# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals (Ex 8.2)

The NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 is based on the midpoint theorem in relation to the sides of a triangle.This theorem serves as the foundation for almost all the questions in this exercise. According to the mid-point theorem, the line segment connecting a triangle’s two midpoints is parallel to the third side. Additionally, the third side of a triangle is divided by a line drawn across the middle of one of its sides and running parallel to another. Seven questions based on the application of this theorem are included in NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2.

Students will be able to correctly apply the midpoint theorem to problem-solving after practising with the NCERT Solutions for Class 9 Maths chapter 8 Exercise 8.2 provided by Extramarks. With the use of appropriate examples and practice questions, NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 effectively explains this subject. The Extramarks mobile application and website offer these NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 in a scrollable PDF format.

The NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 questions have been thoughtfully arranged to convey the knowledge and abilities needed to study the CBSE curriculum. Students are encouraged to practice the questions, examples, and sample problems in this exercise so they can develop a strong understanding of the ideas and quickly master them. Finding precise and thorough answers to all of the questions will be made easier by studying these NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2.

A closed quadrilateral has four sides, four vertices, and four angles. It is a form of polygon. In order to create it, four non-collinear points are joined. Quadrilaterals always have a total internal angle of 360 degrees.

The Latin words quadra, which means four, and latus, which means sides, are combined to form the English word “quadrilateral.” A quadrilateral does not always have to have equal lengths on each of its four sides. As a result, depending on the sides and angles, students can have several sorts of quadrilaterals.

A quadrilateral is a flat shape with four edges and four corners, also known as vertices. The quadrilateral has angles at each of its four vertices, or corners. The angles at the vertices of a quadrilateral ABCD are A, B, C, and D. A quadrilateral has the following sides: AB, BC, CD, and DA.

The diagonals are obtained by joining the opposing vertices of the quadrilateral. The diagonals of the quadrilateral ABCD are shown in the figure below as AC and BD.

Quadrilaterals are frequently undefined and irregular shapes with four sides, such as rectangles, squares, trapezoids, and kites.

Based on the measurements of the angles and side lengths, quadrilateral types are identified. All of these quadrilateral designs have four sides because the word “quad” implies “four,” and their combined angles total 360 degrees. These are the several kinds of quadrilaterals:

• Trapezium
• Parallelogram
• Squares
• Rectangle
• Rhombus
• Kite

• Following is another approach to group different kinds of quadrilaterals:
• Quadrilaterals that are convex have two diagonals that are entirely enclosed within the figure.
• Concave Quadrilaterals: At least one diagonal extends partially or completely beyond the boundaries of the figure.
• Quadrilaterals that intersect: Quadrilaterals that intersect don’t just have two non-adjacent sides cross each other. Quadrilaterals of this type are referred to as crossed or self-intersecting quadrilaterals.

The Properties of a Quadrilateral are mentioned below:

Properties of Square:

• The square’s sides are all of equal length.
• The interior angles of a square are all 90 degrees, and the sides are parallel to one another (i.e., right angle)
• A square’s diagonals are perpendicularly divided by one another.

Properties of Rectangle:

• In a rectangle, the opposing sides are of equal length.
• All of a rectangle’s inner angles are 90 degrees, and its opposite sides are parallel to one another.
• A rectangle’s diagonals cut each other in half.

Properties of Rhombus:

• A rhombus’s four sides are all the same size.
• The rhombus’s opposite sides are parallel to one another, and its opposite angles are equal in size.
• Any two adjacent angles in a rhombus added together equal 180 degrees.
• The diagonals cut each other in half perpendicularly.

Parallelogram Properties of Parallelogram:

• The parallelogram’s opposing side has the same length as its other side.
• Both sides of the object are parallel to one another.
• A parallelogram’s diagonals cut each other in half.
• The contrasting angles are equal in size.
• A parallelogram’s two neighbouring angles added together equal 180 degrees.

Properties of Trapezium:

• Only one pair of a trapezium’s opposite sides are parallel to one another.
• Its two neighbouring sides are supplementary (180 degrees).
•  Its diagonals are divided in the same proportion.

Properties of Kite:

• A kite’s two adjacent sides are the same length.
• A kite’s largest and smallest diagonals are split in half.
• The measure of only one pair of opposite angles is the same.

• If two of the sides of a quadrilateral are parallel to one another, the shape is a trapezoid or trapezium.
• If two pairs of the sides of a quadrilateral are parallel to one another, it is a parallelogram.
• Specialized parallelograms include squares and rectangles. Here are a few unique characteristics- Every angle within is a “right angle” (90 degrees). Each figure has four right angles. The opposite sides of a rectangle are the same length. A square’s sides are all the same length (congruent). A rectangle’s and a square’s opposite sides are parallel.
• If all four sides are the same length and at least two of the pairs of sides are parallel, a quadrilateral is a rhombus.
• An unusual type of quadrilateral called a kite has two sets of neighbouring sides that are equal to one another.

## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals (Ex 8.2) Exercise 8.2

The fundamental ideas of quadrilaterals are covered in Chapter 8 of Class 9 Mathematics. The NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2, which include completed in-text exercises, are included and available on the Extramarks website and mobile application. The academic professionals at Extramarks have solved the problems in the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2, in a thorough, step-by-step manner. Students are recommended to carefully read the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 and diligently practice them in order to get ready for tests and get good grades.

These NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 will aid students in becoming comfortable with the formats of the questions and in honing their problem-solving abilities. Additionally, students can visit Extramarks to practice more than 1340 questions from 64 different books to better prepare for examinations, along with the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2.

Students study a variety of quadrilateral theorems in these NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. Students will be able to answer numerous questions in these NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 based on these theorems. Students should practise all the questions and increase their speed and accuracy when answering them now that they have access to the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 and example questions for Class 9 Maths Ex 8.2.

Below are some points to remember before students dive further into the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2:

The following key ideas have been presented by Extramarks experts to assist students as they study for the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 Quadrilaterals exam:

• If the opposite sides of a quadrilateral are equal, the shape is a parallelogram.
• If the opposite angles of a quadrilateral are equal, the shape is a parallelogram.
• If a quadrilateral is a parallelogram, its diagonals cut each other in half.
• If one pair of opposite sides in a quadrilateral are equal and parallel, the shape is a parallelogram.
• A parallelogram is a quadrilateral created by joining the midpoints of the sides of a quadrilateral in the correct order.
• A parallelogram is split into two congruent triangles by a diagonal.
• In a parallelogram, the two opposing angles are equal.
• A parallelogram’s diagonals cut each other in half.
• Any two consecutive angles’ bisectors in a parallelogram will come together at a right angle.
• If a parallelogram’s diagonal cuts over one of the parallelogram’s angles, it likewise cuts across the other angle.

### Access NCERT Solution for Maths Class 9 Chapter 8 –  Quadrilateral

Through additional skill-building activities that are catered to their grade levels, skills, and interests, NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 brings out the best in children. Every question in Chapter 8, Exercise 8.2, has been answered in the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. The NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 are available for download for students, who can use the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 to perform well in their board examinations. Subject specialists have curated the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2, in a clear, easy, and understandable language. Extramarks experts offer detailed answers to every question in the Class 9 Mathematics textbook in accordance with CBSE board regulations.

### NCERT Solution Class 9 Maths of Chapter 8 All Exercises

To do well on a Mathematics exam, students should use a good preparation plan, which includes memorization of key formulas and concepts together with time-limited practice. NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 provides all the essential formulas and concepts along with suitable examples to help students fully understand them. These NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2  based on parallelograms, the angle sum property, and the mid-point theorem include some of the most significant formulas and ideas. They are listed below:

• A parallelogram is a quadrilateral with equal and parallel pairs of opposing sides.
• A parallelogram’s area is equal to A = bh.
• A parallelogram is split into two congruent triangles by a diagonal.
• In a parallelogram, the opposite sides and angles are both equal.
• A parallelogram’s diagonals cut each other in half.
• The third side of a triangle is parallel to the line segment connecting the midpoints of two of its sides.
• The third side of a triangle is divided by a line that is drawn through the centre of one side and runs parallel to another side.
• Exercise 8.1 contains a number of questions for students to solve. The midpoint Theorem is also discussed.
• The third side of a triangle is parallel to the line segment connecting the midpoints of two of its sides.
• The third side of a triangle is divided by a line drawn across the middle of one of its sides and running parallel to another.
• This section presents two theorems. Students are also given an assignment in which they must state the midpoint theorem converse and decide whether or not the converse is true. Students must practise NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. Typically, students will encounter queries that utilize figures.
• The conclusion lists 10 of the chapter’s most important ideas.

### NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2

NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 centered on the midpoint theorem is applied in Exercise 8.2’s 7 questions, which are based on the explanation of the midpoint theorem. A triangle’s midpoint and its side midpoint are related concepts in the mid-point theorem. There are numerous examples and practice questions offered in the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 to effectively cover this concept.

NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2, are available on the Extramarks website and mobile application. On their smartphone or desktop computer, students can easily learn to use the self-explanatory video tutorial format, which is also available for the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. A thorough step-by-step explanation has been provided for each activity from each chapter, including the NCERT Solutions for Class 9 maths Chapter 8 Exercise 8.2. The most recent CBSE board curriculum and accurate syllabus are taken into consideration when creating the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. These NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 will help students understand the most recent exam format and grade distribution.

Students will learn about several crucial geometry concepts including the angle sum property of a quadrilateral, different forms of quadrilaterals, the mid-point theorem, and characteristics of a parallelogram in the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. The exercise-by-exercise problems and topics associated with this chapter are fully covered in the CBSE Class 9 Mathematics NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 provided on the Extramarks mobile application and website. To assist students in better internalizing the subject, all of the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 are offered in a chronological and easy-to-understand way. There are two exercises in Class 9 Mathematics Chapter 8 Quadrilaterals, one of which requires proving a theorem.

### NCERT Solutions for Class 9

Students can practice all types of questions from each chapter’s exercise with the help of the Class 9 Mathematics NCERT solutions. The NCERT Class 9 Mathematics solutions have been created in a chronological manner to avoid confusion for student use by knowledgeable subject matter experts to make studying more convenient and engaging. To prepare for academic tests and lay a solid foundation for classes at a higher level, it is advisable that the students completely practice all of these NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. In-depth reading of these NCERT Mathematics Class 9 Solutions is advised for students.

Students can improve their academic performance with Extramarks’ online learning materials. The Class 9 Mathematics Solutions are designed as per the latest CBSE syllabus and NCERT curriculum. All of the activities in the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2, are solved in a step-by-step approach to help students better comprehend the topics. By referring to the Class 9 NCERT Solutions for Mathematics, students can solve the problems from the textbooks and increase their confidence. With the aid of the Class 9 Mathematics NCERT Solutions, they are able to advance their time management abilities.

Extramarks also provides NCERT Solutions for all other Class 9 Mathematics Chapter along with the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2.

Below is a list of all the chapters that Extramarks provides NCERT Solutions for

NCERT Solutions for Class 9 Maths Chapter 1 – Number System:

Six exercises in Chapter 1 of Number Systems in Class 9 Mathematics offer detailed answers to each question. Exercise 1.1: Introduction covers subjects including whole numbers, integers, rational numbers, and natural numbers. Finding irrational numbers on the number line is one of the subtopics of Exercise 1.2, “Irrational Numbers,” which also covers real numbers. Students will discover that the decimal expansion of an irrational number is non-terminating and non-recurring and that infinite irrational numbers can be obtained between two given rational numbers in Exercise 1.3 Real Numbers And Their Decimal Expansions. Exercise 1.4: Representing Actual Values on the Number Line includes a sub-topic called “sequential magnification,” which refers to drawing a number line by using a magnifying glass to enlarge the area between two real numbers.

Exercise 1.5 Real Numbers System: Students will practise finding the square root of x for any given positive real number ‘x’ using geometry. They will learn that the sum or difference of two irrational numbers is always irrational and that the result of adding, subtracting, multiplying, or dividing two irrational numbers may be either irrational or rational. Exercise 1.6: Real Numbers and Exponent Laws.

NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials:

There are 5 exercises in Chapter 2 with issues pertaining to every subject addressed in the chapter. The subtopics of Exercise 2.1: Polynomials in One Variable include the definition of a polynomial and zero. A polynomial’s degree, monomials, binomials, and trinomials Exercise 2.2: A’s zeroes Polynomials include the zero or root of the polynomial as well as its value. Exercise 2.3: Final The long division method and the remainder theorem are used to divide a polynomial by another polynomial. The splitting method and factor theorem are both used in Exercise 2.4, Factorization of Polynomials, to factorize quadratic polynomials. Numerous algebraic identities are covered in Exercise 2.5, “Algebraic Identities.”

NCERT Solutions for Class 9 Maths Chapter 3 – Coordinate Geometry:

In Class 9 Maths Chapter 3 Coordinate Geometry, there are three exercises that include questions on the chapter’s subjects. Introduction, Exercise 3.1 Exercise 3.2 Cartesian System covers a variety of related subjects, including the Cartesian Plane, the positive and negative x- and y-axes, quadrants, and the signs of coordinates in the first, second, third, and fourth quadrants, among others. Plotting points in four quadrants and plotting points on the positive and negative x-axis and y-axis are only a few of the subtopics covered in Exercise 3.3, Plotting A Point in the Plane if its Coordinates are given.

NCERT Solutions for Class 9 Maths Chapter 4 – Linear Equations in One Variable:

Students will learn about the linear equation in two variables as well as their auxiliary rules in Chapter 4 of the mathematics textbook for Class 9. This chapter lays the groundwork for pupils to deal successfully with more challenging situations using linear equations in the future. The students will learn how to plot a two-variable linear equations graph.

NCERT Solutions for Class 9 Maths Chapter 5 – Introduction to Euclid’s Geometry:

It may appear that Chapter 5 Introductions to Euclid’s Geometry merely discusses the characteristics of circles, triangles, and quadrilaterals. However, students must start learning this chapter from the very first axiom if you truly want to understand what geometry is in its truest form.

NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles:

Some fundamental definitions and characteristics pertaining to lines and angles in geometry will be introduced in NCERT Chapter 6: Lines and Angles. The foundation for many higher-level concepts in geometry will be laid out in this chapter. Intersecting and non-intersecting lines, pairs of angles, parallel and transversal lines, lines parallel to the same line, the angle sum property of a triangle, summaries, and many other fascinating ideas will be covered in this course. To learn this subject, students are urged to practise using the NCERT solutions for each exercise.

NCERT Solutions for Class 9 Maths Chapter 7 – Triangles:

The triangles are covered in depth in chapter 7 for students. Triangle inequalities, triangle properties, and triangle congruence will all be discussed.

Class 9 Mathematics Chapter 7 Triangles, contains 5 exercises. It is necessary for the students to resolve “to prove” and application-level challenges. Students will learn how to demonstrate the qualities they studied in past sections in this course. This chapter deals with eight theorems. Exercise 7.1 Triangle congruence, Exercise 7.2: Triangular congruence criteria; Exercises 7.3 and 7.4 provide additional requirements for a triangle’s congruence and some features of triangles, respectively. Triangle inequality exercise 7.5.

NCERT Solutions for Class 9 Maths Chapter 8 – Quadrilaterals:

The angle sum property of a quadrilateral, several forms of quadrilaterals, the mid-point theorem, and characteristics of a parallelogram are just a few of the fundamental geometrical concepts that will be covered in the NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2. All of the problems and ideas pertaining to this chapter, broken down into practice, are covered in the CBSE NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 provided on the Extramarks mobile application and website. To assist the learners in better internalizing the topic, every answer is offered in a step-by-step fashion. Two exercises are included in Chapter 8 of the Quadrilaterals textbook for Class 9 students.

Extramarks also provides NCERT Solutions for Class 9 Mathematics chapters 9 to 15.

### CBSE Study Materials for Class 9

Extramarks experts have also provided students with CBSE Study Materials for Class 9. Extramarks has detailed notes on all chapters of all Class 9 subjects along with previous years’ question papers, practice question papers and other study materials.

### CBSE Study Materials

One of the most well-known and esteemed educational boards in the nation is the Central Board of Secondary Education, or CBSE for short. It is a part of the Union Government of India. Numerous public and private schools that are part of the CBSE board use the NCERT curriculum.

Students can learn and prepare for the test more effectively with the help of the CBSE study materials offered by Extramarks professionals that are made in an exciting, engaging, and student-friendly manner. Subject specialists have created the study materials with the most recent CBSE curriculum in mind. The online CBSE study tools, which include the syllabus, books, sample exams, test questions, NCERT solutions, NCERT exemplar solutions, significant questions, and CBSE notes, are beneficial for all students.

Q.1 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}\parallel \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}\parallel \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}\parallel \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.2 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}\parallel \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}\parallel \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}\parallel \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}\parallel \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}\parallel \mathrm{MO}\text{and MP}\parallel \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.3 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}\parallel 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}\parallel 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}\parallel 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}\parallel 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.4 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}\parallel 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}\parallel 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.5 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}\parallel \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}\parallel \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.6 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}\parallel 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}\parallel 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}\parallel 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.7

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

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}\parallel \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}\parallel \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}\parallel \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}$