from equation}\left(ii\right),\text{we get}\\ \text{ar}\left(\text{BDP}\right)-\text{ar}\left(\text{DPC}\right)=\text{ar}\left(\text{ARC}\right)-\text{ar}\left(\text{DRC}\right)\\ ⇒\text{ar}\left(\text{BDC}\right)=\text{ar}\left(\text{ACD}\right)\\ \text{Since, both triangles have same base and equal area}\text{. Then,}\\ \text{AB}\text{DC}\\ ⇒\text{ABCD is a trapezium}\text{.}\\ \text{In this way, we proved that DCPR and ABCD are trapeziums}\text{.}\end{array}$

Q.85 Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Ans

$\begin{array}{l}\text{Given}:\text{Parallelogram ABCD and rectangle ABEF have same}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}base AB and equal area.}\\ \text{To prove: Perimeter of ABCD > Perimeter of ABEF}\\ \text{Proof: Since, both Parallelogram ABCD and rectangle ABEF}\\ \text{have same base AB and equal area}.\text{So, CF}\mathrm{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Now},\text{}\mathrm{AB}=\mathrm{DC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Opposite}\text{sides of rectangle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{EF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Opposite}\text{sides of parallelogram}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CD}=\mathrm{EF}\\ \text{Then},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}+\text{\hspace{0.17em}}\mathrm{CD}=\mathrm{AB}+\mathrm{EF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{Since},\text{the line segment drawn from an external point to a line}\\ \text{is the shortest line segment.}\\ \text{Then,}\mathrm{AD}>\mathrm{AF}\text{}\mathrm{and}\text{}\mathrm{BC}>\mathrm{BE}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}+\mathrm{BC}>\mathrm{AF}+\mathrm{BE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation (i) and equation}\left(\mathrm{ii}\right),\text{we get}\\ \mathrm{AB}+\text{\hspace{0.17em}}\mathrm{CD}+\mathrm{AD}+\mathrm{BC}>\mathrm{AB}+\mathrm{EF}+\mathrm{AF}+\mathrm{BE}\\ ⇒\text{Perimeter of ABCD > Perimeter of ABEF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.86 In the figure given below, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD)= ar(ADE)= ar(AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{ABC,}\mathrm{BD}=\mathrm{DE}=\mathrm{EC}.\\ \text{To prove:}\mathrm{ar}\left(\mathrm{ABD}\right)=\mathrm{ar}\left(\mathrm{ADE}\right)=\mathrm{ar}\left(\mathrm{AEC}\right)\\ \text{Construction: Draw AL}\perp \text{BC}\\ \text{Proof: Since, Area of a triangle}=\frac{1}{2}×\mathrm{perpendicular}×\mathrm{base}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABD}\right)=\frac{1}{2}×\mathrm{BD}×\mathrm{AM}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ADE}\right)=\frac{1}{2}×\mathrm{DE}×\mathrm{AM}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta AEC}\right)=\frac{1}{2}×\mathrm{EC}×\mathrm{AM}\\ \text{Since},\text{}\mathrm{BD}=\mathrm{DE}=\mathrm{EC}\\ \mathrm{So},\text{}\frac{1}{2}×\mathrm{BD}×\mathrm{AM}=\frac{1}{2}×\mathrm{DE}×\mathrm{AM}=\frac{1}{2}×\mathrm{EC}×\mathrm{AM}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABD}\right)=\mathrm{ar}\left(\mathrm{ADE}\right)=\mathrm{ar}\left(\mathrm{AEC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.87 In the figure below, ABCD, DCFE and ABFE are parallelograms. Show that

Ans

$\begin{array}{l}\text{Given}:\text{ABCD},\text{DCFE and ABFE are parallelograms}.\\ \text{To prove}:\text{ar}\left(\text{ADE}\right)\text{}=\text{ar}\left(\text{BCF}\right)\\ \text{Proof}:\text{Since},\text{ABCD is a parallelogram}.\\ \text{So},\text{AD}=\text{BC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\left[\text{Opposite sides of parallelogram}.\right]\\ \text{Since},\text{DCFE is a parallelogram}.\\ \text{So},\text{DE}=\text{CF}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Opposite sides of parallelogram}.\right]\\ \text{Since},\text{ABFE is a parallelogram}.\\ \text{So},\text{AE}=\text{BF}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Opposite sides of parallelogram}.\right]\\ \text{In}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta \text{ADE and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta \text{BCF}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AD}=\text{BC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{DE}=\text{CF}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AE}=\text{BF}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta \text{ADE}\cong \Delta \text{BCF}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{S}\text{.S}\text{.S}\text{.}\right]\\ ar\left(\Delta \text{ADE}\right)=ar\left(\Delta \text{BCF}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Area\text{of congruent triangles is equal}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Hence\text{proved}\text{.}\end{array}$

Q.88 In the figure below, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ.
If AQ intersect DC at P, show that ar(BPC) = ar(DPQ).

Ans

Given : ABCD is a parallelogram and BC is produced upto Q such that AD = CQ.
To prove : ar(BPC) = ar (DPQ)
Construction : Join AC.

$\begin{array}{l}\text{Proof}:\text{Since,}\mathrm{\Delta APC}\text{and}\mathrm{\Delta BPC}\text{have same base and lie between}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}same parallels. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta APC}\right)=\mathrm{ar}\left(\mathrm{\Delta BPC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{In}\text{quadrilateral ACQD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}=\text{CQ and AD}=\mathrm{CQ}\\ \mathrm{So},\text{ACQD is a parallelogram.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta DCQ}\text{and}\mathrm{\Delta ACQ}\text{have same base QC and lie between}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}same parallels. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta DCQ}\right)=\mathrm{ar}\left(\mathrm{\Delta ACQ}\right)\\ \text{Substracting ar}\left(\mathrm{\Delta PCQ}\right)\text{from both sides, we get}\\ \mathrm{ar}\left(\mathrm{\Delta DCQ}\right)-\text{ar}\left(\mathrm{\Delta PCQ}\right)=\mathrm{ar}\left(\mathrm{\Delta ACQ}\right)-\text{ar}\left(\mathrm{\Delta PCQ}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta DPQ}\right)=\text{ar}\left(\mathrm{\Delta APC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{â€‹â€‹ and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BPC}\right)=\text{ar}\left(\mathrm{\Delta DPQ}\right)\text{Hence proved.}\end{array}$

Q.89

$\begin{array}{l}\text{In Fig.9.33, ABC and BDE are two equilateral triangles such}\\ \text{that D is the mid-point of BC. If AE intersects BC at F, show}\\ \text{that}\\ \left(\text{i}\right)\text{ar}\left(\text{BDE}\right)\text{=}\frac{\text{1}}{\text{4}}\text{ar}\left(\text{ABC}\right)\\ \left(\text{ii}\right)\text{ar}\left(\text{BDE}\right)\text{=}\frac{\text{1}}{\text{2}}\text{ar}\left(\text{BAE}\right)\\ \left(\text{iii}\right)\text{ar}\left(\text{ABC}\right)\text{=2ar}\left(\text{BEC}\right)\\ \left(\text{iv}\right)\text{ar}\left(\text{BFE}\right)\text{=ar}\left(\text{AFD}\right)\\ \left(\text{v}\right)\text{ar}\left(\text{BFE}\right)\text{=2ar}\left(\text{FED}\right)\\ \left(\text{vi}\right)\text{ar}\left(\text{FED}\right)\text{=}\frac{\text{1}}{\text{8}}\text{ar}\left(\text{AFC}\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{\Delta ABC}\text{and}\mathrm{\Delta BDE}\text{are two equilateral triangles. D is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}mid-point of BC. AE intersects BC at F.}\\ \mathrm{To}\text{prove:}\left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{BDE}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{BDE}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BAE}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{ABC}\right)=2\mathrm{ar}\left(\mathrm{BEC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\mathrm{ar}\left(\mathrm{BFE}\right)=\mathrm{ar}\left(\mathrm{AFD}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{v}\right)\mathrm{ar}\left(\mathrm{BFE}\right)=2\mathrm{ar}\left(\mathrm{FED}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{vi}\right)\mathrm{ar}\left(\mathrm{FED}\right)=\frac{1}{8}\mathrm{ar}\left(\mathrm{AFC}\right)\\ \mathrm{Construction}:\mathrm{Join}\text{EC and AD. Let mid-points of AB and AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}as G and H respectively. Join G and H, G and D,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}H and D respectively.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta ABC},\text{G and H are the mid-points of AB and AC}\\ \text{respectively. So, by mid-point theorem}\\ \text{GH}\text{BC and GH}=\frac{1}{2}\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{GH}\text{B}\mathrm{D}\text{and GH}=\text{BD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{GHDB}\text{is a parallelogram because one pair of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}opposite sides is equal and parallel.}\\ \mathrm{Similarly},\text{â€‹}\mathrm{GHDC}\text{is a parallelogram because one pair of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}opposite sides is equal and parallel.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta CAB},\text{D and H are the mid-points of CB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and CA respectively. So, by mid-point theorem}\\ \text{DH}\mathrm{A}\text{B and DH}=\frac{1}{2}\mathrm{AB}\text{\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{D}\mathrm{H}\mathrm{A}\text{B and DH}=\text{AG}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DHAG}\text{is a parallelogram because one pair of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}opposite sides is equal and parallel.}\\ \text{Now, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}GB}\mathrm{DE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{alternate}\text{angles are equal i.e., each 60°}\right]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{GB}=\mathrm{DE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\mathrm{DE}=\mathrm{BD}=\frac{1}{2}\mathrm{BC}=\frac{1}{2}\mathrm{AB}=\mathrm{GB}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{\hspace{0.17em}}\mathrm{BEDG}\text{is a parallelogram because one pair}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of opposite sides is equal and parallel.}\\ \mathrm{Since},\text{diagonal divides a parallelogram in two triangles of}\\ \mathrm{equal}\text{\hspace{0.17em}}\mathrm{area}.\\ \text{Then, ar}\left(\mathrm{DGH}\right)=\text{ar}\left(\mathrm{DGB}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{For}\text{}{}^{\mathrm{gm}}\text{\hspace{0.17em}}\mathrm{BDGH}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGH}\right)=\text{ar}\left(\mathrm{AGH}\right)\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{For}\text{}{}^{\mathrm{gm}}\text{\hspace{0.17em}}\mathrm{AHDG}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGH}\right)=\text{ar}\left(\mathrm{HDC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{For}\text{}{}^{\mathrm{gm}}\text{\hspace{0.17em}}\mathrm{GHCD}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGB}\right)=\text{ar}\left(\mathrm{HDC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{For}\text{}{}^{\mathrm{gm}}\text{\hspace{0.17em}}\mathrm{BEDG}\right]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{DGB}\right)=\text{ar}\left(\mathrm{BDE}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{For}\text{}{}^{\mathrm{gm}}\text{\hspace{0.17em}}\mathrm{BEDG}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{ABC}\right)=\text{ar}\left(\mathrm{DGH}\right)+\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{AGH}\right)+\text{ar}\left(\mathrm{HDC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{DGB}\right)+\text{ar}\left(\mathrm{DGB}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=4\text{\hspace{0.17em}ar}\left(\mathrm{DGB}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=4\text{ar}\left(\mathrm{BDE}\right)\\ \therefore \text{ar}\left(\mathrm{BDE}\right)=\frac{1}{4}\text{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{ii}\right)\mathrm{S}\mathrm{ince},\text{}\mathrm{\Delta BDE}\text{and parallelogram BEDG lie on the same base}\\ \text{and between same parallels, so}\\ \text{ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BEDG}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BE}\mathrm{DG}\text{and DG}\mathrm{HA}\\ ⇒\mathrm{BE}\mathrm{AH}\text{and AB}\mathrm{HE}\\ ⇒\mathrm{ABEH}\text{is a parallelogram.}\\ \text{Then, ar}\left(\mathrm{BEDG}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABEH}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{Since,}\mathrm{\Delta ABE}\text{and parallelogram ABEH lie on the same base}\\ \text{and between same parallels, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta ABE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{ABEH}\right)\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right),\text{we have}\\ \text{ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{2}\text{\hspace{0.17em}ar}\left(\mathrm{\Delta ABE}\right)\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{ar}\left(\mathrm{\Delta ABE}\right)=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common base BE and BE}||\text{AC}\right]\\ \mathrm{ar}\left(\mathrm{\Delta ABF}\right)+\mathrm{ar}\left(\mathrm{\Delta BFE}\right)\text{\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{AB}\mathrm{ED}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BED}\right)=\mathrm{ar}\left(\mathrm{\Delta AED}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common base BE and BE}||\text{AC}\right]\\ ⇒\mathrm{ar}\left(\mathrm{\Delta BFE}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BFE}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABF}\right)+\mathrm{ar}\left(\mathrm{\Delta AFD}\right)\text{\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABD}\right)\text{\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta BEC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{AD}\text{is median, so}\\ \mathrm{ar}\left(\mathrm{\Delta ABD}\right)=\text{\hspace{0.17em}}\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\end{array}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABC}\right)=2\mathrm{ar}\left(\mathrm{BEC}\right)\\ \left(\mathrm{iv}\right)\text{}\mathrm{Since},\text{AB}\mathrm{ED}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BED}\right)=\mathrm{ar}\left(\mathrm{\Delta AED}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common base BE and BE}||\text{AC}\right]\\ ⇒\mathrm{ar}\left(\mathrm{\Delta BFE}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)+\mathrm{ar}\left(\mathrm{\Delta EFD}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BFE}\right)=\mathrm{ar}\left(\mathrm{\Delta AFD}\right)\\ \left(\mathrm{v}\right)\mathrm{Let}\text{height of}\mathrm{\Delta BDE}\text{be h and height of}\mathrm{\Delta ABC}\text{be H, then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{2}×\mathrm{BD}×\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\frac{1}{2}×\mathrm{BC}×\mathrm{H}\\ \mathrm{Since},\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BDE}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}×\mathrm{BD}×\mathrm{h}=\frac{1}{4}\left(\frac{1}{2}×\mathrm{BC}×\mathrm{H}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}×\mathrm{BD}×\mathrm{h}=\frac{1}{4}\left(\frac{1}{2}×2\mathrm{BD}×\mathrm{H}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\mathrm{BC}=2\mathrm{BD}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{1}{2}\mathrm{H}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}âˆµ\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{BEF}\right)=\mathrm{ar}\left(\mathrm{AFD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×\mathrm{FD}×\mathrm{H}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×\mathrm{FD}×2\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left(\frac{1}{2}×\mathrm{FD}×\mathrm{h}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \mathrm{ar}\left(\mathrm{BEF}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{ar}\left(\mathrm{\Delta FED}\right)\\ \left(\mathrm{vi}\right)\text{Since, ar}\left(\mathrm{AFC}\right)=\mathrm{ar}\left(\mathrm{AFD}\right)+\mathrm{ar}\left(\mathrm{ADC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{BFE}\right)+\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{from}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\text{and D is}\\ \text{mid-point of BC.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{BFE}\right)+\frac{1}{2}×4\mathrm{ar}\left(\mathrm{BDE}\right)\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{from}\text{}\left(\mathrm{i}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{BFE}\right)+2\mathrm{ar}\left(\mathrm{BDE}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{BFE}\right)+2\left\{\mathrm{ar}\left(\mathrm{BEF}\right)+\mathrm{ar}\left(\mathrm{FED}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{ar}\left(\mathrm{FED}\right)+2\left[2\mathrm{ar}\left(\mathrm{FED}\right)+\mathrm{ar}\left(\mathrm{FED}\right)\right]\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{from}\text{â€‹}\left(\mathrm{v}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{ar}\left(\mathrm{FED}\right)+6\mathrm{ar}\left(\mathrm{FED}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{AFC}\right)=8\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{FED}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{FED}\right)=\frac{1}{8}\text{ar}\left(\mathrm{AFC}\right)\end{array}$

Q.90 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Show that ar(APB) × ar(CPD) = ar(APD) × ar(BPC).

Ans

$\begin{array}{l}\text{Given}:\mathrm{ABCD}\text{is a quadrilateral in which diagonals AC and BD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}intersect at P.}\\ \text{To prove:\hspace{0.17em}}\mathrm{ar}\left(\mathrm{APB}\right)×\mathrm{ar}\left(\mathrm{CPD}\right)=\mathrm{ar}\left(\mathrm{APD}\right)×\mathrm{ar}\left(\mathrm{BPC}\right)\\ \text{Construction}:\text{Draw AM}\perp \text{BD and CN}\perp \text{BD.}\\ \text{Proof:L.H.S.}=\mathrm{ar}\left(\mathrm{APB}\right)×\mathrm{ar}\left(\mathrm{CPD}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{1}{2}×\mathrm{PB}×\mathrm{AM}\right)×\left(\frac{1}{2}×\mathrm{PD}×\mathrm{CN}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{1}{2}×\mathrm{PD}×\mathrm{AM}\right)×\left(\frac{1}{2}×\mathrm{PB}×\mathrm{CN}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{APD}\right)×\mathrm{ar}\left(\mathrm{BPC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.91

$\begin{array}{l}\mathrm{P}\mathrm{and}\mathrm{Q}\mathrm{are}\mathrm{respectively}\mathrm{the}\mathrm{mid}–\mathrm{points}\mathrm{of}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{BC}\\ \mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}\mathrm{and}\mathrm{R}\mathrm{is}\mathrm{the}\mathrm{mid}–\mathrm{point}\mathrm{of}\mathrm{AP},\mathrm{show}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{PRQ}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ARC}\right)\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{RQC}\right)=\frac{3}{8}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{PBQ}\right)=\mathrm{ar}\left(\mathrm{ARC}\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{\Delta ABC},\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively. R is the mid-point of AP.}\\ \mathrm{To}\text{prove:\hspace{0.17em}}\left(\mathrm{i}\right)\text{}\mathrm{ar}\left(\mathrm{PRQ}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ARC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{ar}\left(\mathrm{RQC}\right)=\frac{3}{8}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{PBQ}\right)=\mathrm{ar}\left(\mathrm{ARC}\right)\\ \mathrm{Construction}:\text{Take S as a mid-point of AC and join PS.}\mathrm{Draw}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}CT}\mathrm{AP}\text{and produce PQ upto T as PQ}=\text{QT.Join PC.}\\ \text{Proof: In}\mathrm{\Delta }\text{ABC, P and Q are the mid-points of AB and BC points}\\ \text{of AB and BC respectively}.\text{Hence},\text{by using mid-point}\\ \text{theorem},\text{we}\mathrm{get}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\mathrm{AC}\text{and PQ}=\frac{1}{2}\mathrm{AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\mathrm{AC}\text{and PQ}=\mathrm{AS}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQSA}\text{is a parallelogram.\hspace{0.17em}PQSA is a parallelogram}.\text{We know}\\ \text{that diagonals of a parallelogram bisect it into equal areas of}\\ \text{triangles}.\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta PAS}\right)=\mathrm{ar}\left(\mathrm{\Delta PAQ}\right)=\mathrm{ar}\left(\mathrm{\Delta SQP}\right)=\mathrm{ar}\left(\mathrm{\Delta SQA}\right)\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\mathrm{SC}\text{and PQ}=\mathrm{SC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQCS}\text{is a parallelogram.\hspace{0.17em}PQCS is a parallelogram}.\text{We know}\\ \text{that diagonals of a parallelogram bisect it into equal areas of}\\ \text{triangles}.\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta PQS}\right)=\mathrm{ar}\left(\mathrm{\Delta QSC}\right)\\ \mathrm{Similarly},\text{QSCT and PSQB are also parallelogram.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta QSC}\right)=\mathrm{ar}\left(\mathrm{\Delta CTQ}\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta PSQ}\right)=\mathrm{ar}\left(\mathrm{\Delta QBP}\right)\\ \mathrm{Thus},\text{}\mathrm{ar}\left(\mathrm{\Delta PAS}\right)=\mathrm{ar}\left(\mathrm{\Delta SQP}\right)=\mathrm{ar}\left(\mathrm{\Delta SQA}\right)=\mathrm{ar}\left(\mathrm{\Delta PAQ}\right)=\mathrm{ar}\left(\mathrm{\Delta QSC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta CTQ}\right)=\mathrm{ar}\left(\mathrm{\Delta CTQ}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ âˆµ\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PAS}\right)+\mathrm{ar}\left(\mathrm{\Delta PQS}\right)+\mathrm{ar}\left(\mathrm{\Delta QSC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=4\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\\ ⇒\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left(\mathrm{i}\right)\mathrm{In}\text{}\mathrm{\Delta PAQ},\text{â€‹ QR is median. So,}\\ \text{ar}\left(\mathrm{\Delta PQR}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta PAQ}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\mathrm{QP}\text{is a median in}\mathrm{\Delta QBA}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \left(\mathrm{ii}\right)\mathrm{In}\text{â€‹}\mathrm{\Delta ABC},\text{P and Q are the mid-points of AB and BC res}\\ \text{So, by mid-point theorem}\\ \text{PQ}=\frac{1}{2}\mathrm{AC}\text{and PQ}\text{AC}\\ ⇒\text{2PQ}=\mathrm{AC}\text{and PQ}\text{AC}\\ ⇒\text{PT}=\mathrm{AC}\text{and PT}\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Q}\text{is mid-point of PT.}\right]\\ ⇒\mathrm{PACT}\text{is a parallelogram.}\\ \mathrm{ar}\left(\mathrm{PACT}\right)=\mathrm{ar}\left(\mathrm{PACQ}\right)+\mathrm{ar}\left(\mathrm{\Delta QTC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{PACQ}\right)+\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{ar}\left(\mathrm{PACT}\right)=\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iv}\right)\end{array}$

$\begin{array}{l}\mathrm{ar}\left(\mathrm{\Delta ARC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta PAC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\mathrm{CR}\text{is a median.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×\frac{1}{2}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\mathrm{CP}\text{is a median.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2×\frac{1}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \mathrm{ar}\left(\mathrm{\Delta ARC}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{v}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{iii}\right)\right]\\ \mathrm{Now},\text{â€‹}\\ \mathrm{ar}\left(\mathrm{\Delta RQC}\right)=\mathrm{ar}\left(\mathrm{PACQ}\right)-\mathrm{ar}\left(\mathrm{\Delta ARC}\right)-\text{ar}\left(\mathrm{\Delta PQR}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{ar}\left(\mathrm{PACQ}\right)-2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)-\text{ar}\left(\mathrm{\Delta PQR}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{v}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{PACT}\right)-\mathrm{ar}\left(\mathrm{\Delta QTC}\right)-3\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\mathrm{ar}\left(\mathrm{\Delta QTC}\right)-3\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{iv}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)-3\text{\hspace{0.17em}ar}\left(\mathrm{\Delta PQR}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{3}{8}\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right)\right]\\ \mathrm{ar}\left(\mathrm{\Delta RQC}\right)=\frac{3}{8}\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\mathrm{R}.\mathrm{H}.\mathrm{S}.=\mathrm{ar}\left(\mathrm{\Delta ARC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{PACQ}\right)-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\mathrm{ar}\left(\mathrm{PACT}\right)-\mathrm{ar}\left(\mathrm{\Delta QTC}\right)\right\}-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\right\}-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{}\mathrm{and}\text{}\left(\mathrm{iv}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\right\}-\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)-\mathrm{ar}\left(\mathrm{\Delta QRC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{1}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)-\frac{3}{8}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\text{}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{ar}\left(\mathrm{\Delta PBQ}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\\ \mathrm{ar}\left(\mathrm{\Delta PBQ}\right)=\mathrm{ar}\left(\mathrm{\Delta ARC}\right)\end{array}$

Q.92

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{figure}\mathrm{below},\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{A}.\mathrm{BCED},\\ \mathrm{ACFG}\mathrm{and}\mathrm{ABMN}\mathrm{are}\mathrm{squares}\mathrm{on}\mathrm{the}\mathrm{sides}\mathrm{BC},\mathrm{CA}\mathrm{and}\mathrm{AB}\\ \mathrm{respectively}.\mathrm{Line}\mathrm{segment}\mathrm{AX}\perp \mathrm{DE}\mathrm{meets}\mathrm{BC}\mathrm{at}\mathrm{Y}.\\ \mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta MBC}\cong \mathrm{\Delta ABD}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{BYXD}\right)=2\mathrm{ar}\left(\mathrm{MBC}\right)\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{BYXD}\right)=\mathrm{ar}\left(\mathrm{ABMN}\right)\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}}\mathrm{\Delta FCB}\cong \mathrm{\Delta ACE}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{CYXE}\right)=2\mathrm{ar}\left(\mathrm{FCB}\right)\\ \left(\mathrm{vi}\right)\mathrm{ar}\left(\mathrm{CYXE}\right)=\mathrm{ar}\left(\mathrm{ACFG}\right)\\ \left(\mathrm{vii}\right)\mathrm{ar}\left(\mathrm{BCED}\right)=\mathrm{ar}\left(\mathrm{ABMN}\right)+\mathrm{ar}\left(\mathrm{ACFG}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{In}\text{}\mathrm{\Delta MBC}\text{and}\mathrm{\Delta ABD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{MB}=\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Sides}\text{of square ABMN}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{MBC}=\angle \mathrm{ABD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\angle \mathrm{MBC}=90\mathrm{°}+\angle \mathrm{ABC}=\angle \mathrm{ABD}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\text{Sides of square BCED}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta MBC}\cong \mathrm{\Delta ABD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{(ii)ΔMBC and parallelogram BYXD lie on the same base BD and}\\ \text{lie between same parallels BD and AX. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BYXD}\right)\\ ⇒2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\mathrm{ar}\left(\mathrm{BYXD}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{BYXD}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)\\ \text{(iii)ΔMBC and parallelogram ABMN lie on the same base MB and}\\ \text{lie between same parallels MB and NC. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABMN}\right)\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta MBC}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BYXD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{\hspace{0.17em}}\left(\mathrm{ii}\right)\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\mathrm{ar}\left(\mathrm{ABMN}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{BYXD}\right)\\ ⇒\text{}\mathrm{ar}\left(\mathrm{ABMN}\right)=\mathrm{ar}\left(\mathrm{BYXD}\right)\\ \text{(iv)In}\mathrm{\Delta FCB}\text{and}\mathrm{\Delta ACE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{F}\mathrm{C}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Sides}\text{of square ACFG}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{FCB}=\angle \mathrm{ACE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\angle \mathrm{FCB}=90\mathrm{°}+\angle \mathrm{ACB}=\angle \mathrm{ACE}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{CE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Sides}\text{of square BCED}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta FCB}\cong \mathrm{\Delta ACE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{v}\right)\mathrm{S}\mathrm{ince},\text{\hspace{0.17em}}\mathrm{\Delta FCB}\cong \mathrm{\Delta ACE}\\ \text{So, ar}\left(\mathrm{\Delta FCB}\right)=\mathrm{ar}\left(\mathrm{\Delta ACE}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{}\mathrm{\Delta ACE}\text{and parallelogram CYXE lie on the same base CE and}\\ \text{lie between same parallels CE and AX. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta ACE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{CYXE}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\text{CYXE}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta ACE}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\text{CYXE}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \left(\mathrm{vi}\right)\mathrm{\Delta }\mathrm{FCB}\text{and parallelogram ACFG lie on the same base CF and}\\ \text{lie between same parallels CF and BG. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ACFG}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ACFG}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\text{CYXE}\right)=2\text{\hspace{0.17em}ar}\left(\mathrm{\Delta FCB}\right)\\ \mathrm{So},\text{â€‹\hspace{0.17em}}\mathrm{ar}\left(\text{CYXE}\right)=\mathrm{ar}\left(\mathrm{ACFG}\right)\\ \left(\mathrm{vii}\right)\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{BCED}\right)=\mathrm{ar}\left(\mathrm{BYXD}\right)+\mathrm{ar}\left(\mathrm{CYXE}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}ar}\left(\mathrm{ABMN}\right)+\text{\hspace{0.17em}ar}\left(\mathrm{ACFG}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{}\left(\mathrm{iii}\right)\text{and}\left(\mathrm{vi}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{He}\text{nce proved.}\end{array}$

Q.93 Fill in the blanks:
(i) The centre of a circle lies in_______ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in______ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a _____ of the circle.
(iv) An arc is a_______ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and ________of the circle.
(vi) A circle divides the plane, on which it lies, in_______ parts.

Ans

(i) The centre of a circle lies in interior of the circle.
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semi-circle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.

Ans

(i) True. Since, all the points on the circumference are equidistant from the centre of the circle, and this distance is called the radius of the circle.
(ii) False. Since, there are infinite points on the circumference. Then, we can draw infinite number of chords of same length. Hence, a circle has infinite number of equal chords.
(iii) False. Three equal arcs in a circle will not be major arc because there is no minor arc.
(iv) True. Since, a chord equal to twice of the radius is called diameter of the circle.

(v) False. Sector is the region bounded by an arc and two radii of the circle. OAB is the sector of the circle in the figure.

(vi) True. A circle is a two dimensional figure as it has not thickness, so it is called plane figure.

Q.95 Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Ans

$\begin{array}{l}\text{Given}:\text{Two congruent circles with centre O and O’ i}\text{.e}\text{., OA}=\text{OP,}\\ \text{OB}=\text{O’Q and}\angle AOB=\angle PO‘Q\text{.}\\ \text{To prove: chord AB}=\text{chord PQ}\\ \text{Proof}:\text{In}\Delta AOB\text{and}\Delta PO‘Q\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}OA=O‘P\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Radii of congruent circles}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle AOB=\angle PO‘Q\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Given}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}OB=O‘Q\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Radi}i\text{of congruent circles}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\therefore \Delta AOB\cong \Delta PO‘Q\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{S}\text{.A}\text{.S}\text{.}\right]\\ \text{Therefore,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AB=PQ\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \text{Thus},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{chord AB}=chord\text{PQ}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.96 Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Ans

Different pairs of circles are given below:

There are zero common points in figure (a) and (d).

There is one common point in figure (b). There are two common points in figure (c).

Thus, maximum number of common points is 2.

Q.97 Suppose you are given a circle. Give a construction to find its centre.

Ans

Given: Let there are three point A, B and C on the circumference.

To find: Centre of circle.

Steps of construction:

(i) Join A and B, B and C.
(ii) Draw perpendicular bisector of AB and BC.

The intersection point O, of perpendicular bisector is called centre of circle.

Q.98 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Ans

$\begin{array}{l}\text{Given}:\text{Two circles with centre O and O}’\text{intersect at A and B}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AB is common chord of both circles}.\\ \text{To prove}:\text{Centres O and O}’\text{lie on the perpendicular bisector of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AB i}.\text{e}.,\text{OO}’\text{is perpendicular bisector of AB}.\\ \text{Proof}:\text{In}\mathrm{\Delta OAO}‘\text{and}\mathrm{\Delta OBO}‘\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OA}=\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{radii of same circle.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OO}‘=\mathrm{O}‘\mathrm{O}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{O}‘\mathrm{A}=\mathrm{O}‘\mathrm{B}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{radii of same circle.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta OAO}‘\cong \mathrm{\Delta OBO}‘\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOO}‘=\angle \mathrm{BO}‘\mathrm{O}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AO}‘\mathrm{O}=\angle \mathrm{BOO}‘\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta OAP}\text{and}\mathrm{\Delta OBP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OA}=\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{radii of same circle.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOP}=\angle \mathrm{BO}‘\mathrm{P}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OP}=\mathrm{OP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta OAP}\cong \mathrm{\Delta OBP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\end{array}$ $\begin{array}{l}So,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AP=BP\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle OPA=\angle OPB\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ But\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle OPA+\angle OPB=180°\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle OPA=\angle OPB=90°\\ ⇒\text{OP is perpendicular bisector of AB}\text{.}\\ ⇒OO‘\text{is perpendicular bisector of AB}\text{.}\\ O\text{and O’ lies on perpendicular bisector of AB}\text{. Hence proved}\text{.}\end{array}$

Q.99 Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Ans

Let the assumed figure is as given below:

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}Given:Radius OA}=\text{5 cm, O’A}=\text{3 cm and OO’}=4\text{}\mathrm{cm}.\\ \text{To find: AB}\\ \text{Let OC}=\text{x cm and O’C}=\left(4-\mathrm{x}\right)\text{cm}\\ \text{In}\mathrm{\Delta ACO},\text{}\angle \mathrm{ACO}=90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}By Pythagoras theorem,}\\ {\text{OA}}^{\text{2}}={\mathrm{AC}}^{2}+{\mathrm{OC}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}{5}^{2}={\mathrm{AC}}^{2}+{\mathrm{x}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{AC}}^{2}=25-{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{In}\mathrm{\Delta ACO}‘,\text{}\angle \mathrm{ACO}‘=90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}By Pythagoras theorem,}\\ {\text{O’A}}^{\text{2}}={\mathrm{AC}}^{2}+\mathrm{O}‘{\mathrm{C}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}{3}^{2}={\mathrm{AC}}^{2}+{\left(4-\mathrm{x}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{AC}}^{2}=9-{\left(4-\mathrm{x}\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25-{\mathrm{x}}^{2}=9-{\left(4-\mathrm{x}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25-{\mathrm{x}}^{2}=9-\left(16-8\mathrm{x}+{\mathrm{x}}^{2}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25-{\mathrm{x}}^{2}=9-16+8\mathrm{x}-{\mathrm{x}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25=-7+8\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}25+7=8\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{32}{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=4\text{cm}\\ \text{Substituting value of x in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{AC}}^{2}=25-{4}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25-16\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=9\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{AC}=\sqrt{9}=3\\ \text{Since},\text{OO’ is perpendicular bisector of AB.}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=2×\mathrm{AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=2×3=6\text{cm}\\ \text{Thus, the length of common chord is 6 cm.}\end{array}$

Q.100 If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Ans

$\begin{array}{l}\text{Given:Chord AB}=\text{Chord CD, both chords intersect at R.}\\ \text{To prove:AR}=\text{RC and RB}=\text{RD}\\ \text{Construction}:\text{Draw OP}\perp \text{AB and OQ}\perp \text{CD. Join OR.}\\ \text{Proof:In}\mathrm{\Delta }\text{OPR and}\mathrm{\Delta }\text{OQR,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}OP}=\text{OQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Equal}\text{chords are equidistant.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OPR}=\angle \mathrm{OQR}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OR}=\mathrm{OR}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \mathrm{\Delta OPR}\cong \mathrm{\Delta OQR}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \text{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PR}=\text{QR\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PB}=\text{QD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Perpendicular}\text{from centre bisects the}\\ \text{chord.}\end{array}\right]\\ \text{Subtracting equation}\left(\mathrm{i}\right)\text{from equation}\left(\mathrm{ii}\right),\text{â€‹ we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PB}-\mathrm{PR}=\text{QD}-\mathrm{QR}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} RB}=\text{RD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=\text{CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iv}\right)\left[\mathrm{Given}\right]\\ \text{Equation(iv)−equation(iii)},\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB\hspace{0.17em}}-\text{RB}=\text{CD}-\text{RD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AR}=\mathrm{RC}\\ \text{and RB}=\text{RD.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence Proved.}\end{array}$

Q.101 If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Ans

$\begin{array}{l}\text{Given:Chord AB}=\text{Chord CD, both chords intersect at R.}\\ \text{To prove:}\angle \mathrm{ORA}=\angle \mathrm{ORC}\\ \text{Construction}:\text{Draw OP}\perp \text{AB and OQ}\perp \text{CD. Join OR.}\\ \text{Proof:In}\mathrm{\Delta }\text{OPR and}\mathrm{\Delta }\text{OQR,}\\ \text{OP}=\text{OQ}\left[\text{Equal chords are equidistant.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OPR}=\angle \mathrm{OQR}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{OR}=\mathrm{OR}\left[\text{Common}\right]\\ \therefore \mathrm{\Delta OPR}\cong \mathrm{\Delta OQR}\left[\mathrm{By}\text{R.H.S.}\right]\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ORA}=\angle \mathrm{ORC}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Hence Proved.}\end{array}$

Q.102 If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Ans

$\begin{array}{l}Given:\text{Two concentric circles with centre O}\text{. A common chord}\\ \text{intersect these circles at AD and BC}\text{.}\\ To\text{prove: AB}=\text{CD}\\ \text{Contruction: Draw OP}\perp \text{AD}\end{array}$

$\begin{array}{l}\text{Proof:Since, perpendicular from centre to chord of circle bisects}\\ \text{the chord}\text{. So,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AP}=\text{PD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}BP=PC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{Subtracting}\text{equation}\left(i\right)\text{from equation}\left(ii\right),\text{we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}BP-\text{AP}=PC-\text{PD}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AB=CD\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.103 Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Ans

$\begin{array}{l}\text{Given:Radius of circular park is 5 m, distance between Reshma}\\ \text{and Salma}\left(\mathrm{AB}\right)\text{and distance between Salma and Mandeep}\\ \text{}\left(\mathrm{BC}\right)\text{is equal to 6 m each}\\ \text{To Find: Distance between Reshma and Mandeep i.e., AC.}\\ \text{Proof:In}\mathrm{\Delta ODC},\text{}\angle \mathrm{ODC}=90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{by Pythagoras Theorem,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}OC}}^{\text{2}}={\text{OD}}^{\text{2}}{\text{+ DC}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{DC}}^{2}={\text{OC}}^{\text{2}}-{\text{OD}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{DC}}^{2}={5}^{2}-{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta BDC},\text{}\angle \mathrm{BDC}=90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{by Pythagoras Theorem,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC}}^{\text{2}}={\text{BD}}^{\text{2}}{\text{+ DC}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{DC}}^{2}={\text{BC}}^{\text{2}}-{\text{BD}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{DC}}^{2}={6}^{2}-{\left(5-\mathrm{x}\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}{5}^{2}-{\mathrm{x}}^{2}={6}^{2}-{\left(5-\mathrm{x}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25-{\mathrm{x}}^{2}=36-25+10\mathrm{x}-{\mathrm{x}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25-11=10\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{14}{10}=1.4\\ \text{Substituting value of x in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{DC}}^{2}={5}^{2}-{\left(1.4\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25-1.96\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=23.04\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DC}=\sqrt{23.04}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4.8\text{\hspace{0.17em}}\mathrm{m}\\ \text{Thus, AC}=2\text{\hspace{0.17em}}\mathrm{DC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2×4.8\text{\hspace{0.17em}}\mathrm{m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9.6\text{\hspace{0.17em}}\mathrm{m}\\ \text{Therefore},\text{the distance between Reshma and Mandeep is 9.6 m.}\end{array}$

Q.104 A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Ans

$\begin{array}{l}\text{Let A, B and C be the positions of Ankur, Syed and David}\\ \text{respectively}\text{.Radius of circle is 20 m}\text{.}\\ \text{AD is median of triangle}\text{. O is centre of circle as well as}\\ \text{intersection point of medians too}\text{. Let each side of}\Delta ABC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{be 2x}\text{.}\\ \text{Since, cetroid divides a median in 2:1, then}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{AO}}{OD}=\frac{2}{1}⇒\frac{20}{OD}=\frac{2}{1}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}OD=\frac{20}{2}=10\text{\hspace{0.17em}}m\\ \text{Therefore, AD=20}+\text{10=30}\\ \text{ar}\left(\Delta ABC\right)=\frac{1}{2}×BC×AD\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}×2x×30\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=30x{\text{m}}^{\text{2}}\\ \text{In}\Delta BOD,\text{}\angle ODB=90°\\ So,\text{by Pythagoras theorem,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}O{B}^{2}=O{D}^{2}+B{D}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(20\right)}^{2}={\left(10\right)}^{2}+{x}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}=400-100\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=300\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{300}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=10\sqrt{3}\\ So,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{BC}=2x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=2×10\sqrt{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=20\sqrt{3}\\ \text{Thus, the length of the string of each phone is 20}\sqrt{3}\text{m}\text{.}\end{array}$

Q.105 In figure below, A,B and C are three points on a circle with center O such that ∠BOC = 30° and ∠AOB = 60°.
If D is a point on the circle other than the arc ABC, find ∠ADC.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{circle, with centre O,}\angle BOC=30°\text{and}\text{\hspace{0.17em}}\angle AOB=60°.\\ \text{To find:}\text{\hspace{0.17em}}\angle ADC\\ \text{Sol:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}âˆµ\angle AOC=60°+30°\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=90°\\ \text{Since, the angle subtended by an arc at the centre is double the}\\ \text{angle subtended by it at any point on the remaining part of the}\\ \text{circle}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\therefore \angle AOC=2\text{\hspace{0.17em}}\angle ADC\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}90°=2\text{\hspace{0.17em}}\angle ADC\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle ADC=\frac{90°}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=45°\\ \text{Thus},\text{the required angle is 45°}\text{.}\end{array}$

Q.106 A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Ans

$\begin{array}{l}\text{Given:Let O is the centre of circle and chord AB is equal to the}\\ \text{radius of circle.}\\ \text{To find:}\angle \mathrm{ACB}\text{and}\angle \mathrm{ADB}.\\ \text{Proof: In}\mathrm{\Delta }\text{AOB, OA}=\text{AB}=\text{OB.}\\ \text{So,}\angle \mathrm{AOB}=60\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Angle of equilateral triangle.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{theorem}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×60\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=30\mathrm{°}\\ \text{Since},\text{â€‹ADBC is a cyclic quadrilateral. So,}\\ \text{}\angle \mathrm{ACB}+\angle \mathrm{ADB}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}30\mathrm{°}+\angle \mathrm{ADB}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADB}=180\mathrm{°}-30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=150\mathrm{°}\end{array}$

Q.107 In the figure below, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Ans

$\begin{array}{l}\text{Given:In circle with centre O,}\angle \mathrm{PQR}=100\mathrm{°}.\\ \text{To find:}\angle \mathrm{OPR}\\ \text{Solution: Since, The angle subtended by an arc at the centre is}\\ \text{double the angle subtended by it at any point on the remaining}\\ \text{part of the circle.}\\ \therefore \text{Reflex}\angle \mathrm{POR}=2\text{}\angle \mathrm{PQR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=2×100\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=200\mathrm{°}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{POR}+\text{Reflex}\angle \mathrm{POR}=360\mathrm{°}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{POR}+200\mathrm{°}=360\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{POR}=360\mathrm{°}-200\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=160\mathrm{°}\\ \text{In}\mathrm{\Delta OPR},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OP}=\mathrm{OR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ORP}=\angle \mathrm{OPR}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Angles}\text{opposite to equal sides}\\ \text{are equal.}\end{array}\right]\\ \angle \mathrm{ORP}+\angle \mathrm{OPR}+\angle \mathrm{POR}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{angle sum property.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OPR}+\angle \mathrm{OPR}+160\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}2\text{\hspace{0.17em}}\angle \mathrm{OPR}=180\mathrm{°}-160\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}2\text{\hspace{0.17em}}\angle \mathrm{OPR}=20\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OPR}=10\mathrm{°}\\ \text{Thus},\text{the}\angle \mathrm{OPR}\text{is 10°.}\end{array}$

Q.108 In the figure below, ∠ABC = 69°, ∠ACB=31°, find ∠BDC.

Ans

$\begin{array}{l}\text{Given}:\text{In circle,}\angle \text{ABC}=\text{69}\mathrm{°},\angle \text{ACB}=\text{31}\mathrm{°}.\\ \text{To find:}\angle \text{BDC}\\ \text{Solution: In}\mathrm{\Delta ABC},\\ \text{}\angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}+69\mathrm{°}+31\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=180\mathrm{°}-100\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=80\mathrm{°}\\ \text{Since},\text{angles in same segment of a circle are equal.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BDC}=\angle \mathrm{BAC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=80\mathrm{°}\end{array}$

Q.109 In the figure below A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Ans

$\begin{array}{l}\text{Given:In a circle, chords AC and BD intersect at E.}\angle \mathrm{BEC}=130\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and}\angle \mathrm{ECD}=20\mathrm{°}.\\ \text{To find:}\angle \mathrm{BAC}.\\ \text{Solution:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CEB}+\angle \mathrm{CED}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}130\mathrm{°}+\angle \mathrm{CED}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\angle \mathrm{CED}=180\mathrm{°}-130\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=50\mathrm{°}\\ \text{In}\mathrm{\Delta DEC},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ECD}+\angle \mathrm{CED}+\angle \mathrm{CDE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20\mathrm{°}+50\mathrm{°}+\angle \mathrm{CDE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CDE}=180\mathrm{°}-70\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=130\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BDC}=130\mathrm{°}\\ \mathrm{Since},\text{angles in same segment of a circle are equal.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{BDC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=130\mathrm{°}\\ \text{Thus},\text{the required angle}\angle \mathrm{BAC}\text{is 130°.}\end{array}$

Q.110 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Ans

$\begin{array}{l}\text{Given:ABCD is a cyclic quadrilateral}\text{.}\angle \text{DBC}=\text{7}0°,\angle \text{BAC}=\text{3}0°.\\ To\text{find:}\angle \text{BCD,}\text{\hspace{0.17em}}\angle \text{ECD}\left(if\text{AB}=\text{BC}\right)\\ \text{Since, angles in the same segments of the circle are equal}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{So,}\angle \text{DAC}=\angle \text{DBC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=70°\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{DAB}=\angle \text{DAC}+\angle \text{BAC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=70°+30°\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=100°\\ \text{Since},\text{sum of opposite angles in a cyclic quadrilateral is 180°}\text{.}\\ \text{Then,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{DAB}+\angle \text{DCB}=180°\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}100°+\angle \text{DCB}=180°\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{DCB}=180°-100°\\ \text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{BCD}=80°\\ If\text{AB}=\text{BC}\end{array}$

$\begin{array}{l}\text{Then},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{BCA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides are}\\ \text{equal.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}30\mathrm{°}=\angle \mathrm{BCA}\\ \text{Since},\text{â€‹ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BCD}=80\mathrm{°}\\ \mathrm{Then},\mathrm{ }\angle \mathrm{BCA}+\angle \mathrm{ACD}=80\mathrm{°}\\ \mathrm{ }30\mathrm{°}+\angle \mathrm{ECD}=80\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ECD}=80\mathrm{°}-30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50\mathrm{°}\end{array}$

Q.111 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans

$\begin{array}{l}\text{Given:ABCD is a cyclic quadrilateral}\text{.AC and BD are the}\\ \text{}\text{\hspace{0.17em}}\text{diameters of the circle}\text{.}\\ \text{To prove: ABCD is a rectangle}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Proof:}\text{\hspace{0.17em}}\text{Since, angle is semicircle is right angle}\text{.}\\ \text{So,}\angle ABC=90°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[AC\text{is diameter}\text{.}\right]\\ \text{}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle ADC=90°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[AC\text{is diameter}\text{.}\right]\\ \text{Now,}\angle BAD=90°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[BD\text{is diameter}\text{.}\right]\\ \text{}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle BCD=90°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[BD\text{is diameter}\text{.}\right]\\ \text{Since},\text{each angles of ABCD is 90°, so ABCD is a rectangle}\text{.}\end{array}$

Q.112 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Ans

$\begin{array}{l}\text{Given:Let ABCD be trapezium in which AD}=\text{BC.}\\ \text{\hspace{0.17em}To prove:\hspace{0.17em}ABCD is a cyclic quadrilateral.}\\ \text{Construction:\hspace{0.17em}Draw EC}\mathrm{AD}\text{and produce AB upto E.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: Since, AE}\text{DC and AD}\text{EC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}So, AECD is a prallelogram.}\\ \text{Then, AD}=\text{EC but AD}=\text{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} EC}=\text{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}And}\angle \mathrm{CEA}=\angle \mathrm{CDA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Opposite angles of a}\\ \text{parallelogram are equal.}\end{array}\right]\\ \text{Inâ€‹}\mathrm{\Delta }\text{CEB, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}=\angle \mathrm{CEB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Opposite angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}+\angle \mathrm{CBA}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Linear pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CEB}+\angle \mathrm{CBA}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[âˆµ\angle \mathrm{CBE}=\angle \mathrm{CEB}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CEA}+\angle \mathrm{CBA}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CDA}+\angle \mathrm{CBA}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[âˆµ\angle \mathrm{CEA}=\angle \mathrm{CDA}\right]\\ âˆµ\angle \mathrm{CDA}+\angle \mathrm{CBA}\text{\hspace{0.17em}}+\angle \mathrm{ABC}+\angle \mathrm{ADC}\text{\hspace{0.17em}}=360\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Angle sum property in}\\ \text{quadrilateral.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}180\mathrm{°}+\angle \mathrm{ABC}+\angle \mathrm{ADC}\text{\hspace{0.17em}}=360\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABC}+\angle \mathrm{ADC}\text{\hspace{0.17em}}=360\mathrm{°}-180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABC}+\angle \mathrm{ADC}=180\mathrm{°}\\ \text{Since},\text{sum of opposite angles is 180°, so ABCD is a cyclic}\\ \text{quadrilateral.}\end{array}$

Q.113 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the figure below). Prove that ∠ACP=∠QCD.

Ans

$\begin{array}{l}\text{Given:Intersection points of two circles are B and C}\text{. Two line}\\ \text{segments ABD and PBQ}\text{are drawn to intersect the circles}\\ \text{at A},\text{D and P},\text{Q respectively}\\ \text{To prove:}\angle \text{ACP}=\angle \text{QCD}\\ \text{Proof:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Since, angles in the same segments of a circle are equal}\text{.}\\ \text{}\angle \text{ACP}=\angle \text{ABP}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{QCD}=\angle \text{QBD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{ABP}=\angle \text{QBD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(iii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\mathrm{Vertical}\mathrm{opposite}\mathrm{angles}\right]\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{ACP}=\angle \text{QCD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{From equation}\left(i\right)\text{and equation}\text{(ii) and equation (iii)}\text{]}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.114 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{ABC}\text{is a triangle. Two circles are drawn with}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}diameter AB and AC respectively.}\\ \mathrm{To}\text{prove: Point of intersection(D) of circles lie on BC.}\end{array}$

$\begin{array}{l}\text{Proof:\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}=90\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Angle}\text{in semi circle is 90°.}\right]\\ \text{and}\angle \mathrm{ADB}=90\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Angle}\text{in semi circle is 90°.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}\text{\hspace{0.17em}}+\angle \mathrm{ADC}=90\mathrm{°}+90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\angle \mathrm{BDC}=180\mathrm{°}\\ ⇒\mathrm{BDC}\text{is a straight line.}\\ ⇒\mathrm{D}\text{lies on BC i.e., third side of triangle.}\\ \text{Thus, point of intersection of both circles lie on third side}\\ \text{of triangle.}\end{array}$

Q.115 ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD=∠CBD.

Ans

$\begin{array}{l}\text{Given}:\Delta ABC\text{and}\Delta DAC\text{have equal hypotenous AC}\text{.}\\ \text{To prove:}\angle \text{CAD}=\angle \text{CBD}\\ \text{Proof:In quadrilateral ABCD,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle B+\angle D=180°\text{}\\ \text{So, quadrilateral ABCD is cyclic}\text{.}\\ \text{Then,}\angle \text{CAD}=\angle \text{CBD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}Angles\text{in same segment of}\\ \text{circle are equal}\text{.}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Hence Proved}\text{.}\end{array}$

Q.116 Prove that a cyclic parallelogram is a rectangle.

Ans

$\begin{array}{l}\text{Given:ABCD is a cyclic parallelogram}\text{.}\\ \text{To prove:ABCD is a rectangle}\text{.}\\ \text{Proof:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle A+\angle C=180°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{ABCD is a cyclic quadrilateral}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle A=\angle C\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Oppositeâ€‹ angles of a parallelogram}\text{.}\right]\\ \text{Then},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle A=\angle C=90°\\ \text{Since},\text{a parallelogram will be a rectangle if any angle is right}\\ \text{angle}\text{. Thus, ABCD is a rectangle}\text{.}\end{array}$

Q.117 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Ans

$\begin{array}{l}\text{Given:Two circles with centres P and Q are intersecting at A and B}\text{.}\\ \text{To prove:}\angle PAQ=\angle PBQ\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Proof:}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In}\Delta APQ\text{and}\Delta BPC\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AP=BP\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Radii of same circle}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AQ=BQ\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Radii of same circle}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}PQ=PQ\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Commo}n\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta APC\cong \Delta BPC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{S}\text{.S}\text{.S}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle PAQ=\angle PBQ\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \text{Thus},\text{centres P and Q subtend equal angle at A and B}\text{.}\end{array}$

Q.118 Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Ans

$\begin{array}{l}\text{Let OP}=\text{x m and OQ}=\left(\text{6}-\text{x}\right)\text{m}.\\ \text{In right}\Delta \text{AOP},\text{by Pythagoras Theorem},\\ {\text{OA}}^{\text{2}}={\text{AP}}^{\text{2}}+{\text{OP}}^{\text{2}}\end{array}$

$\begin{array}{l}{\text{OA}}^{\text{2}}={\left(\frac{\mathrm{AB}}{2}\right)}^{2}+{\left(\mathrm{x}\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Since},\text{perpendicular from centre}\\ \text{bisect the chord.}\end{array}\right]\\ {\text{OA}}^{\text{2}}={\left(\frac{5}{2}\right)}^{2}+{\mathrm{x}}^{2}\\ {\text{OA}}^{\text{2}}=\frac{25}{4}+{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{In}\mathrm{\Delta CQO},\text{}\angle \mathrm{Q}=90\mathrm{°}\\ \mathrm{So},\text{by Pythagoras theorem, we have}\\ {\text{OC}}^{\text{2}}={\text{CQ}}^{\text{2}}+{\text{OQ}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\left(\frac{\mathrm{CD}}{2}\right)}^{2}+{\left(6-\mathrm{x}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\frac{11}{2}\right)}^{2}+{\left(6-\mathrm{x}\right)}^{2}\\ \mathrm{}{\text{OC}}^{\text{2}}=\frac{121}{4}+36-12\mathrm{x}+{\mathrm{x}}^{2}\text{\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ {\text{Since,\hspace{0.17em}\hspace{0.17em}OA}}^{\text{2}}={\text{OC}}^{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Radius}\text{of same circle.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{25}{4}+{\mathrm{x}}^{2}=\frac{121}{4}+36-12\mathrm{x}+{\mathrm{x}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\frac{25}{4}=\frac{121}{4}+36-12\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}12\mathrm{x}=\frac{121}{4}+36-\frac{25}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{96}{4}+36\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=24+36\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{60}{12}=5\\ \text{Substituting value of x in equation}\left(\mathrm{i}\right),\text{we get}\\ {\text{OA}}^{\text{2}}=\frac{25}{4}+{5}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{25}{4}+25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OA}=\sqrt{\frac{125}{4}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5\sqrt{5}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5×2.24}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5.6\text{\hspace{0.17em}}\mathrm{m}\text{\hspace{0.17em}\hspace{0.17em}}\left(\text{approx}\right)\\ \text{Thus},\text{the radius of circle is 5.6 m}\left(\text{approx}\right)\text{.}\end{array}$

Q.119 The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the center?

Ans

Let AB and CD are two parallel chords of 6 cm and 8 cm respectively. The distance of AB is 4 cm from center. Let distance of CD from center O is x cm.

$\begin{array}{l}\text{In right}\Delta \text{AOP},\text{by Pythagoras Theorem},\\ {\text{OA}}^{\text{2}}={\text{AP}}^{\text{2}}+{\text{OP}}^{\text{2}}\\ {\text{OA}}^{\text{2}}={\left(\frac{AB}{2}\right)}^{2}+{\left(4\right)}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}\text{Since},\text{perpendicular from centre}\\ \text{bisect the chord}\text{.}\end{array}\right]\\ {\text{OA}}^{\text{2}}={\left(\frac{6}{2}\right)}^{2}+{4}^{2}\\ {\text{OA}}^{\text{2}}=9+16\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}OA=\sqrt{25}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}OA=5\text{cm}\text{.}\\ \text{In}\Delta CQO,\text{}\angle Q=90°\\ \text{So, by Pythagoras theorem, we have}\\ {\text{OC}}^{\text{2}}={\text{CQ}}^{\text{2}}+{\text{OQ}}^{\text{2}}\\ {\left(5\right)}^{2}={\left(\frac{CD}{2}\right)}^{2}+{x}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[âˆµOC=OA=5\text{\hspace{0.17em}}cm\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}25={\left(\frac{8}{2}\right)}^{2}+{x}^{2}\\ \text{\hspace{0.17em}}{\text{x}}^{\text{2}}=25-16\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{9}=3\\ \text{Thus},\text{the distance of chord CD from centre O is 3 cm}\text{.}\end{array}$

Q.120 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Ans

$\begin{array}{l}\text{Given: AD}=\text{CE,}\angle \text{ABC is in exterior of circle with centre O.}\\ \text{To prove:}\angle \mathrm{ABC}=\frac{1}{2}\left(\angle \mathrm{DOE}-\angle \mathrm{AOC}\right)\\ \text{Construction: Join AC, OA, OD, OC and OE.}\\ \text{Proof:\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta AOD}\text{and}\mathrm{\Delta COE},\\ \mathrm{ }\mathrm{OA}=\mathrm{OC}\mathrm{ }\left[\mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{circle}\right]\\ \mathrm{ }\mathrm{OD}=\mathrm{OE}\mathrm{ }\left[\mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{circle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\mathrm{CE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOD}\cong \mathrm{\Delta COE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OAD}=\angle \mathrm{OCE}\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{and}\text{\hspace{0.17em}}\angle \mathrm{ODA}=\angle \mathrm{OEC}\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{But}\text{\hspace{0.17em}}\angle \mathrm{OAD}=\angle \mathrm{ODA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides of a}\\ \text{triangle are equal.}\end{array}\right]\\ \text{Then},\text{\hspace{0.17em}}\angle \mathrm{OAD}=\angle \mathrm{ODA}=\angle \mathrm{OCE}=\angle \mathrm{OEC}=\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{let}\right)\\ \mathrm{In}\text{}\mathrm{\Delta OAC},\text{OA}=\text{OC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OCA}=\angle \mathrm{OAC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Angle}\text{opposite to equal sides are equal}\\ \text{in a triangle.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{y}\left(\mathrm{let}\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOC}=180\mathrm{°}-2\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Angle}\text{sum property in a triangle}\right]\\ \mathrm{In}\text{}\mathrm{\Delta ODE},\text{OD}=\text{OE}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OED}=\angle \mathrm{ODE}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Angle}\text{opposite to equal sides are equal}\\ \text{in a triangle.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{z}\left(\mathrm{let}\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DOE}=180\mathrm{°}-2\mathrm{z}\text{\hspace{0.17em}\hspace{0.17em}}\left[\text{Angle}\mathrm{sum}\mathrm{property}\mathrm{in}\mathrm{a}\mathrm{triangle}\right]\\ \mathrm{Now},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DOE}-\angle \mathrm{AOC}=\left(180\mathrm{°}-2\mathrm{z}\right)-\left(180\mathrm{°}-2\mathrm{y}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left(\mathrm{y}-\mathrm{z}\right)\\ \frac{1}{2}\left(\angle \mathrm{DOE}-\angle \mathrm{AOC}\right)=\mathrm{y}-\mathrm{z}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=180\mathrm{°}-\left(\mathrm{x}+\mathrm{y}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles}\right]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BCA}=180\mathrm{°}-\left(\mathrm{x}+\mathrm{y}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles}\right]\\ \mathrm{In}\text{}\mathrm{\Delta BAC},\text{\hspace{0.17em}}\\ \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{BCA}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\angle \mathrm{ABC}=180\mathrm{°}-\angle \mathrm{BAC}-\angle \mathrm{BCA}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=180\mathrm{°}-\left\{180\mathrm{°}-\left(\mathrm{x}+\mathrm{y}\right)\right\}-\left\{180\mathrm{°}-\left(\mathrm{x}+\mathrm{y}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\angle \mathrm{B}=2\left(\mathrm{x}+\mathrm{y}\right)-180\mathrm{°}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}2\left(\mathrm{x}+\mathrm{y}\right)-\angle \mathrm{B}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{In}\text{}\mathrm{\Delta BDE},\text{\hspace{0.17em}}\\ \angle \mathrm{BDE}+\angle \mathrm{BED}+\angle \mathrm{DBE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DBE}=180\mathrm{°}-\angle \mathrm{BDE}-\angle \mathrm{BED}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}=2\left(\mathrm{x}+\mathrm{y}\right)-\angle \mathrm{B}-\left(\mathrm{x}+\mathrm{z}\right)-\left(\mathrm{x}+\mathrm{z}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\text{From equation}\left(\mathrm{ii}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}2\angle \mathrm{B}=2\left(\mathrm{x}+\mathrm{y}\right)-2\left(\mathrm{x}+\mathrm{z}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}=\left(\mathrm{y}-\mathrm{z}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABC}=\frac{1}{2}\left(\angle \mathrm{DOE}-\angle \mathrm{AOC}\right)\text{\hspace{0.17em} Hence proved.}\end{array}$

Q.121 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Ans

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}}Let\text{ABCD be a rhombus and circle is drawn with}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{diameter AB}\text{.}\\ \text{To prove:}\text{\hspace{0.17em}}\text{Circle passes through point O}\text{.}\\ \text{Proof: In circle with diameter AB and angle in semicircle is}\\ \text{right angle}\text{.So,}\angle AOB=90°\\ \text{Since, diagonals of a rhombus bisect at 90°}\text{. So,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle AOB=90°\\ ⇒\text{Point O, lies on circumference}\text{. Thus, circle passes through}\\ \text{the point of intersection of diagonals of rhombus}\text{.}\end{array}$

Q.122 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Ans

$\begin{array}{l}\text{Given:Let ABCD be a parallelogram and circle passing through}\\ \text{vertex A, B and C intersects at E on side CD.}\\ \text{To prove:AD}=\text{AE}\\ \text{Proof:Since, ABCD is a parallelogram. So,}\\ \text{}\angle \mathrm{B}=\angle \mathrm{D}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\begin{array}{l}\text{Opposite angles of parallelogram are}\\ \text{equal.}\end{array}\right]\\ \text{Since},\mathrm{ABCE}\text{is a cyclic quadrilateral, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}+\angle \mathrm{AEC}=180\mathrm{°}\text{\hspace{0.17em}}\dots \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Sum of opposite angles in a cylic}\\ \text{quadrilateral is 180°.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AED}\text{\hspace{0.17em}}+\angle \mathrm{AEC}\text{\hspace{0.17em}}=180\mathrm{°}\text{\hspace{0.17em}}...\left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Linear pair of angles]}\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}+\angle \mathrm{AEC}=\angle \mathrm{AED}\text{\hspace{0.17em}}+\angle \mathrm{AEC}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}=\angle \mathrm{AED}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iv}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iv}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\angle \mathrm{D}=\angle \mathrm{AED}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADE}=\angle \mathrm{AED}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ\angle \mathrm{D}=\angle \mathrm{ADE}\right]\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AE}=\mathrm{AD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Opposite sides of equal angles are equal}\\ \text{in}\mathrm{\Delta ADE}\text{.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.123 AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.

Ans

$\begin{array}{l}\text{Given:In circle with centre O, AC and BD are two chords which}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{bisect each other}\text{.}\\ \text{To prove:}\left(i\right)\text{AC and BD are diameters,}\left(ii\right)ABCD\text{is a rectangle}\text{.}\\ \text{Proof:In}\Delta \text{AOB and}\Delta \text{COD}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AO}=\text{OC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Given\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle AOC=\angle BOD\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Vertical\text{opposite angles}\right]\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{BO}=\text{OD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Given\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta \text{AOB}\cong \Delta \text{COD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{SAS}\right]\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AB=DC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In}\Delta \text{AOD and}\Delta \text{BOC}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AO}=\text{OC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Given\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle AOD=\angle BOC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Vertical\text{opposite angles}\right]\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{DO}=\text{OB}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Given\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta \text{AOD}\cong \Delta \text{BOC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{SAS}\right]\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AD=BC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \text{Since},\text{opposite sides of quadrilateral ABCD are equal,so ABCD}\\ \text{is a prallelogram}\text{.}\\ \text{Then},\text{}\angle \text{A}=\angle \text{C}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}Opposite\text{angles of a parallelogram}\\ \text{are equal}\text{.}\end{array}\right]\\ \mathrm{But}\text{ABCD is a cyclic quadrilateral}\text{.So,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle A+\angle C=180°\dots \left(ii\right)\\ \text{From equation}\left(i\right)\text{and}\left(ii\right),\text{we have}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle A=90°\\ \text{If any angle in a parallelogram is right angle, then it is a}\\ \text{rectangle}\text{.}\\ \text{Here, in parallelogram ABCD,}\text{\hspace{0.17em}}\angle A=90°,\text{so ABCD is a rectangle}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.124

$\begin{array}{l}\mathrm{Bisectors}\mathrm{of}\mathrm{angles}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}\mathrm{intersect}\mathrm{its}\\ \mathrm{circumcircle}\mathrm{at}\mathrm{D},\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{respectively}.\mathrm{ }\mathrm{Prove}\mathrm{that}\mathrm{the}\\ \mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{DEF}\mathrm{are}90\mathrm{°}-\frac{1}{2}\mathrm{A},90\mathrm{°}-\frac{1}{2}\mathrm{B}\\ \mathrm{and}90\mathrm{°}-\frac{1}{2}\mathrm{C}.\end{array}$

Ans

$\begin{array}{l}\text{Given}:\mathrm{ABC}\text{is a triangle.AD, BE and CF are bisector of}\angle \mathrm{A},\text{}\angle \mathrm{B}\text{}\\ \text{and}\angle \mathrm{C}\text{respectively.}\\ \text{To prove:\hspace{0.17em}}\angle \mathrm{D}=90\mathrm{°}-\frac{1}{2}\mathrm{A},\text{}\angle \mathrm{E}=90\mathrm{°}-\frac{1}{2}\mathrm{B},\angle \mathrm{F}=90\mathrm{°}-\frac{1}{2}\mathrm{C}\end{array}$

$\begin{array}{l}\text{Proof:}\angle \mathrm{BEF}=\angle \mathrm{BCF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Angles}\text{in same segment of circle}\\ \text{are equal.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{C}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BED}=\angle \mathrm{BAD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Angles}\text{in same segment of circle}\\ \text{are equal.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{A}}{2}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{E}=\angle \mathrm{BEF}+\angle \mathrm{BED}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\mathrm{C}}{2}+\frac{\mathrm{A}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(\mathrm{C}+\mathrm{A}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{2}\left(180\mathrm{°}-\mathrm{B}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}-\frac{1}{2}\mathrm{B}\\ \mathrm{Similarly},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{F}=90\mathrm{°}-\frac{1}{2}\mathrm{C}\text{and}\angle \mathrm{D}=90\mathrm{°}-\frac{1}{2}\mathrm{A}.\\ \text{Thus},\text{}\angle \mathrm{D}=90\mathrm{°}-\frac{1}{2}\mathrm{A},\text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{E}=90\mathrm{°}-\frac{1}{2}\mathrm{B}\text{and}\angle \mathrm{F}=90\mathrm{°}-\frac{1}{2}\mathrm{C}.\end{array}$

Q.125 Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Ans

$\begin{array}{l}\text{Given}:\text{Two congruent circles intersect at A and B}\text{. PAQ}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{intersects circles at P and Q respectively}\text{.}\\ \text{To prove: BP}=\text{BQ}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Proof:Since, AB is common chord in two congruent triangles}\text{.}\\ \text{So, the angles subtended by common arc at any point}\\ \text{on the circumference of both circles will be equal}\text{.}\\ \text{}\therefore \angle BPA=\angle BQA\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle BPQ=\angle BQP\\ \text{In}\Delta BPQ,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle BPQ=\angle BQP\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}BQ=BP\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}\text{Opposite sides of equal angles}\\ \text{are equal}\text{.}\end{array}\right]\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}BP=BQ\end{array}$

Q.126 In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Ans

$\begin{array}{l}\text{Given}:\text{AD is bisector of}\angle \text{A in}\mathrm{}\text{}\mathrm{\Delta ABC}.\\ \mathrm{To}\text{prove: AD and pependicular of BC intersect at point D on}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}circumcircle.}\\ \text{Proof:In}\mathrm{\Delta BOM}\text{and}\mathrm{\Delta COM}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{radius}\text{of circle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OMB}=\angle \mathrm{OMC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{MB}=\mathrm{CM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta BOM}\cong \mathrm{\Delta COM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOM}=\angle \mathrm{COM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOC}=2\text{\hspace{0.17em}}\angle \mathrm{BOM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{The angle subtended by an arc at the centre is double the angle}\\ \text{subtended by it at any point on the remaining part of the circle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOC}=2\text{\hspace{0.17em}}\angle \mathrm{BAC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{Similarly},\text{fo}\mathrm{r}\text{arc BD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOD}=2\text{\hspace{0.17em}}\angle \mathrm{BAD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{Since},\text{AD is bisector of}\angle \mathrm{A},\text{then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAD}=\frac{1}{2}\angle \mathrm{BAC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×\frac{1}{2}\angle \mathrm{BOC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAD}=\frac{1}{2}\angle \mathrm{BOM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\angle \mathrm{BOD}=\frac{1}{2}\angle \mathrm{BOM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{iii}\right)\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOD}=\angle \mathrm{BOM}\\ ⇒\mathrm{perpendicular}\text{bisector of BC and bisector of}\angle \mathrm{A}\text{intersect}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}circumcircle at the point D.}\end{array}$

Q.127 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Ans

$\begin{array}{l}\text{Given}:\text{Two congruent circles with centre O and O’ i}\text{.e}\text{., OA}=\text{O’P,}\\ \text{OB}=\text{O’Q and}\angle AOB=\angle PO‘Q\text{.}\\ \text{To prove: chord AB}=\text{chord PQ}\\ \text{Proof}:\text{In}\Delta AOB\text{and}\Delta PO‘Q\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}OA=O‘P\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}[Radii of congruent circles]}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle AOB=\angle PO‘Q\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \mathrm{ }\mathrm{OB}=\mathrm{O}‘\mathrm{Q} \mathrm{ } \left[\mathrm{Radii}\mathrm{of}\mathrm{congruent}\mathrm{circles}\right]\\ \mathrm{ }\therefore \mathrm{\Delta AOB}\cong \mathrm{\Delta PO}‘\mathrm{Q} \mathrm{ }\left[\mathrm{By}\mathrm{S}.\mathrm{A}.\mathrm{S}\mathrm{.}\right]\\ \text{Therefore},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}AB=PQ\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \text{Thus},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{chord AB}=\mathrm{chord}\text{PQ}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.128 In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Ans

Number of hits a boundary = 6

Total number of balls played= 30
Number of ball she did not hit = 30 – 6
= 24

Probability of not hitting boundary = (24/30)
= (4/5)
= 0.8

Thus, the probability that she did not hit a boundary.

Q.129 1500 families with 2 children were selected randomly, and the following data were recorded:

 Number of girls in a family 2 1 0 Number of families 475 814 211

Compute the probability of a family, chosen at random, having
(i) 2 girls (ii) 1 girl (iii) No girl
Also check whether the sum of these probabilities is 1.

Ans

$\begin{array}{l}\text{Number of families}=\text{15}00\\ \left(\text{i}\right)\text{\hspace{0.17em}}\mathrm{}\text{P}\left(\text{2 girls}\right)\text{\hspace{0.17em}}=\frac{\mathrm{Number}\text{of families having two girls}}{\mathrm{Total}\text{families}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{475}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{19}{60}\\ \left(\text{ii}\right)\text{P}\left(\text{1 girl}\right)=\frac{\mathrm{Number}\text{of families having one girl}}{\mathrm{Total}\text{families}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{}\frac{814}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{407}{750}\\ \left(\text{iii}\right)\text{P}\left(\text{No girl}\right)=\frac{\mathrm{Number}\text{of families having one girl}}{\mathrm{Total}\text{families}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\text{}\frac{211}{1500}\\ \mathrm{}\text{The sum of probabilities}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{475}{1500}\text{}+\text{}\frac{814}{1500}+\frac{211}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1500}{1500}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{1}\end{array}$

Q.130 In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:

Find the probability that a student of the class was born in August.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of student in class IX}=\text{40}\\ \text{Number of students born in August}=6\\ \mathrm{P}\left(\mathrm{A}\text{student born in August}\right)=\frac{\text{Number of students born in August}}{\mathrm{Number}\text{of student in class IX}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{6}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{20}\\ \text{Thus, the probability of a student born in August is}\frac{3}{20}.\end{array}$

Q.131 Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Ans

$\begin{array}{l}\mathrm{Total}\text{\hspace{0.17em}\hspace{0.17em}n}\mathrm{umber}\text{of tossing of two coins}=\text{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Frequency of coming up two heads}=72\\ \mathrm{P}\left(2\text{heads coming up}\right)=\frac{\text{Frequency of coming up two heads}}{\mathrm{Total}\text{\hspace{0.17em}\hspace{0.17em}n}\mathrm{umber}\text{of tossing of two coins}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{72}{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{9}{25}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=0.36\\ \mathrm{Thus},\text{the probability of coming up two heads is 0.36.}\end{array}$

Q.132 An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

Monthly income Vehicles per family (in Rs) Vehicles per family

 0 1 2 Above 2
Less than 7000

7000 – 10000

10000 – 13000

13000 – 16000

16000 or more

 10 0 1 2 1 160 305 535 469 579 25 27 29 59 82 0 2 1 25 88

Suppose a family is chosen. Find the probability that the family chosen is
(i) earning Rs 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs 7000 per month and does not own any vehicle.
(iv) earning Rs 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.

Ans

$\begin{array}{l}\mathrm{Total}\text{number of families}=2400\\ \left(\mathrm{i}\right)\text{â€‹ Number of families earning Rs 1}0000\text{}–\text{13}000\text{per month}\\ \text{and}\mathrm{}\text{owning exactly 2 vehicles}=29\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{exactly}\text{two vehicles}\right)=\frac{29}{2400}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}Number of families earning Rs 16}000\text{or more per month}\\ \text{and}\mathrm{}\text{owning exactly 1 vehicle}=579\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{exactly}\text{two vehicles}\right)=\frac{579}{2400}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}Number of families earning less than Rs 7}000\text{per month}\\ \text{and}\mathrm{}\text{does not own any vehicle}=10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{does not own any vehicle}\right)=\frac{10}{2400}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{240}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}Number of families earning Rs 13}000–\text{16}000\text{per month}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and}\mathrm{}\text{more than 2 vehicles}=25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{more than 2 vehicles}\right)=\frac{25}{2400}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{96}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}Number of families owning not more than 1 vehicle}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=10+1+2+1+160+305+535\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}+469+579\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2062\\ \text{\hspace{0.17em}}\mathrm{P}\left(\text{owning not more than 1 vehicle}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2062}{2400}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1031}{1200}\end{array}$

Q.133 A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows:
0 – 20, 20 – 30, . . ., 60 – 70, 70 – 100. Then she formed the following table:

 Marks Number of students 0 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – above 7 10 10 20 20 15 8

(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.

Ans

$\begin{array}{l}\mathrm{Total}\text{number of students in a class}=7+10+10+20+20+15+\text{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=9\text{0}\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}Number of students obtained marks less than 20%}=7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{a student obtained less than 2}0\mathrm{%}\text{marks}\right)=\frac{7}{90}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}Number of students obtained marks 6}0\text{or above}=15+8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=23\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{a student obtained marks 6}0\text{or above}\right)=\frac{23}{90}\end{array}$

Q.134 To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.

 Opinion Number of students like dislike 135 65

Find the probability that a student chosen at random

(i) likes statistics, (ii) does not like it.

Ans

$\begin{array}{c}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{number of students}=135+65\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=200\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of students who likes statistics}=135\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{who}\text{likes statistics}\right)=\frac{135}{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{27}{40}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of students who dislike statistics}=65\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{who}\text{likes statistics}\right)=\frac{65}{200}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{13}{40}\end{array}$

Q.135 The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12.
What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within (1/2) km from her place of work?

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Total}\text{number of engineers}=\text{40}\\ \left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of engineers living in less than 7 km from her place of work}=9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{engineers living in less than 7 km\hspace{0.17em}from her place}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{9}{40}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of engineers living in more than or equal to 7 km from her place of work}=31\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\text{engineers living in more than or equal to 7 km\hspace{0.17em}from her place}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{31}{40}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{Number}\text{of engineers living within}\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}km from her place of work}=0\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{P}\left(\text{engineers living within\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}km\hspace{0.17em}from her place}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}}=\frac{0}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}}=0\end{array}$

Q.136 Activity : Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

Ans

Let us consider the data obtained from this activity took place any time are given below:

 Types of vehicle Number of vehicle Two – Wheeler 40 Three – Wheeler 35 Four – Wheeler 25 Total vehicles 100

$\begin{array}{l}\mathrm{Total}\text{number of vehicles}=\text{100}\\ \text{\hspace{0.17em}Number of two-wheelers}=\text{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}\left(\mathrm{Two}-\mathrm{wheeler}\right)=\frac{\text{Number of two-wheelers}}{\mathrm{Total}\text{number of vehicles}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{40}{100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{2}{5}\end{array}$

Q.137 Activity : Ask all the students in your class to write a 3-digit number. Choose any student from the room at random.
What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Ans

Let there are 30 students in a class and the 3-digits numbers written by them are given below:
100, 101, 271, 284, 290, 319, 152, 128, 342, 505, 504, 405, 519, 203, 901, 982, 802, 107, 308, 310, 412, 516, 418, 712, 672, 503, 615, 725, 528, 817

$\begin{array}{l}\text{The 3}-\text{digits numbers which are divisible by 3 are}:\\ \text{342,\hspace{0.17em}5}0\text{4,\hspace{0.17em}4}0\text{5,\hspace{0.17em}519,\hspace{0.17em}516,\hspace{0.17em}612,\hspace{0.17em}615,\hspace{0.17em}}\mathrm{}\text{528}\\ \text{Total numbers divisible by 3}=\text{8}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P}\left(\text{Number divisible by 3}\right)=\frac{\text{Total numbers divisible by 3}}{\mathrm{Total}\text{written}\mathrm{Numbers}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{8}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{4}{15}\\ \text{Thus, the probability of a selected number to be divisible by 3}\\ \text{is}\frac{4}{15}.\end{array}$

Q.138 Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Ans

$\begin{array}{l}\mathrm{Number}\text{of flour bags = 11}\\ \text{Number of flour bags contains more than 5 kg of flour}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7\\ \mathrm{P}\left(\mathrm{bag}\text{contains more than 5 kg flour}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{\text{bags containing more than 5 kg flour}}{\mathrm{Total}\text{number of flour}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{7}{11}\\ \text{Thus},\text{the required probability of a selected bag having more than 5 kg flour is}\left(\text{7}/\text{11}\right).\end{array}$

Q.139 Frequency distribution table of the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days is given below.

 Concentration of SO2 (in ppm) Number of days 0.00 – 0.04 4 0.04 – 0.08 9 0.08 – 0.12 9 0.12 – 0.16 2 0.16 – 0.20 4 0.20 – 0.24 2 Total 30

Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{number of days}=\text{30}\\ \text{Number of days having the concentration of sulphur dioxide in}\\ \text{the interval}0.\text{12}-0.\text{16}=2\\ \mathrm{P}\left(\mathrm{day}\text{having}0.\text{12}-0.{\text{16 concentration of SO}}_{\text{2}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{2}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{15}\end{array}$

Q.140 Frequency distribution table of the blood groups of 30 students of a class is given below.

 Number of students A 9 B 6 AB 3 O 12 Total Students 30

Use this table to determine the probability that a student of this class, selected at random, has blood group AB.

Ans

$\begin{array}{l}\mathrm{Total}\text{number of students}=\text{30}\\ \text{Number of students having blood group AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{Blood}\text{group AB}\right)=\frac{\begin{array}{l}\mathrm{Number}\text{of students having blood}\\ \text{group AB}\end{array}}{\mathrm{Total}\text{number of students}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{3}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0.1\\ \text{Thus, the required probability is 0.1.}\end{array}$

Q.141 Construct an angle of 90° at the initial point of a given ray and justify the construction.

Ans

Given:

A ray OA with initial point O is given.

To construct:

An angle of 90°

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O.

(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.

(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.

(v) Join O and N by a ray OB.

(vi) ∠BOA is required angle of 90°.

Justification:

(i) Since, ∠M2OA = ∠M3OM2 = 60°

(ii) Ray OB is bisector of ∠M3OM2 i.e.

∠BOM2 = (1/2) ∠M3OM2

= (1/2) 60°

= 30°

(iii) ∠BOA = ∠BOM2 + ∠M2OA

= 30° + 60°

= 90°

Hence, it is justified that ∠BOA = 90°.

Q.142 Construct an angle of 45° at the initial point of a given ray and justify the construction.

Ans

Given:

A ray with initial point O is given.

To construct:

An angle of 45°

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O.

(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.

(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.

(v) Join O and N by a ray OB.

(vi) ∠BOA is an angle of 90°.

(vii) Now, mark arcs of any radius with centre M1 and Q respectively, which intersect at P.

(viii) Draw ray OC through P, we get ∠COA which is equal to 45°.

(ix) Thus, ∠COA is required angle.

Justification:

∠COA = (1/2) ∠BOA
= (1/2) 90°
= 45°

Q.143 Construct the angles of the following measurements: (i) 30° (ii) 22.5° (iii) 15°

Ans

(i)

Given:

A ray with initial point O is given.

To construct:

An angle of 30°

Steps of Construction:

(i) Draw a ray OA with initial point O.
(ii) Mark an arc of any radius with centre O, which intersects ray OA at the point M1.
(iii) Mark another arc with centre M1 and same radius, which intersects earlier arc at M2.
(iv) Now, we draw arcs with any radius and centres M1 and M2 respectively. These arcs intersect at N.

(v) Join O and N with the help of ray OB.

(vi) ∠BOA is required angle of 30°.

(ii)

Given:

A ray with initial point O

To construct:

An angle of 22.5°

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O.

(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.

(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.

(v) Join O and N by a ray OB.

(vi) ∠BOA is an angle of 90°.

(vii) Now, mark arcs of any radius with centre M1 and X respectively, which intersect at P.

(viii) Draw ray OC through P, we get ∠COA which is equal to 45°.

(ix) Now we draw bisector of ∠COA and we get ∠DOA, which is equal to 22.5°.

(x) Thus, ∠DOA is required angle.

(iii)

Given:

A ray with initial point O is given.

To construct:

An angle of 15°.

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O, which intersects ray OA at the point M1.

(iii) Mark another arc with centre M1 and same radius, which intersects earlier arc at M2.

(iv) Now, we draw arcs with any radius and centres M1 and M2 respectively. These arcs intersect at N.

(v) Join O and N with the help of ray OB.

(vi) ∠BOA is angle of 30°.

(vii) Now, draw bisector OC of ∠BOA.

(viii) ∠COA is required angle of 15°.

Q.144 Construct the following angles and verify by measuring them by a protractor:
(i) 75° (ii) 105° (iii) 135°

Ans

(i)

Given:

A ray with initial point O is given.

To construct:

An angle of 75°

Steps of Construction:

(i) Draw ray OP and construct angle of 90°.

(ii) Mark arcs with centre P and R respectively with any radius. These arcs intersect at Q.

(iii) Draw ray OC throw point Q.

(iv) ∠COA is required angle.

(ii)

Given:

A ray with initial point O is given.

To construct:

An angle of 105°

Steps of Construction:

(v) Draw ray OP and construct angle of 90°.

(vi) Mark arcs with centre P and Q respectively with any radius. These arcs intersect at R.

(vii) Draw ray OC throw point R.

(viii) ∠COA is required angle.

(iii)
Given:

A ray with initial point O is given.

To construct:

An angle of 135°

Steps of Construction:

(i) Draw ray OP and construct angle of 90°.

(ii) Mark arcs with centre P and Q respectively with any radius. These arcs intersect at R.

(iii) Draw ray OC throw point R.

(iv) ∠COA is required angle.

Q.145 Construct an equilateral triangle, given its side and justify the construction.

Ans

Given:

A side of triangle is given.

To construct:

An equilateral triangle ABC

Steps of Construction:

(i) Draw a side of given length.

(ii) Draw two arcs of radius AB with centres A and B respectively. These arcs intersect at C.

(iii) Join AC and BC.

(iv) ΔABC is required triangle.

Q.146 Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Ans

Given:

In triangle ABC, BC = 7 cm, ∠B = 75° and
AB + AC = 13 cm

To construct:

A triangle ABC

Steps of Construction:

(i) Cut a line segment BC = 7cm from a ray BX.

(ii) Draw ray BY at 75° and cut BD = 13 cm from ray BY.

(iii) Join DC and draw perpendicular bisector PQ of DC, which intersect BD at A.

(iv) Join AC.

(v) DABC is required triangle.

Q.147 Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

Ans

Given:

In triangle ABC, BC = 8 cm, ∠B = 45° and
AB – AC = 3.5 cm

To construct:

A triangle ABC

Steps of Construction:

(i) Cut a line segment BC = 8 cm from a ray BX.

(ii) Draw ray BY at 45° and cut BD = 3.5 cm from ray BD.

(iii) Join DC and draw perpendicular bisector of DC, which intersect BD at A.

(iv) Join AC.

(v) ΔABC is required triangle.

Q.148 Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.

Ans

Given: In triang