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.

 

NCERT Solutions Class 9 Mathematics Download

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. 
  • The NCERT solutions are very easy to access and you can download for free. 

 

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 8 – Quadrilaterals

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

 

Download NCERT Solutions for Class 9 Mathematics 

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

Since, y:z=3:7Let y=3a and z=7aSince, ABCD and CDEFso, ABEFx=z[Alternate interior angles.]∵      ABCDx+y=180°[Cointerior angles]z+y=180°[∵x=z]     7a+3a=180°           10a=180°      a=180°10=18°So, x=7a=7(18°)=126°

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

Ans

Given:ABCD,  EFCD and GED=126°.To find: AGE, GEF and FGEABCDAGE=GED    =126°GED=GEF+FED 126°=GEF+90°GEF=126°90°    =36°AGE+FGE=180°   126°+FGE=180°       FGE=180°126°=54°

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


Ans

Given:PQST, PQR=110°,  RST=130°To prove: QRSConstruction: Draw line l paralllel to ST through R.Since, STl, thenTSR+SRW=180°           [Cointerior​ angles]    130°+SRW=180°

            SRW=180°130°    =50°Since,  PQST and STlSo,     PQlPQR=SRW[Alternate interior angles.]   110°=QRS+SRW    =QRS+50°QRS=110°50°    =60°

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

Ans

Given:ABCD,  APQ=50° and PRD = 127°To find: x and y Since, ABCD So, APR=PRD        50°+y=127°y=127°50°    =77°And,       APQ=PQR     50°=xThus, x=50° and y=77°.

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

Given:PQRS, AB is incident ray to PQ and BC is incident ray       to RS.To Prove: ABCDConstruction: Draw MBPQ and NCRS.Proof: ABM=MBC...(i)      [incident angle= reflected angle]       BCN=NCD...(ii)      [incident angle= reflected angle]

Since, perpendiculars to parallel lines are parallel.So,   MBNC     MBN=BCN12ABC=12BCD    ABC=BCDSince, alternate interior angles are equal, so  ABCD.

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

Given:In ΔPQR, Ext.SPR=135° and Ext.PQT=110°To Find: PRQ        PQR+PQT=180°    PQR+110°=180°                  PQR=180°110°                  PQR=70°Ext.  SPR=PQR+PRQ 135°=70°+PRQ        PRQ=135°70°   =65°

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

Given:InΔXYZ,  X=62°, XYZ=54°, YO  and ZO are the      bisectors of XYZ andXZY  respectively.To find:OZY andYOZIn ΔXYZ,  XYZ+YXZ+ YZX=180°                  54°+62°+ YZX=180°   YZX=180°116°      =64° OZY=12 YZX      =12×64°      =32° OYZ=12 XYZ      =12×54°      =27°  InΔXYZ,       OYZ+ OZY+YOZ=180°        27°+32°+YOZ=180°YOZ=180°59°      =121°

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

Ans

   Given:ABDE,BAC=35° and CDE=53°To find: DCE       Sol: Since, ABCD, soDEA=BAE[Alternate​​ interior angles]DEA=35°CED=35°InΔCDE,CDE+DCE+CED=180°    [By angle sum property.]53°+DCE+35°=180°    DCE=180°88°    DCE=92°

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

  Given:PRT=40°,RPT=95° and  TSQ=75°To find: SQT             In ΔPRT,PRT+PTR+RPT=180°         40°+PTR+95°=180°   PTR=180°135°       =45°   QTS=PTR[Vertical opposite angels]   QTS=45°In ΔQST,        QST+SQT+QTS=180°         75°+PTR+45°=180°   PTR=180°120°       =60°

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

  Given: PQPS, PQ||SR,  SQR=28°  and  QRT=65°To find: x and y       Sol:Since,   PQSTPQR=PRT         x+28°=65°x=65°28°     =37°In ΔPQS,QPS+PSQ+PQR=180°        [By angle sum property.]     90°+y+37°=180°     y=180°127°         =53°

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

  Given: The side QR of ΔPQR is produced to a point S. The                   bisectors ofPQR and PRS meet at point T.To Prove: QTR=12QPRProof:Since,PQT=TQR=x(let)              andPRT=TRS=y(let)In​ ΔPQR,                    Ext.PRS=PQR+QPR                  2y=2x+QPR      2y2x=QPRyx=12QPR...(i)In​ ΔQTR,                    Ext.TRS=TQR+QTR        y=x+QTRyx=QTR...(ii)From equation(i) and equation(ii), we have       QTR=12QPRHence Proved.

Q.12

In quadrilateral ACBD,AC=AD and AB bisects A(see figure below).Show that ΔABCΔABD. What canyou say about BC and BD.

Ans

      Given: ABCD is a quadrilateral, AC=BD, AB bisects A.To prove: ΔABCΔABD        Proof:In ΔABC and ΔABDAC=AD            [Given]    BAC=ABD         [Given]AB=AB            [Common]ΔABCΔABD         [By SAS]BC=BD            [By CPCT]

Q.13

ABCD is a quadrilateral in which AD=BC and DAB=CBA(see figure below).Prove that(i)ΔABDΔBAC(ii)BD=AC(iii)ABD=BAC.

Ans

Given: ABCD is a quadrilateral, AD=BC, DAB=CBA.To prove:(i)ΔABDΔBAC(ii)BD=AC(iii)ABD=BACProof:(i)In ΔABD and ΔBACAB=BA[Common]BAD=ABC[Given]AD=BC[Given]ΔABDΔBAC[By SAS](ii)BD=AC[By CPCT](iii)ABD=BAC[By CPCT]

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

Ans

Given:AD and BC are equal perpendiculars to a line segment AB.To Prove: CD bisects AB i.e., CO=OD      Proof:In​ ΔBOCΔAOD     CBO=DAO       [Given]     BOC=AOD       [Vertical opposite angles]  BC=AD            [Given]ΔBOCΔAOD        [By AAS]       BO=OA           [By C.P.C.T.]CD bisects line segment AB.

Q.15

l and m are two parallell ines intersected by another pair ofparallel lines p and q (see figure below).Show that ΔABCΔCDA.

Ans

Given: line l and m are parallel and p⇀q⇀.To prove: ΔABCΔCDAProof: In ΔABCandΔCDABAC=DCA      [Alternate interior angles]     AC=CA          [Common]BCA=DAC      [Alternate interior angles]ΔABCΔCDA      [By​ ASA]

Q.16

Line  l  is the bisector of an angle  ∠A and ∠B is any point on  l.BP and BQ are perpendiculars  from B to the arms of ∠A(see figure below). Show that:(i)ΔAPB ΔAQB (ii)BP=BQ or B is equidistant from the armsofA.

Ans

Given: Line l is bisector of A, BP and BQ are perpendiculars to       the arms of angle.To prove: (i)ΔAPBΔAQB(ii) BP=BQ or B is equidistant from the arms  of∠A.Proof:(i) In ΔABQ and ΔAPBBAQ=BAP      [Given]     AB=AB          [Common]AQB=APB       [each 90°]ΔABQΔAPB      [By​ AAS](ii)      BQ=BP         [By C.P.C.T.]      BP=BQ

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

Ans

      Given : In ΔABC and ΔADE, AC=AE, AB=AD and BAD=EACTo prove:BC=DE        Proof: In ΔABC and ΔADE     AB=AD         [Given]BAC=DAE     [BAD+DAC=DAC+EAC]     AC=AE         [Given]ΔABCΔADE     [By​ SAS]      BC=DE        [By C.P.C.T.]

Q.18

AB is a line segment and P is its midpoint.D and E arepoints on the same side of AB such that BAD=ABEand EPA=DPB (see figure below).Show that(i)ΔDAPΔEBP(ii)AD=BE

Ans

      Given: P is mid-point of AB, BAD=ABE​ and EPA=DPBTo prove:(i)ΔDAPΔEBP(ii)      AD=BE        Proof:(i) In ΔDAPandΔEBP                 DPA=EPB[APE+EPD=BPD+DPE]      AP=PB[Given]                 DAP=EBP[Given]ΔDAPΔEPB[By​ SAS](ii)     AD=BE[By C.P.C.T.]

Q.19

In right triangle ABC, right angled at C,M is the midpointof hypotenuse AB.C is joined to M and produced to a pointD such that DM=CM.Point D is joined to point B(see figure below).Show that:(i)ΔAMCΔBMD(ii)ΔDBC is a right angle.(iii)ΔDBCΔACBivCM=12AB

Ans

   Given: In right ΔABC, C=90° and M is mid-point on    hypotenuse AB, CM=DM.To Prove: (i)ΔAMCΔBMD     (ii)DBC is a right angle.     (iii)ΔDBCΔACB     (iv)CM=12AB     Proof:(i) In ΔAMC andΔBMDAM=BM            [Given]    AMC=BMD        [Vertical opposite angles]CM=DM            [Given] ΔAMCΔBMD        [S.A.S.]AC=DB            [By C.P.C.T.](ii)   ACM=BDM        [By C.P.C.T.]   ACD=BDCACDB            [Alternate angles are equal.]   ACB+DBC=180°  [Co-interior angles]         90°+DBC=180°  DBC=180°90°      =90°So,DBC is a right angle.(iii)InΔDCB and ΔACB        DB=AC         [Proved]ACB=DBC     [Each 90°]      BC=BC          [Common]       ΔDCBΔACB      [S.A.S.](iv)      DC=AB         [By C.P.C.T.]    2CM=AB        [CM=DM]      CM=12AB

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

Given: In ΔABC, AB=AC and bisectors of B and C intersect        at O.To Prove: (i) OB=OC (ii) AO bisectsAProof:(i) In ΔABC, AB=AC So,   ACB=ABC       [Opposite angles of equal sides are equal.]    12ACB=12ABC      OCB=OBC        [Given, as OB and OC are angle bisectors.]  OB=OC             [Opposite angles of equal sides are equal.] (ii) In ΔOAB and ΔOAC,  OA=OA            [Common]  AB=AC            [Given]  OB=OC           [Proved above]             ΔOAB ΔOAC       [By S.S.S.]Hence,    OAB=OAC        [By C.P.C.T.]Thus, OA bisects A.

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

Given: In ΔABC, BD=DC and ADBC.To Prove: AB=ACProof: In ΔABD and ΔACD    AD=AD                [Common]        ADB=ADC            [Given]     BD=DC                 [Given]      ΔABD ΔACD            [By S.A.S.]            AB=AC                 [By C.P.C.T.]Therefore, ΔABC is isosceles triangle.

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

Given:In ΔABC,AC=AB, BEAC and CFAB.To prove: BE=CF        Proof:InΔABE and ΔACF    BAE=CAF         [Common angle]    BEA=CFA         [Each 90°]AB=AC            [Given]    ΔABEΔACF          [By A.A.S.]BE=CF            [By C.P.C.T.]                            Hence proved.

Q.23

ABC is a triangle in which altitudes BE and CF to sides ACand AB are equal (see figure below).Show that(i)ΔABEΔACF(ii)AB=AC,i.e.,ABC is an isosceles triangle.

Ans

Given:InΔABC,BE=CF, BEAC and CFAB.To prove: (i)ΔABEΔACF            (ii)AB=AC,i.e.,ABC is an isosceles triangle.        Proof:(i)InΔABE andΔACF    BAE=CAF        [Common angle]    BEA=CFA        [Each 90°]BE=CF           [Given]    ΔABEΔACF          [By A.A.S.](ii)AB=AC                  [By C.P.C.T.]i.e.,ABC is an isosceles triangle.                              Hence proved.

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

Ans

  Given:InΔABC,AB=AC and inΔDBC,DB=DC.To prove: ABD=ACD        Proof:InΔABCAB=AC            [Given]    ACB=ABC  ...(i)    [Opposite angles of equal sides are equal.]InΔDBCDB=DC            [Given]    DCB=DBC  ...(ii)  [Opposite angles of equal sides are equal.]Adding equation(i) and equation(ii),we getACB+DCB=ABC+DBC      ACD=ABD            ABD=ACD                                    Hence proved.

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

Given:In​ ΔABC, AB=AC, side BA​ is produced to D such that      AD=AB.To prove:BCD is a right angle.Proof:In ΔABC,     AB=AC        [Given]ACB=ABC     [Opposite angles of equal sides are equal.]    =x  (let)            In ΔACD,     AC=AD        [Given]ADC=ACD     [Opposite angles of equal sides are equal.]    =y  (let)Now, In ΔBCDBCD+CBD+BDC=180°                   (x+y)+x+y=180°           2(x+y)=180°    (x+y)=180°2    BCD=90°âˆ†BCD is right triangle.

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

Ans

  Given :In ΔABC, A=90° and AB=AC.To find :  B and C        Sol : In ΔABC,                    AB=AC           [Given]So,        ACB=ABC=x    (let)Since,   A+B+C=180°     90°+x+x=180°2x=180°90°  x=90°2=45°So,B=45° and C=45°.

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

Ans

Given : ΔABC is an equilateral triangle.To prove : A=B=C=60°Proof: In ΔABC,        AB=AC        [Given]

           C=B    ...(i)     [Angles opposite to equal sides are equal.]AB=BC            [Given]          C=A    ...(ii)    [Angles opposite to equal sides are equal.]By angle sum property in a triangle,A+B+C=180°C+C+C=180°        [From equation (i) and equation(ii)]3C=180°  C=180°3=60°So, A=60° and B=60°Therefore, each angle of an equilateral triangle is 60° each.

Q.28

ΔABC and ΔDBC are two isosceles triangle son the samebase BC and vertices A and D are on the same side of BC(see the figure below).If AD is extended to intersect BC at P, showthat (i)ΔABDΔACD(ii)ΔABPΔACP(iii)AP bisects A as well as D.(iv)AP is the perpendicular bisector of BC.

Ans

  Given:ΔABC and ΔDBC are two isosceles triangles on the same       base BC and vertices A and D are on the same side of BC.To prove:(i)ΔABDΔACD(ii)ΔABPΔACP(iii)AP bisects A as well as D.(iv)AP is the perpendicular bisector of BC.      Proof:(i)In ΔABD   and ΔACD      AB=AC        [Given]      DB=DC        [Given]    AD=AD        [Common]ΔABDΔACD    [By S.S.S.]        BAD=CAD    [By C.P.C.T.] (ii)In ΔABP  andΔACP      AB=AC          [Given]         BAP=CAP       [Proved above]     AP=AP          [Common]ΔABPΔACP      [By S.A.S.]        BAP=CAP      [By C.P.C.T.]      BP=CP        [By C.P.C.T.](iii)∵        BAP=CAP      [By C.P.C.T.]AP bisects A.         In ΔBDP  andΔCDP      BD=CD        [Given]      DP=DP        [Common]      BP=CP        [By C.P.C.T.]                 ΔBDPΔCDP     [By S.S.S.]        BDP=CDP     [By C.P.C.T.]DP bisects  BDC i.e., D.Thus, AP bisects A as well as D.(iv) APB=APC         [By C.P.C.T.]and  APB+APC=180°    APB+APB=180°  2APB=180°    APB=180°2=90°And BP=CPSo,​ AP is perpendicular bisector of BC.

Q.29 AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.

Ans

(i)       Given: AD is altitude in isosceles ΔABC in which AB=AC.To​ prove: (i) AD bisects BC (ii) AD bisects  A       Proof:(i)In ΔABD and ΔACD AB=AC           [Given]    ADB=ADC        [Each​ 90°] AD=AD           [Common]    ΔABDΔACD        [By R.H.S.]So,​       BD=CD           [By C.P.C.T.]Therefore, AD bisects BC.(ii)    BAD=CAD       [By C.P.C.T.]Therefore, AD bisects A.       Hence proved.

Q.30

Two sides AB and BC and median AM of one triangle ABC arerespectively equal to sides PQ and QR and median PN ofΔPQR (see figure below).Show that:(i)ΔABMΔPQN(ii)ΔABCΔPQR

Ans

Given:In ΔABC and ΔPQR, AB=PQ,BC=QR and AM=PN.To prove:(i)ΔABMΔPQN(ii)ΔABCΔPQRProof:(i)In ΔABM and ΔPQN AB=PQ[Given]BM=QN[12BC=12QR]AM=PN[Given]ΔABCΔPQR[By S.S.S.]ABM=PQN[By C.P.C.T.]B=Q(ii)In ΔABC and ΔPQRAB=PQ[Given]B=Q[Proved above]BC=QR[Given]ΔABCΔPQR[By S.A.S.]  Hence Proved.

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

Given:In​ ΔABC, BEAC and CFAB and BE = CFTo​ prove: ΔABC is isosceles triangle.        Proof:(i)In​ ΔBCE and ΔCBF     BEC=CFB        [Each 90°]BC=BC           [Common]BE=CF            [Given]    ΔBCEΔCBF        [By R.H.S.](ii) AB = AC [By CPCT]i.e ABC is an isosceles triangle

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

Ans

Given:In​ ΔABC, AB=AC and APBCTo​ prove: ΔABC is isosceles triangle.        Proof:In​ ΔABP and ΔACPAPB=APC       [Each 90°] AB=AC          [Given] AP=AP          [Common]          ΔABPΔACP       [By R.H.S.]ABP=ACP      [By C.P.C.T.]           B=C                               Hence Proved.

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

Ans

Given:In ΔABC, B=90°To prove: AC is the longest side of ΔABC.Proof:In ΔABC,A+B+C=180°A+90°+C=180°       A+C=180°90°       A+C=90°Hence, the other two angles have to be acute (i.e.,less than 90º).  B is the largest angle of the triangle.  B>A and B>CAC>BC and AC>AB[In any triangle, the side opposite to the larger (greater) angle is longer.]Therefore,​ AC is the largest side of the triangle.

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

Given: In ΔABC, AB and AC are produced upto P and Q respectively. PBC<QCB.To prove: AC>ABProof: Since, PBC+ABC=180°     [Linear Pair of angles]     PBC=180°ABC      ...(i)        and  QCB+ACB=180°          [Linear Pair of angles]     QCB=180°ACB     ...(ii)It​​ is given that,               PBC<QCB180°ABC<180°ACBABC<ACB    ACB<ABC         AB<AC             [Side opposite to smaller angle is smaller.]         AC>AB                         Hence​ proved.

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

Ans

Given:In​ ΔOAB, B<A and in ΔOCD, C<D.To prove: AD<BCProof: In​ ΔOAB, B<A            [Given]OA<OB      ...(i)[Side opposite to smaller angle is smaller.]In​ ΔOCD, C<D            [Given]OD<OC      ...(ii)[Side opposite to smaller angle is smaller.]Adding relation(i)​ and (ii), we get         OA+OD<OB+OCAD<BC           Hence Proved.

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

Given:In​ quadrilateral ABCD, AB is the smallest side and CD is the largest side.To prove: A>C and B>D.

Proof :  In ΔABC,BC>AB          [Given]   BAC>ACB ...(i) [Angle opposite to greater side isgreater.] In ΔACD,CD>AD               [Given]   CAD>ACD  ...(ii) [Angle opposite to greater side isgreater.]Adding relation(i) and relation(ii), we getBAC+CAD>ACB+ACD      BAD>BCD  A>C    ...(iii) In ΔABD,AD>AB                [Given]   ABD>ADB  ...(iv) [Angle opposite to greater side isgreater.] In ΔBCD,CD>BC               [Given]   CBD>BDC  ...(v) [Angle opposite to greater side isgreater.]Adding relation(iv) and relation(v), we getABD+CBD>ADB+BDC     ABC>ADCB>D   ...(vi)From relation (iii)​ and relation(vi), we haveA>C and B>D.      Hence proved.

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

Ans

 Given:In​ ΔPQR, PR > PQ and PS bisects QPR.To prove: PSR>PSQ       Proof: In ΔPQR, PS bisects QPRi.e.,              QPS=RPS    ...(i)In ΔPQS,      Ex.PSR=PQS+QPS     Ex.PSR=PQR+QPS           PQR=PSRQPS    ...(ii)In ΔPRS,       Ex.PSQ=PRS+SPR     Ex.PSQ=PRQ+RPS           PRQ=PSQRPS           PRQ=PSQQPS    ...(iii)[From equation(i)]and            PR>PQ                  PQR>PRQ    [Opposite angle of greater sideis greater.]      PSRQPS>PSQQPS  [From equation(ii) and (iii)]                      PSR>PSQ            Hence proved.

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

Ans

Given: A line l and a point A which is away from line l.To prove: Perpendicular line segment from A to line l is the shortest.Contruction: Draw ABl, AC and AD.Proof: In ΔABC, B=90°            Then, A+C=90°              B>C              AC>AB       [Side opposite to greater angle is greater.]Similarly,     B>D              AD>AB       [Side opposite to greater angle is greater.]Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

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 ∆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 ∆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 ∆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 ∆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

( i )Area( ΔAOB )= 3 4 ( side ) 2 = 3 4 ( 5 ) 2 = 25 3 4 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaykW7daqadaqaaiaadMgaaiaawIcacaGLPaaacaWGbbGaamOCaiaadwgacaWGHbWaaeWaaeaacqGHuoarcaWGbbGaam4taiaadkeaaiaawIcacaGLPaaacqGH9aqpdGaMaUaaaeacyc4aiGjGkaaabGaMakacyciIZaaaleqcyciaaOqaiGjGcGaMaIinaaaadGaMagWaaeacycOaiGjGdohacGaMaoyAaiacyc4GKbGaiGjGdwgaaiacycOLOaGaiGjGwMcaamacycihaaWcbKaMagacycOaiGjGikdaaaaakeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpdaWcaaqaamaakaaabaGaaG4maaWcbeaaaOqaaiaaisdaaaWaaeWaaeaacaaI1aaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGPaVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9maalaaabaGaaGOmaiaaiwdadaGcaaqaaiaaiodaaSqabaaakeaacaaI0aaaaiaabccacaqGJbGaaeyBamaaCaaaleqabaGaaeOmaaaaaaaa@AC3B@

Area of hexagonal – shaped rangoli=6×Area( ΔAOB ) =6× 25 3 4 cm 2 = 75 3 2 cm 2 Area of equilateral triangle having its side as 1 cm = 3 4 ( side ) 2 = 3 4 ( 1 ) 2 = 3 4 cm 2 Number of equilateral triangles of 1 cm side that can be filled in hexagonal-shaped rangoli = ( 75 3 2 ) ( 3 4 ) =150

MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D93E@ Starshaped rangoli has 12 equilateral triangles of side 5 cm in it.Area of equilateral triangle=34(side)2=34(5)2=2534 cm2Area of star-shaped rangoli=12×2534 cm2=755 cm2Area of equilateral triangle having its side as 1 cm=34(side)2=34(1)2=34 cm2Number of equilateral triangles of 1 cm side that can be filled in hexagonal-shaped rangoli=(753 cm2)(34)=300Therefore, star-shaped rangoli has more equilateral triangles in it.

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°

The second angle of quadrilateral
= 5(12°)
= 60°

The third angle of quadrilateral
= 9(12°)
= 108°

The fourth angle of quadrilateral
= 13(12°)
= 156°

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

Ans

Given: Let ABCD be a parallelogram, in which AC=BD.To prove: ABCD is a rectangle.Proof: In ΔADC and ΔBCD,  AD=BC          [Opposite sides of parallelogram are equal.]              DC=CD          [Common]              AC=BD          [Given]       ΔADCΔBCD     [By S.S.S.]        ADC=BCD     [By C.P.C.T.]but     ADC+BCD=180°   [Cointerior angles]       ADC=BCD=90°Since, one angle of parallelogram is 90°.So, ABCD is a rectangle.         Hence proved.

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

Ans

Given: Let ABCD be a quadrilateral, in which AO=OC, BO=OD. AC and BD bisect each other at 90°.To prove: ABCD is a rhombus.       Proof:​  In ΔAOB and ΔCOB                AO=OC          [Given]             AOB=COB      [Each 90°]                OB=OB          [Common]            ΔAOBΔCOB      [By SAS]              AB=BC   ...(i)   [By C.P.C.T.]        In ΔBOC and ΔDOC               BO=OD         [Given]          BOC=DOC     [Each 90°]             OC=OC          [Common]             BOCΔDOC       [By SAS]              BC=DC   ...(ii)  [By C.P.C.T.]            In ΔCOD and ΔDOA                CO=OA          [Given]             COD=AOD         [Each 90°]                OD=OD         [Common]             ΔCODΔDOA      [By SAS]              DC=DA   ...(iii)[By C.P.C.T.]From equation(i),(ii) and (iii), we get              AB=BC=CD=DASince, all sides of quadrilateral ABCD are equal, soABCD is a rhombus.              Hence proved.

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

Ans

Given: Let ABCD be a square.To prove:AC=BD, AO=OC, BO=OD and ACBD.        Proof: In ΔABC and ΔDCB           AB=DC                    [Sides of square]        ABC=DCB              [Each 90°]           BC=BC                    [Common]        ΔABCΔDCB              [By S.A.S.]                   AC=BD           [By C.P.C.T.]In ΔAOB and ΔCOD           AB=CD           [Sides of square]       AOB=COD    [Vertical opposite angles]       BAO=DCO    [Alternate interior angles]       ΔAOBΔCOD     [By S.A.S.]           AO=OC           [By C.P.C.T.]          BO=OD           [By C.P.C.T.]In ΔAOB and ΔCOB           AB=CB           [Sides of square]           OB=OB           [Common]           AO=OC           [Proved above]           ΔAOBΔCOB     [By S.A.S.]           AOB=COB     [By C.P.C.T.]and      AOB+COB=180°                 AOB=90°Thus, diagonals AC and BD are equal and bisect each other at 90°.         Hence proved.

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

Given:Let ABCD be a quadrilateral. AC=BD and ACBD,    OA=OC and OB=OD.To prove: ABCD is a square.        Proof:In ΔAOB and ΔCOB   AO=OC [Given]     AOB=COB [Each 90°]            OB=OB [Common]      ΔAOBΔCOB [By S.A.S.]            AB=BC   ...(i)      [By C.P.C.T.]  In ΔBOC and ΔCOD   BO=OD [Given]     BOC=DOC [Each 90°]            OC=OC [Common]      ΔBOCΔCOD [By S.A.S.]            BC=CD   ...(ii)        [By C.P.C.T.]In ΔCOD and ΔDOA   CO=OA [Given]     DOC=DOA [Each 90°]            OD=OD [Common]      ΔCODΔDOA [By S.A.S.]            CD=DA   ...(iii)     [By C.P.C.T.]From equation(i),(ii) and (iii), we haveAB=BC=CD=DAThus, ABCD is a square.Hence proved.

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

Ans

     Given: ABCD is a parallelogram in which diagonal AC    bisects A.To prove:(i) it bisects  C also,    (ii) ABCD is a rhombus        Proof:(i)Since, ADBC so, DAC=BCA ...(i) [Alternate interior angles]and ABDCso,  BAC=DCA ...(ii) [Alternate interior angles]But       BAC=DAC  ...(iii) [Given]So,      DCA=BCA [From equation(i) and (ii)]AC bisects C.(ii) From equation(i) and equation(iii), we have       BAC=BCA  BC=BA [Sides opposite to equal angles are equal.]Since, adjacent sides of parallelogram are equal, so ABCDis a rhombus. Hence proved.

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

   Given:ABCD is a rhombus.To prove: AC bisects A as well as C.  BD bisects B as well as D.       Proof:In ΔABC, AB=BCSo, BAC=BCA ...(i) [Opposite angles of equal sidesare equal.]Since, ABDCSo,             BAC=DCA. ..(ii) [Alternate interior angles.]From equation(i) and equation(ii), we have        BCA=DCAdiagonal AC bisects C.Since, ADBCSo,             DAC=BCA (iii) [Alternate interior angles.]From equation(i) and equation(iii), we have        BAC=DACdiagonal AC bisects A.Thus, AC bisects A as well as C.Similarly, BC bisects B as well as D.Hence proved.

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

   Given:ABCD is a rectangle. Diagonal AC bisects A as well   as C.To prove:(i) ABCD is a square.(ii)Diagonal BD bisects B as well as D.        Proof:(i)Since, 1=2 and 3=4∵ABCD     2=3 [Alternate interior angles]     1=3    CD=AD [Opposite sides of equal angles are equal.] Since,adjecent sides of a rectangle are equal, so ABCD is a square.(ii)As ABCD is a square. So,    AB=AD      ADB=ABD ...(i) [Opposite angles of equal sides are equal.]Since,ABCDSo,    ABD=CDB ...(ii) [Alternate interior angles]      ADB=CDB MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2BFD@ BD bisects D.Similarly, we can prove that BD bisects B.Therefore, BD bisects B as well as D.    Hence proved.

Q.51

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ . Show that:(i)ΔAPDΔCQB(ii)AP=CQ (iii)ΔAQBΔCPD (iv)AQ=CP (v)APCQ is a parallelogram

Ans

Given:ABCD is a parallelogram and DP=BQ.To prove: (i)ΔAPDΔCQB            (ii)AP=CQ             (iii)ΔAQBΔCPD             (iv)AQ=CP             (v)APCQ is a parallelogram Proof: In ΔAPD and ΔCQB              AD=BC               [Opposite sides of parallelogram.]           ADP=CBQ           [Alternater interior angles.]                 DP=BQ               [Given]             ΔAPDΔCQB           [By S.A.S.](ii)              AP=CQ               [By C.P.C.T.](iii) In ΔAQB and ΔCPD                 AB=CD               [Opposite sides of parallelogram.]           ABQ=CDP           [Alternater interior angles.]                 BQ=DP               [Given]             ΔAQBΔCPD           [By S.A.S.](iv)          AQ=PC               [By C.P.C.T.](v)Since, AP=CQ and AQ=PC So, APCQ is a parallelogram. [Opposite sides are parallel.]                                          Hence proved.

Q.52

ABCD isaparallelogramandAPandCQareperpendicularsfromverticesAandCondiagonalBDseethefigurebelow.ShowthatiΔAPBΔCQDii   AP=CQ

Ans

Given:ABCD is a parallelogram and APBD and CQBD.To prove:iΔAPBΔCQDiiAP=CQProof:i In ΔAPB and ΔCQD               ABP=CDQ           Alternate interior angles               APB=CQD           [Each 90°]                AB=CD                Opposite sides of parallelogram.              ΔAPBΔCQD           By A.A.S.iiAP=CQ              By C.P.C.T.Hence proved.

Q.53

In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the figure below). Show that(i)quadrilateral ABED is a parallelogram(ii)quadrilateral BEFC is a parallelogram(iii)AD||CF and AD=CF(iv)quadrilateral ACFD is a parallelogram(v)AC=DF(vi)ΔABCΔDEF.

Ans

   Given:In ΔABC and ΔDEF,AB=DE, AB||DE, BC=EF and BC||EF.To prove:(i)quadrilateral ABED is a parallelogram(ii)quadrilateral BEFC is a parallelogram(iii)AD||CF and AD=CF(iv)quadrilateral ACFD is a parallelogram(v)AC=DF(vi)ΔABCΔDEF.       Proof:(i) Since, AB= DE and ABDESo, ABED is a parallelogram.[One pair of opposite sides is parallel and equal.]    (ii)Since, BC=EF and BC||EFSo, BEFC is a parallelogram.[One pair of opposite sides is parallel and equal.]  (iii)Since, ABED is a prallelogram.So, AD=BE and ADBE  ...(i)[One pair of opposite sides is parallel and equal.]Since, BEFC is a prallelogram.So, BE=CF and BECF  ...(ii)[One pair of opposite sides is parallel and equal.]From equation(i) and equation(ii), we haveAD=CF and ADCF(iv)Since,  AD=CF and ADCFSo, ADFC is a prallelogram.[One pair of opposite sides is parallel and equal.](v)Since, ADFC is a prallelogram. So, AC=DF[Opposite sides of parallelogram.](vi) In ΔABC and ΔDEF   AB=DE[Given]BC=EF[Given]AC=DF[Proved above]   ΔABCΔDEF[ByS.S.S.]Hence proved.

Q.54

ABCD is a trapezium in which AB||CD and AD=BC(see the figure below).Show that(i)A=B(ii)C=D(iii)ΔABCΔBAD(iv)diagonal AC=diagonal BD

Ans

       Given:In trapezium ABCD, ABCD and AD=BC.To prove: (i)A=B  (ii)C=D(iii)ΔABCΔBAD(iv)diagonal AC=diagonal BDConstruction: Draw CEDA. Join AC.   Proof:Since, ABCD AECDand CEDASo, AECD is a prallelogram.Then, AD=CEBut, AD=BCCE=BC    CBE=CEB [Opposite sides of equal angles are equal in ΔCEB.]         CBE+CBA=180°   ...(i)[Linear pair of angles]         CEA+DAE=180°  [Cointerior angles]         CBE+DAE=180°    ...(ii)[∵CBE=CEA]From equation(i) and equation  (ii), we have         CBE+CBA=CBE+DAE      CBA=DAEB=AA=B(ii) Since, ABDC So, A+D=180°   and          B+C=180°A+D=B+C           D=C[∵A=B](iii) In ΔABC and ΔBADAB=BA[Common]A=B[Proved above]AD=BC[Given]ΔABCΔBAD[By S.A.S.](iv) AC=BD[By C.P.C.T.]Thus, diagonal AC = diagonal BD.Hence proved.

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

Given:ABCD is a quadrilateral in which P, Q, R and S  are  mid-points of the sides AB, BC, CD and DA.To prove:(i) SR||AC and SR=12AC            (ii) PQ=SR            (iii) PQRS is a parallelogram.Proof:  (i) In ΔADC,S and R are the mid-points of DA and DC                respectively. So, by mid-point theorem                     SRAC and SR=12AC         ...(i)            (ii) In ΔABC,P and Q are the mid-points of AB and BC                 respectively. So, by mid-point theorem                     PQAC and PQ=12AC        ...(ii)               From equation (i) and equation(ii), we get                       PQ=SR          (iii) From equation(i) and equaton(ii),                  PQ=SR and PQSR                 PQSR is a prallelogram. [One pair of opposite sides is equal and parallel.]                                                Hence proved.

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

Given:ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.To prove:PQRS is a rectangle.Construction: Draw AC and BD.Proof: In ΔABC, P and Q are the mid-points of AB and BC           respectively.So, by mid-point theorem          PQ=12AC and PQAC        ...(i)            In ΔADC, S and R are the mid-points of AD and DC         respectively. So, by mid-point theorem      SR=12AC and SRAC          ...(ii)      From equation(i) and equation(ii), we have          PQ=SR and PQSR So, PQRS is a parallelogram.          [If one pair of opposite sides of a quadrilateral is paralleland equal, then it is a parallelogram.]            In ΔABD, P and S are the mid-points of AB and AD         respectively. So, by mid-point theorem         PS=12BD and PSBD   ...(iii) From equation(i) and equation(iii), we have PNMO and MPON So, PMON is a parallelogram. P=MON     [Opposite angles of parallelogram are equal.]         =90° Since, P=90° in parallelogram PQRS, so PQRS is a rectangle.

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

Given:ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.To prove:PQRS is a rhombus.Construction: Draw AC and BD.They are the diagonals of the quadrilateral.Proof: In ΔABC, P and Q are the mid-points of AB and BC respectively. So, by mid-point theoremPQ=12AC and PQAC(i)In ΔADC, S and R are the mid-points of AD and DC respectively. So, by mid-point theoremSR=12AC and SRAC(ii) From equation(i) and equation(ii), we have PQ=SR and PQSR So, PQRS is a parallelogram.[If one pair of opposite sides of a quadrilateral is paralleland equal, then it is a parallelogram.]In ΔABD, P and S are the mid-points of AB and ADrespectively. So, by mid-point theorem PS=12BD and PSBD(iii) Since, AC=BD[Diagonals of rectangle are equal.]PQ=PS[12AC=12BD] Since, adjecent sides of a parallelogram ABCD are equal.So, PQRS is rhombus. Hence proved.

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

Given: ABCD is a trapezium in which AB||DC, BD is a diagonaland E is the midpoint of AD. A line through E is prallel to AB.To prove: F is the mid-point of BC i.e., BF=CF.Proof: In ΔABD, E is mid-point of AD and EOAB So, by converse of mid-point theorem,OD=OBIn ΔBCD, O is mid-point of BD and EFCD So, by converse of mid-point theorem,CF=FBTherefore, F is the mid-point of BC. Hence proved.

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

Given: In parallelogram ABCD, E and F are the mid-points of           sides AB and CD respectively.To prove: DP=PQ=QBProof: In ΔDQC, F is mid-point of DC and PFQC,        So, by converse of mid-point theorem, we have                    DP=PQ          ...(i)       In ΔABP, E is mid-point of AB and EQAP,       So, by converse of mid-point theorem, we have                    PQ=QB          ...(ii)       From equation(i) and equation(ii), we have                    DP=PQ=QBThis implies that line segments AF and EC trisect the diagonal BD.                                        Hence proved.

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

Ans

Given: Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.To prove:PR and QS bisect each other.Construction: Draw AC.Proof: In ΔABC, P and Q are the mid-points of AB and BC respectively. So, by mid-point theoremPQ=12AC and PQAC(i)In ΔADC, S and R are the mid-points of AD and DC respectively. So, by mid-point theorem SR=12AC and SRAC(ii) From equation(i) and equation(ii), we have PQ=SR and PQSR So, PQRS is a parallelogram. [If one pair of opposite sides of a quadrilateral is paralleland equal, then it is a parallelogram.] Since, diagonals of a parallelogram bisect each other, so PR and QS bisect each other.  Hence proved.

Q.61

ABC is a triangle right angled at C. A line through themidpoint M of hypotenuse AB and parallel to BC intersectsAC at D. Show that(i)D is the mid-point ofAC(ii)MD⊥ACiiiCM=MA=12AB

Ans

Given:In ΔABC, C=90° and M is mid-point of AB, MDBC.To prove: (i) D is the mid-point of AC             (ii)MD⊥AC            (iii)CM=MA=12ABProof:(i)In ΔABC, MDBC and M is mid-point of AB, then         by converse of mid-point theorem,            AD=DC i.e., D is the mid-point of AC.      (ii) Since, MDBC          So, MDA=BCA         [Corresponding angles]                          =90°          Therefore, MDAC.      (iii)  In ΔMDC and ΔMDA             DC=DA             [Proved above]        MDC=MDA        [Each 90° as MDAC]           MD=MD               [Common]        ΔMDCΔMDA         [By S.A.S.]         CM=MA             [By C.P.C.T.]          Since, AM=12AB       So,CM=MA=12AB                  Hence proved.

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

In parallelogram ABCD, CD=AB=16 cm [Opposite sides of a parallelogram are equal] We know that Area of a parallelogram=Base×Corresponding altitude Area of parallelogram ABCD=CD × AE=AD×CF 16 cm × 8 cm = AD × 10 cmAD=16×810=12.8mThus, the length of ADis2.8cm.

Q.64

If E, F, G and H are respectively the midpoints of the sidesofa parallelogram ABCD show that ar (EFGH)=12ar (ABCD).

Ans

Given: ABCD is a parallelogram in which E, F, G and H are mid-points of AB, BC, CD and DA respectively.To Prove: ar(EFGH)=12ar(ABCD)Contruction: Join HF.Proof: Since, ABCD is a parallelogram.So, ADBC and AD=BC12AD=12BC and ADBCAH=BF and AHBFSo, ABFH is a parallelogram.Then, ABHF.Since ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,Area (ΔHEF)=12Area (ABFH)(i)Similarly, it can be proved thatArea (ΔHGF)=12Area (DCFH)(ii)On adding equations (1) and (2), we getArea (ΔHEF)+Area (ΔHGF)=12Area (ABFH)+12Area (DCFH) =12{Area (ABFH)+Area (DCFH)} Area (ΔEFGH)=12Area (ABCD)

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 ∆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, ∆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

In the given figure, P is a point in the interior of aparallelogram ABCD.Show that (i)ar(APB)+ar(PCD)=12ar(ABCD) (ii)ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

Ans

Given:ABCD is a parallelogram. P is any point inside of it.To prove: (i)ar(APB)+ar(PCD)=12ar(ABCD)(ii)ar(APD)+ar(PBC)=ar(APB)+ar(PCD) Construction:​ Draw EFAB and QRAD through point P.Proof: In parallelogram ABCD, ADBC[Opposite sides of parallelogram.] AEBFand ABEF[By construction]Thus, ABFE is a prallelogram.Since, ΔAPB and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.  area(ΔAPB)=12area(ABFE)(i)Now EFCD because EFAB and ABCD.So, ΔDPC and parallelogram ABFE are lying on the same baseDC and between the same parallel lines AB and EF. area(ΔDPC)=12area(ABFE)(ii)Adding equation(i) and equation(ii), we getarea(ΔAPB)+area(ΔDPC)=12area(ABFE)+12area(ABFE)=12area(ABCD)(A)(ii)In parallelogram ABCD, ABDC[Opposite sides of parallelogram.] AQDRand  ADQR[By construction]Thus, AQRD is a prallelogram.Since, ΔAPD and parallelogram AQRD are lying on the same base AD and between the same parallel lines AD and QR. area(ΔAPD)=12area(AQRD)(iii)Now QRBC because QRBC and ADBC.So, ΔBPC and parallelogram QBCR are lying on the same baseBC and between the same parallel lines QR and BC. area(ΔBPC)=12area(QBCR)(iv)Adding equation(iii) and equation(iv), we getarea(ΔAPD)+area(ΔBPC)=12area(AQRD)+12area(QBCR)= 12area(ABCD)(B)From equatin (A) and (B), we havearea(ΔAPB)+area(ΔDPC)=area(ΔAPD)+area(ΔBPC).

Q.67

In the given figure, PQRS and ABRS are parallelograms andX is any point on side BR.Show that (i) ar(PQRS)=ar(ABRS)(ii) arΔPXS=12arPQRS

Ans

Given: PQRS and ABRS are parallelograms and X is any point on           side BR.To prove: (i) ar(PQRS)=ar(ABRS)            (ii)  arΔPXS=12arPQRSProof: (i) Since, the parallelogram PQRS and ABRS lie on the          same base SR and also, these lie in between the same parallel lines SR and PB.      Area (PQRS) = Area (ABRS) ... (i) (ii) Since, ΔAXS and parallelogram ABRS lie on the same base and are between the same parallel lines AS and BR.  Area(ΔAXS)=12area(ABRS)     ...(ii)From equation(i) and equation (ii), we have  Area(ΔAXS)=12area(PQRS).   Hence proved.

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

In the figure, point A divides the field into three parts. These parts are triangular in shape: ΔPSA, ΔPAQ, and ΔSRA Area of ΔPSA+Area of ΔPAQ+Area of ΔSRA = Area of PQRS (i) We know that if a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.Area(ΔPAQ)=12Area (PQRS) (ii) From equation (i) and equation (ii), we get Area of ΔPSA+12Area of ΔPAQ+Area of ΔSRA =Area of PQRSArea of ΔPSA+Area of ΔSRA =Area of PQRS12Area of ΔPAQ=12Area of ΔPAQTherefore,we can say that the farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and SRA or wheat in triangular parts PSA and SRA and pulses in triangular parts PAQ.

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

Ans

Given : In ΔABC, AD is median and E is any point on AD.To prove : ar(ABE)=ar(ACE)Proof: In ΔABC, AD is median. So, ar(ΔABD)=ar(ΔACD)     ...(i)[Median divides a triangle into two triangles of equal areas.]In ΔEBC, ED is median. So, ar(ΔEBD)=ar(ΔECD)     ...(ii)[Median divides a triangle into two triangles of equal areas.]Subtracting equation(ii) from equation(i), we getar(ΔABD)ar(ΔEBD)=ar(ΔACD)ar(ΔECD)             Area (ΔABE)= Area (ΔACE)

Q.70

In a triangle ABC, E is the midpoint of median AD. Showthat arBED=14arABC.

Ans

Given:In​ ΔABC, AD is the median and E is the mid-point of AD.To prove : arBED=14arABCProof : In ΔABC, AD is median. So, ar(ΔABD)=12ar(ΔABC)      ..(i)[Median divides a triangle into two triangles of equal areas.]In ΔBDA, BE is median. So, ar(ΔBED)=12ar(ΔABD)      ...(ii)[Median divides a triangle into two triangles of equal areas.]From equation(i) and equation(ii), we have      ar(ΔBED)=12{12ar(ΔABC)}      ar(ΔBED)=14ar(ΔABC)                                    Hence proved.

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

Given:ΔABC and ΔADB have same base AB and OC=OD.To prove: ar(ΔABC)=ar(ΔABD)Proof:In ΔADC, AO is median, so        area(ΔAOC)=area(ΔAOD)     ...(i)[Median divides a triangle into two triangles of equal areas.]In ΔBDC, BO is median, so       area(ΔBOC)=area(ΔBOD)       ...(ii)[Median divides a triangle into two triangles of equal areas.]Adding equation(i) and equation(ii), we getarea(ΔAOC)+area(ΔBOC)=area(ΔAOD)+area(ΔBOD)area(ΔABC)=area(ΔDAB)​       Hence proved.

Q.73

D, E and F are respectively the midpoints of the sidesBC, CA and AB of a ΔABC.Show that(i)BDEF is a parallelogram.(ii) arDEF=14arABC(iii) arBDEF=12arABC

Ans

Given:In ΔABC, D, E and F are the mid-points of BC, CA and AB respectively.To prove: (i)BDEF is a parallelogram.             (ii)ar(DEF)=14ar(ABC)            (iii)ar(BDEF)=12ar(ABC)Proof:(i) In ΔABC, F and E are the mid-points of AB and AC              respectively, so by mid-point theorem, we have                 FE=12BC and FEBC              FE=BD and FEBD     [∵D is the mid-point of BC.]BDEF is a parallelogram because one pair of opposite sides is parallel and equal.(ii) Since,     FE=12BC and FEBC      So,                   FE=DC and FEDCThus, EFDC is a parallelogram. [Since, one pair of opposite sides is equal and parallel.]Similarly, DEAF is a parallelogram.Since, diagonal of a parallelogram divides it into two congruenttriangles. So,ar(ΔAFE)=ar(ΔDFE), ar(ΔBFD)=ar(ΔDFE) and ar(ΔDFE)=ar(ΔDCE) ar(ΔAFE)=ar(ΔDFE)=ar(ΔBFD)=ar(ΔDFE)ar(ΔDFE)=14ar(ΔABC)[∵ar(ΔABC)=ar(ΔAFE)+ar(ΔDFE)+ar(ΔBFD)+ar(ΔDFE)](iii) ar(BDEF)=ar(ΔBFD)+ar(ΔDFE)                    =14ar(ΔABC)+14ar(ΔABC)                    =12ar(ΔABC)

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

Given:In quadrilateral ABCD, AC and BD are diagonals which             intersect at O, OB = OD and AB = CD.To prove:  (i) ar (DOC)= ar (AOB)               (ii) ar (DCB)= ar (ACB)             (iii) DA||CB or ABCD is a parallelogram.Construction: Draw BMAC and DNAC.Proof: (i)In ΔBOM and ΔDON           BOM=DON          [Vertical opposite angles]           BMO=DNO          [each 90°]              OB=OD               [Given]       ΔBOMΔDON        [By​ A.A.S]            BM=DN              [By C.P.C.T.]and   ar(ΔBOM)=ar(ΔDON)   ...(i)In ΔBMA and ΔDNC                 BM=DN           [Proved above]          BMA=DNC      [each 90°]              AB=CD          [Given]         ΔBMAΔDNC      [By R.H.S.]   ar(ΔBMA)=ar(ΔDNC)   ...(ii)Adding equation(i) and equation(ii), we getar(ΔBOM)+ar(ΔBMA)=ar(ΔDON)+ar(ΔDNC)       ar(ΔAOB)=ar(ΔDOC)     ar(ΔDOC)=ar(ΔAOB)(ii) Since, ar(ΔDOC)=ar(ΔAOB)Adding ar(ΔBOC) both sides, we getar(ΔDOC)+ar(ΔBOC)=ar(ΔAOB)+ar(ΔBOC)                  ar(ΔDCB)=ar(ΔABC)(iii) ar(ΔDCB)=ar(ΔABC) If two triangles having same base and equal area, thenthey lie between same parallels.So, BCADor DABCAs  ΔAOBΔCOD  OAB=OCDCAB=ACDABCD          [Alternate angles are equal.]So, ABCD is a parallelogram because opposite sideare parallel.

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

Given : In  ΔABC, D and E are the points on AB and AC            respectively and ar(DBC)=ar(EBC).To prove : DEBCProof : Since, ar(DBC)=ar(EBC)     and both triangles have same base BC.So, DEBC   [∵If two triangles have equal area and same base, then they lie between same parallels.]

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

Given:In ΔABC, XYBC, BEAC and CFAB.To prove: ar(ABE)= ar(ACF)Proof:Since, parallelograms BCYE and BCFX have same base BC and lie between same parallels.So, ar(BCYE)=ar(BCFX)     ...(i)ΔAEB and parallelogram BCYE have same base and lie between same parallels. So,         ar(ΔABE)=12ar(BCYE)    ..(ii)Similarly,         ar(ΔACF)=12ar(BCFX)    ...(ii)From equation(i), we have 12ar(BCYE)=12ar(BCFX)From equation(ii) and equation(iii), we have         ar(ΔABE)=ar(ΔACF)                                     Hence Proved.

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

Given:ABCD and PBQR are parallelogram. AQCP.To prove: ar(ABCD)=ar(PBQR)Construction: Join AC and PQ.Proof: Since, AQCP ΔACP and ΔQPC have same base CP and both triangles are between same parallels. So,        ar(ΔACP)=ar(ΔQPC)Subtracting ar(ΔBCP) from both sides, we getar(ΔACP)ar(ΔBCP)=ar(ΔQPC)ar(ΔBCP)        ar(ΔABC)=ar(ΔQBP)      2×ar(ΔABC)=2×ar(ΔQBP)        ar(ABCD)=ar(PBQR)[∵Diagonal divides a parallelogram in two triangles of equal area.]

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

Given:In trapezium ABCD, ABCD.To prove: ar (AOD) = ar (BOC)Proof: Area of two triangles is equal if they have same base and lie between same parallels.Here, ABCD, ΔADC and ΔBDC have same base.    ar(ΔADC)=ar(ΔBDC)     ar(ΔAOD)+ar(ΔDOC)=ar(ΔBDC)+ar(ΔDOC)     ar(ΔAOD)=ar(ΔBDC)

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

Given: ABCDE is a pentagon and BFAC.To prove : (i) ar(ACB)=ar(ACF)             (ii) ar(AEDF)=ar(ABCDE)Proof: (i) Since,ACBF, ΔACB and ΔACF have same base AC.          So,       ar(ACB)=ar(ACF)[Areas of two triangles are equal if they have same baseand lie between same parallels.](ii) We proved above:                      ar(ACB)=ar(ACF)Adding ar(AEDC) both sides, we getar(ACB)+ar(AEDC)=ar(ACF)+ar(AEDC)           ar(ABCDE)=ar(AEDF)or            ar(AEDF)=ar(ABCDE)                            Hence proved.

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

Given:Let ABCD be the shape of the plot of land of Itwaari and AED is his adjacent plot. ED be the side of his adjoining land.        To prove: ar(ΔAOB)=ar(ΔEOD)Construction: The proposal may be implemented as follows. Join diagonal BD and draw a line parallel to BD through point A. Let it meet the extended side CD of ABCD at point E. Join BE and AD. Let them intersect each other at O. Then,portion ΔAOB can be cut from the original field so that the new shape of the field will be ΔBCE.

Proof: Since, ΔEDB and ΔADB have same base and lie between same parallels. So,       ar(ΔEDB)=ar(ΔADB)Subtracting ar(ΔBOD) from both sides, we getar(ΔEDB)ar(ΔBOD)=ar(ΔADB)ar(ΔBOD)                ar(ΔEOD)=ar(ΔBOA)Adding ar(BCDO) both sides, we getar(ΔEOD)+ar(BCDO)=ar(ΔBOA)+ar(BCDO)                  ar(ΔBCE)=ar(ABCD)Thus, we can replace land in the shape of ΔBOA by the land in the shape of ΔEOD to implement the proposal of Itwaari.

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

Given: ABCD is a trapezium in which ABCD and XYAC.To prove: ar(ADX)=ar(ACY)Construction:Join CX.Proof:Since, ABCD, then ar(ΔADX)=ar(ΔACX)    ...(i)     [Both Δs have same base andlie between same parallels.]and ACXY, then ar(ΔACX)=ar(ΔACY)    ...(ii)From equation(i) and equqation(ii), we have ar(ΔADX)=ar(ΔACY)                                 Hence proved.

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


Ans

Given:In given figure, AP||BQ||CR

To prove:ar(AQC)=ar(PBR)Proof:Since,  AP||BQ|| So, ar(ΔBPQ)=ar(ΔABQ)         ...(i)[Both triangles have same base and lie between same parallels.]Since,  BQCRSo,  ar(ΔBQR)=ar(ΔBQC)               ...(ii)[Both triangles have same base and lie between same parallels.]Adding equation(i) and equation (ii), we getar(ΔBPQ)+ar(ΔBQR)=ar(ΔABQ)+ar(ΔBQC)ar(ΔBPR)=ar(ΔAQC) ar(ΔAQC)=ar(ΔBPR)              Hence proved.

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

Given:ABCD is a quadrilateral and its diagonals AC and BD intersect at O and ar(AOD)=ar(BOC).To prove: ABCD is a trapezium.Proof:It​ is given that           ar(AOD)=ar(BOC)Adding ar(AOB) both sides, we getar(AOD)+ar(AOB)=ar(BOC)+ar(AOB)               ar(ABD)=ar(ABC)Since, both triagles have same base AB and having equal area.So they will lie between same parallels.Thus, ABCD  ABCD is a trapezium.

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

Given:In given figure, ar(DRC)=ar(DPC) and ar(BDP)=ar(ARC).To prove: The quadrilaterals ABCD and DCPR are trapeziums.Proof: It is given that ar(DRC)=ar(DPC)orar(DPC)=ar(DRC)(i) Since, both triangles have same base and equal area. Then, DCRPDCPR is a trapezium.Given that ar(BDP)=ar(ARC)(ii)Subtracting equation(i) from equation(ii), we getar(BDP)ar(DPC)=ar(ARC)ar(DRC) ar(BDC)=ar(ACD) Since, both triangles have same base and equal area. Then, ABDCABCD is a trapezium.In this way, we proved that DCPR and ABCD are trapeziums.

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

Given:Parallelogram ABCD and rectangle ABEF have same            base AB and equal area.To prove: Perimeter of ABCD > Perimeter of ABEFProof: Since, both Parallelogram ABCD and rectangle ABEF have same base AB and equal area. So, CFAB            Now, AB=DC              [Opposite sides of rectangle]           and     AB=EF              [Opposite sides of parallelogram]                 CD=EFThen,     AB+CD=AB+EF       ...(i)Since, the line segment drawn from an external point to a line is the shortest line segment.Then, AD>AF and BC>BE        AD+BC>AF+BE       ...(ii)From equation (i) and equation(ii), we getAB+CD+AD+BC>AB+EF+AF+BEPerimeter of ABCD > Perimeter of ABEF                                                 Hence proved.

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

Given:In ΔABC, BD=DE=EC.To prove:ar(ABD)=ar(ADE)=ar(AEC)Construction: Draw ALBCProof: Since, Area of a triangle=12×perpendicular×base                        ar(ΔABD)=12×BD×AM                         ar(ΔADE)=12×DE×AM                         ar(ΔAEC)=12×EC×AMSince, BD=DE=ECSo, 12×BD×AM=12×DE×AM=12×EC×AM        ar(ABD)=ar(ADE)=ar(AEC)                                             Hence proved.

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

ar (ADE) = ar (BCF).

Ans

Given: ABCD, DCFE and ABFE are parallelograms. To prove: ar( ADE ) = ar( BCF ) Proof: Since, ABCD is a parallelogram. So, AD = BC [ Opposite sides of parallelogram. ] Since, DCFE is a parallelogram. So, DE = CF[ Opposite sides of parallelogram. ] Since, ABFE is a parallelogram. So, AE = BF [ Opposite sides of parallelogram. ] InΔADE andΔBCF AD=BC[ Opposite sides of parallelogram. ] DE=CF[ Opposite sides of parallelogram. ] AE=BF[ Opposite sides of parallelogram. ] ΔADEΔBCF[ By S.S.S. ] ar( ΔADE )=ar( ΔBCF )[ Area of congruent triangles is equal. ] Hence proved. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabEeacaqGPbGaaeODaiaabwgacaqGUbGaaiOoaiaabccacaqGbbGaaeOqaiaaboeacaqGebGaaiilaiaabccacaqGebGaae4qaiaabAeacaqGfbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGbbGaaeOqaiaabAeacaqGfbGaaeiiaiaabggacaqGYbGaaeyzaiaabccacaqGWbGaaeyyaiaabkhacaqGHbGaaeiBaiaabYgacaqGLbGaaeiBaiaab+gacaqGNbGaaeOCaiaabggacaqGTbGaae4Caiaac6caaeaacaqGubGaae4BaiaabccacaqGWbGaaeOCaiaab+gacaqG2bGaaeyzaiaacQdacaqGGaGaaeyyaiaabkhadaqadaqaaiaabgeacaqGebGaaeyraaGaayjkaiaawMcaaiaabccacqGH9aqpcaqGGaGaaeyyaiaabkhadaqadaqaaiaabkeacaqGdbGaaeOraaGaayjkaiaawMcaaaqaaiaabcfacaqGYbGaae4Baiaab+gacaqGMbGaaiOoaiaabccacaqGtbGaaeyAaiaab6gacaqGJbGaaeyzaiaacYcacaqGGaGaaeyqaiaabkeacaqGdbGaaeiraiaabccacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabchacaqGHbGaaeOCaiaabggacaqGSbGaaeiBaiaabwgacaqGSbGaae4BaiaabEgacaqGYbGaaeyyaiaab2gacaGGUaaabaGaae4uaiaab+gacaGGSaGaaeiiaiaabgeacaqGebGaaeiiaiabg2da9iaabccacaqGcbGaae4qaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaabccadaWadaqaaiaab+eacaqGWbGaaeiCaiaab+gacaqGZbGaaeyAaiaabshacaqGLbGaaeiiaiaabohacaqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabchacaqGHbGaaeOCaiaabggacaqGSbGaaeiBaiaabwgacaqGSbGaae4BaiaabEgacaqGYbGaaeyyaiaab2gacaGGUaaacaGLBbGaayzxaaaabaGaae4uaiaabMgacaqGUbGaae4yaiaabwgacaGGSaGaaeiiaiaabseacaqGdbGaaeOraiaabweacaqGGaGaaeyAaiaabohacaqGGaGaaeyyaiaabccacaqGWbGaaeyyaiaabkhacaqGHbGaaeiBaiaabYgacaqGLbGaaeiBaiaab+gacaqGNbGaaeOCaiaabggacaqGTbGaaiOlaaqaaiaabofacaqGVbGaaiilaiaabccacaqGebGaaeyraiaabccacqGH9aqpcaqGGaGaae4qaiaabAeacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMc8+aamWaaeaacaqGpbGaaeiCaiaabchacaqGVbGaae4CaiaabMgacaqG0bGaaeyzaiaabccacaqGZbGaaeyAaiaabsgacaqGLbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqGWbGaaeyyaiaabkhacaqGHbGaaeiBaiaabYgacaqGLbGaaeiBaiaab+gacaqGNbGaaeOCaiaabggacaqGTbGaaiOlaaGaay5waiaaw2faaaqaaiaabofacaqGPbGaaeOBaiaabogacaqGLbGaaiilaiaabccacaqGbbGaaeOqaiaabAeacaqGfbGaaeiiaiaabMgacaqGZbGaaeiiaiaabggacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabYgacaqGSbGaaeyzaiaabYgacaqGVbGaae4zaiaabkhacaqGHbGaaeyBaiaac6caaeaacaqGtbGaae4BaiaacYcacaqGGaGaaeyqaiaabweacaqGGaGaeyypa0JaaeiiaiaabkeacaqGgbGaaeiiaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVpaadmaabaGaae4taiaabchacaqGWbGaae4BaiaabohacaqGPbGaaeiDaiaabwgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabYgacaqGSbGaaeyzaiaabYgacaqGVbGaae4zaiaabkhacaqGHbGaaeyBaiaac6caaiaawUfacaGLDbaaaeaacaqGjbGaaeOBaiaaykW7caaMc8UaeyiLdqKaaeyqaiaabseacaqGfbGaaeiiaiaabggacaqGUbGaaeizaiaaykW7caaMc8UaeyiLdqKaaeOqaiaaboeacaqGgbaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caqGbbGaaeiraiabg2da9iaabkeacaqGdbGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVpaadmaabaGaam4taiaadchacaWGWbGaam4BaiaadohacaWGPbGaamiDaiaadwgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabYgacaqGSbGaaeyzaiaabYgacaqGVbGaae4zaiaabkhacaqGHbGaaeyBaiaab6caaiaawUfacaGLDbaaaeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaabseacaqGfbGaeyypa0Jaae4qaiaabAeacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8+aamWaaeaacaWGpbGaamiCaiaadchacaWGVbGaam4CaiaadMgacaWG0bGaamyzaiaabccacaqGZbGaaeyAaiaabsgacaqGLbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqGWbGaaeyyaiaabkhacaqGHbGaaeiBaiaabYgacaqGLbGaaeiBaiaab+gacaqGNbGaaeOCaiaabggacaqGTbGaaeOlaaGaay5waiaaw2faaaqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaeyqaiaabweacqGH9aqpcaqGcbGaaeOraiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7daWadaqaaiaad+eacaWGWbGaamiCaiaad+gacaWGZbGaamyAaiaadshacaWGLbGaaeiiaiaabohacaqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabchacaqGHbGaaeOCaiaabggacaqGSbGaaeiBaiaabwgacaqGSbGaae4BaiaabEgacaqGYbGaaeyyaiaab2gacaqGUaaacaGLBbGaayzxaaaabaGaaGPaVlaaykW7caaMc8UaeyiLdqKaaeyqaiaabseacaqGfbGaeyyrIaKaeyiLdqKaaeOqaiaaboeacaqGgbGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8+aamWaaeaacaWGcbGaamyEaiaabccacaqGtbGaaeOlaiaabofacaqGUaGaae4uaiaab6caaiaawUfacaGLDbaaaeaacaWGHbGaamOCamaabmaabaGaeyiLdqKaaeyqaiaabseacaqGfbaacaGLOaGaayzkaaGaeyypa0Jaamyyaiaadkhadaqadaqaaiabgs5aejaabkeacaqGdbGaaeOraaGaayjkaiaawMcaaiaaykW7caaMc8UaaGPaVlaaykW7daWadaqaaiaadgeacaWGYbGaamyzaiaadggacaqGGaGaae4BaiaabAgacaqGGaGaae4yaiaab+gacaqGUbGaae4zaiaabkhacaqG1bGaaeyzaiaab6gacaqG0bGaaeiiaiaabshacaqGYbGaaeyAaiaabggacaqGUbGaae4zaiaabYgacaqGLbGaae4CaiaabccacaqGPbGaae4CaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiBaiaab6caaiaawUfacaGLDbaaaeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caWGibGaamyzaiaad6gacaWGJbGaamyzaiaabccacaqGWbGaaeOCaiaab+gacaqG2bGaaeyzaiaabsgacaqGUaaaaaa@B9E4@

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.

Proof : Since, ΔAPC and ΔBPC have same base and lie between            same parallels. So,               ar(ΔAPC)=ar(ΔBPC)           ...(i)In quadrilateral ACQD,               AD=CQ and AD=CQSo, ACQD is a parallelogram.     ΔDCQ and ΔACQ have same base QC and lie between        same parallels. So,        ar(ΔDCQ)=ar(ΔACQ)Substracting ar(ΔPCQ) from both sides, we getar(ΔDCQ)ar(ΔPCQ)=ar(ΔACQ)ar(ΔPCQ)      ar(ΔDPQ)=ar(ΔAPC)           ...(ii)From equation(i)​​ and equation (ii), we get           ar(ΔBPC)=ar(ΔDPQ) Hence proved.

Q.89

In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show thatiarBDE=14arABCiiarBDE=12arBAEiiiarABC=2arBECivarBFE=arAFDvarBFE=2arFEDviarFED=18arAFC

Ans

Given:ΔABC and ΔBDE are two equilateral triangles. D is             mid-point of BC. AE intersects BC at F.To prove:(i)ar(BDE)=14ar(ABC)            (ii)ar(BDE)=12ar(BAE)            (iii)ar(ABC)=2ar(BEC)            (iv)ar(BFE)=ar(AFD)            (v)ar(BFE)=2ar(FED)            (vi)ar(FED)=18ar(AFC)Construction:Join EC and AD. Let mid-points of AB and AC                   as G and H respectively. Join G and H, G and D,                   H and D respectively.           Proof:  In ΔABC, G and H are the mid-points of AB and AC respectively. So, by mid-point theorem GHBC and GH=12BC                      GHBD and GH=BD                  GHDB is a parallelogram because one pair of                     opposite sides is equal and parallel.Similarly,​ GHDC is a parallelogram because one pair of                    opposite sides is equal and parallel.                 In ΔCAB, D and H are the mid-points of CB                  and CA respectively. So, by mid-point theorem DHAB and DH=12AB                  DHAB and DH=AG                  DHAG is a parallelogram because one pair of                     opposite sides is equal and parallel.Now,     GBDE       [alternate angles are equal i.e., each 60°]and            GB=DE       [∵DE=BD=12BC=12AB=GB]              So,BEDG is a parallelogram because one pair               of opposite sides is equal and parallel.Since, diagonal divides a parallelogram in two triangles of equalarea.Then, ar(DGH)=ar(DGB)     [For gmBDGH]             ar(DGH)=ar(AGH)      [For gmAHDG]             ar(DGH)=ar(HDC)     [For gmGHCD]             ar(DGB)=ar(HDC)     [For gmBEDG]and     ar(DGB)=ar(BDE)    [For gmBEDG]    ar(ABC)=ar(DGH)+ar(DGB)+ar(AGH)+ar(HDC)                   =ar(DGB)+ar(DGB)+ar(DGB)+ar(DGB)                   =4 ar(DGB)                   =4ar(BDE)ar(BDE)=14ar(ABC)(ii) Since, ΔBDE and parallelogram BEDG lie on the same base and between same parallels, so ar(ΔBDE)=12ar(BEDG)   ...(i)∵    BEDG and DGHABEAH and ABHEABEH is a parallelogram.Then, ar(BEDG)=12ar(ABEH)   ...(ii)Since, ΔABE and parallelogram ABEH lie on the same base and between same parallels, so       ar(ΔABE)=12ar(ABEH)  ...(iii)From equation(i),(ii) and (iii), we have ar(ΔBDE)=12 ar(ΔABE)

(iii) ar(ΔABE)=ar(ΔBEC)     [Common base BE and BE||AC]ar(ΔABF)+ar(ΔBFE)=ar(ΔBEC)    ...(i)Since, ABED    ar(ΔBED)=ar(ΔAED)     [Common base BE and BE||AC]ar(ΔBFE)+ar(ΔEFD)=ar(ΔAFD)+ar(ΔEFD)              ar(ΔBFE)=ar(ΔAFD)    ...(ii)From equation (i) and (ii), we have   ar(ΔABF)+ar(ΔAFD)=ar(ΔBEC)                ar(ΔABD)=ar(ΔBEC)               12ar(ΔABC)=ar(ΔBEC)     [AD is median, so ar(ΔABD)=12ar(ΔABC)]                   ar(ABC)=2ar(BEC)(iv) Since, ABED    ar(ΔBED)=ar(ΔAED)      [Common base BE and BE||AC]ar(ΔBFE)+ar(ΔEFD)=ar(ΔAFD)+ar(ΔEFD)              ar(ΔBFE)=ar(ΔAFD)(v)Let height of ΔBDE be h and height of ΔABC be H, then         ar(ΔBDE)=12×BD×h         ar(ΔABC)=12×BC×HSince,ar(ΔBDE)=14ar(ΔABC)          12×BD×h=14(12×BC×H)          12×BD×h=14(12×2BD×H)    [∵BC=2BD]                    h=12HNow,    ∵  ar(BEF)=ar(AFD)         [From  (iv)]                      =12×FD×H                      =12×FD×2h                      =2(12×FD×h)           ar(BEF)   =2ar(ΔFED)(vi) Since, ar(AFC)=ar(AFD)+ar(ADC)                          =ar(BFE)+12ar(ABC)      [from  (iv) and D is mid-point of BC.]                          =ar(BFE)+12×4ar(BDE)  [from (i)]                          =ar(BFE)+2ar(BDE)                          =ar(BFE)+2{ar(BEF)+ar(FED)}                            =2ar(FED)+2[2ar(FED)+ar(FED)]                                                                  [from​ (v)]                          =2ar(FED)+6ar(FED)                ar(AFC)=8ar(FED)             ar(FED)=18ar(AFC)

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

Given:ABCD is a quadrilateral in which diagonals AC and BD            intersect at P.To prove: ar(APB)×ar(CPD)=ar(APD)×ar(BPC)Construction: Draw AMBD and CNBD.Proof:L.H.S.=ar(APB)×ar(CPD)                  =(12×PB×AM)×(12×PD×CN)                  =(12×PD×AM)×(12×PB×CN)                  =ar(APD)×ar(BPC)                  =R.H.S.       Hence proved.

Q.91

P and Q are respectively the midpoints of sides AB and BCof a triangle ABC and R is the midpoint of AP ,show thati arPRQ=12arARCii arRQC=38arABCiii arPBQ=arARC

Ans

Given:In ΔABC, P and Q are the mid-points of AB and BC            respectively. R is the mid-point of AP.To prove: i arPRQ=12arARC            ii arRQC=38arABC           (iii)ar(PBQ)=ar(ARC)Construction: Take S as a mid-point of AC and join PS. Draw                   CTAP and produce PQ upto T as PQ=QT.Join PC.Proof: In ΔABC, P and Q are the mid-points of AB and BC points of AB and BC respectively. Hence, by using mid-point theorem,we get            PQAC and PQ=12AC           PQAC and PQ=AS      PQSA is a parallelogram. PQSA is a parallelogram. We know that diagonals of a parallelogram bisect it into equal areas of triangles.      ar(ΔPAS)=ar(ΔPAQ)=ar(ΔSQP)=ar(ΔSQA)Now,     PQSC and PQ=SC      PQCS is a parallelogram. PQCS is a parallelogram. We know that diagonals of a parallelogram bisect it into equal areas of triangles.     ar(ΔPQS)=ar(ΔQSC)Similarly, QSCT and PSQB are also parallelogram.       ar(ΔQSC)=ar(ΔCTQ)and  ar(ΔPSQ)=ar(ΔQBP)Thus, ar(ΔPAS)=ar(ΔSQP)=ar(ΔSQA)=ar(ΔPAQ)=ar(ΔQSC)                      =ar(ΔCTQ)=ar(ΔCTQ)       ...(i)∵ar(ΔABC)=ar(ΔPBQ)+ar(ΔPAS)+ar(ΔPQS)+ar(ΔQSC)                 =ar(ΔPBQ)+ar(ΔPBQ)+ar(ΔPBQ)+ar(ΔPBQ)                 =4ar(ΔPBQ) ar(ΔPBQ)=14ar(ΔABC)       ...(ii)(i) In ΔPAQ,​ QR is median. So, ar(ΔPQR)=12ar(ΔPAQ)                 =12ar(ΔPBQ)                [∵QP is a median in ΔQBA]                 =12×14ar(ΔABC)         [From equation(ii)]                 =18ar(ΔABC)      ...(iii)(ii) In​ ΔABC, P and Q are the mid-points of AB and BC res So, by mid-point theorem PQ=12AC and PQAC 2PQ=AC and PQAC PT=AC and PTAC         [Q is mid-point of PT.]PACT is a parallelogram.ar(PACT)=ar(PACQ)+ar(ΔQTC)             =ar(PACQ)+ar(ΔPBQ)             [From equation(i)]ar(PACT)=ar(ΔABC)        ...(iv)

ar(ΔARC)=12ar(ΔPAC)        [∵CR is a median.]              =12×12ar(ΔABC)    [∵CP is a median.]              =14ar(ΔABC)              =2×18ar(ΔABC)ar(ΔARC)=2 ar(ΔPQR)   ...(v)     [From equation(iii)]Now,​ar(ΔRQC)=ar(PACQ)ar(ΔARC)ar(ΔPQR)              =ar(PACQ)2 ar(ΔPQR)ar(ΔPQR)                                         [From equation(v)]              =ar(PACT)ar(ΔQTC)3 ar(ΔPQR)              =ar(ΔABC)ar(ΔQTC)3 ar(ΔPQR)                                       [From equation(iv)]              =ar(ΔABC)ar(ΔPBQ)3 ar(ΔPQR)                                       [From equation(i)]             =ar(ΔABC)14ar(ΔABC)38ar(ΔABC)                           [From equation(ii) and equation(iii)]ar(ΔRQC)=38ar(ΔABC)(iii) R.H.S.=ar(ΔARC)            = ar(PACQ)ar(ΔPRQ)ar(ΔQRC)

           ={ar(PACT)ar(ΔQTC)}ar(ΔPRQ)ar(ΔQRC)           ={ar(ΔABC)ar(ΔPBQ)}ar(ΔPRQ)ar(ΔQRC)                               [From equation(i)and (iv)]           ={ar(ΔABC)14ar(ΔABC)}ar(ΔPRQ)ar(ΔQRC)           =34ar(ΔABC)18ar(ΔABC)38ar(ΔABC)                             [From equation(ii) and equation(iii)]           =14ar(ΔABC)           =ar(ΔPBQ)    [From equation(ii)]           =L.H.S.Thus,ar(ΔPBQ)=ar(ΔARC)

Q.92

In the figure below, ABC is a right triangle right angled at A. BCED,ACFG and ABMN are squares on the sides BC, CA and ABrespectively. Line segment AXDE meets BC at Y.Show that:(i)     ΔMBCΔABD(ii)   ar(BYXD)=2ar(MBC)(iii)  ar(BYXD)=ar(ABMN)(iv)ΔFCBΔACE(v)ar(CYXE)=2ar(FCB)(vi)ar(CYXE)=ar(ACFG)(vii)ar(BCED)=ar(ABMN)+ar(ACFG)

Ans

(i)In ΔMBC and ΔABD       MB=AB             [Sides of square ABMN]   MBC=ABD         [MBC=90°+ABC=ABD]       BC=BD             [Sides of square BCED]  ΔMBCΔABD        [By S.A.S.](ii)ΔMBC and parallelogram BYXD lie on the same base BD and lie between same parallels BD and AX. So,       ar(ΔMBC)=12ar(BYXD)2 ar(ΔMBC)=ar(BYXD)or      ar(BYXD)=2 ar(ΔMBC)(iii)ΔMBC and parallelogram ABMN lie on the same base MB and lie between same parallels MB and NC. So,        ar(ΔMBC)=12ar(ABMN)But  ar(ΔMBC)=12ar(BYXD)      [From(ii)]  12ar(ABMN)=12ar(BYXD) ar(ABMN)=ar(BYXD)(iv)In ΔFCB and ΔACE      FC=AC             [Sides of square ACFG]   FCB=ACE           [FCB=90°+ACB=ACE]      BC=CE                  [Sides of square BCED]  ΔFCBΔACE        [By S.A.S.](v) Since,ΔFCBΔACE So, ar(ΔFCB)=ar(ΔACE)    ...(i) ΔACE and parallelogram CYXE lie on the same base CE and lie between same parallels CE and AX. So,      ar(ΔACE)=12ar(CYXE)or   ar(CYXE)=2 ar(ΔACE)or   ar(CYXE)=2 ar(ΔFCB)     [From equation(i)](vi) ΔFCB and parallelogram ACFG lie on the same base CF and lie between same parallels CF and BG. So,      ar(ΔFCB)=12ar(ACFG)or   ar(ACFG)=2 ar(ΔFCB)∵     ar(CYXE)=2 ar(ΔFCB)So,​ ar(CYXE)=ar(ACFG)(vii)ar(BCED)=ar(BYXD)+ar(CYXE)                  = ar(ABMN)+ ar(ACFG)    [From (iii) and (vi)]                               Hence proved.

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.

Q.94 Write True or False: Give reasons for your answers.
(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

Given:Two congruent circles with centre O and O’ i.e., OA=OP, OB=O’Q and AOB=POQ.To prove: chord AB=chord PQProof: In ΔAOB and ΔPOQOA=OP[Radii of congruent circles]AOB=POQ[Given]OB=OQ[Radii of congruent circles]ΔAOBΔPOQ[By S.A.S.]Therefore,AB=PQ[By C.P.C.T.]Thus,chord AB=chord PQ  Hence proved.

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

Given: Two circles with centre O and O intersect at A and B.                AB is common chord of both circles.To prove: Centres O and O lie on the perpendicular bisector of                       AB i.e., OO is perpendicular bisector of AB.Proof: In ΔOAO and ΔOBO             OA=OB               [radii of same circle.]           OO=OO            [Common]          OA=OB              [radii of same circle.]         ΔOAOΔOBO       [By S.S.S.]         AOO=BOO      [By C.P.C.T.]         AOO=BOO      [By C.P.C.T.]       In ΔOAP and ΔOBP            OA=OB               [radii of same circle.]    AOP=BOP       [Proved above]             OP=OP               [Common]      ΔOAPΔOBP         [By S.A.S.] So,AP=BP[ By C.P.C.T. ] OPA=OPB[ By C.P.C.T. ] ButOPA+OPB=180° OPA=OPB=90° OP is perpendicular bisector of AB. OO is perpendicular bisector of AB. O and O’ lies on perpendicular bisector of AB. Hence proved. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@3EDC@

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:

  Given:Radius OA=5 cm, O’A=3 cm and OO’=4 cm.To find: AB Let OC=x cm and O’C=(4x) cm In ΔACO, ACO=90°        By Pythagoras theorem, OA2=AC2+OC2                52=AC2+x2      AC2=25x2       ...(i) In ΔACO, ACO=90°          By Pythagoras theorem, O’A2=AC2+OC2                 32=AC2+(4x)2        AC2=9(4x)2        ...(ii)From equation(i) and equation(ii), we have       25x2=9(4x)2    25x2=9(168x+x2)    25x2=916+8xx2              25=7+8x       25+7=8x           x=328                 =4 cmSubstituting value of x in equation(i), we get              AC2=2542                    =2516                    =9           AC=9=3Since, OO’ is perpendicular bisector of AB.So,             AB=2×AC                      =2×3=6 cmThus, the length of common chord is 6 cm.

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

Given:Chord AB= Chord CD, both chords intersect at R.To prove:AR=RC and RB=RDConstruction: Draw OPAB and OQCD. Join OR.Proof:In ΔOPR and ΔOQR,            OP=OQ       [Equal chords are equidistant.]         OPR=OQR   [Each 90°]               OR=OR        [Common]      ΔOPRΔOQR   [By R.H.S.]So,       PR=QR   ...(i)   [By C.P.C.T.]Since,      PB=QD   ...(ii)   [Perpendicular from centre bisects the chord.]Subtracting equation (i) from equation(ii),​ we get         PBPR=QDQR     RB=RD     ...(iii)∵      AB=CD    ...(iv)[Given]Equation(iv)−equation(iii), we get       AB RB=CDRD          AR=RCand RB=RD.      Hence Proved.

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

Given:Chord AB= Chord CD, both chords intersect at R.To prove:ORA=ORCConstruction: Draw OPAB and OQCD. Join OR.Proof:In ΔOPR and ΔOQR,OP=OQ[Equal chords are equidistant.]   OPR=OQR [Each 90°]          OR=OR[Common]ΔOPRΔOQR [By R.H.S.]So,     ORA=ORC [By C.P.C.T.]Hence Proved.

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

Given: Two concentric circles with centre O. A common chord intersect these circles at AD and BC. To prove: AB=CD Contruction: Draw OPAD MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B685@

Proof:Since, perpendicular from centre to chord of circle bisects the chord. So,AP=PD(i)   and   BP=PC(ii)Subtracting equation(i) from equation(ii), we getBPAP=PCPDAB=CD Hence proved.

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

Given:Radius of circular park is 5 m, distance between Reshma and Salma(AB) and distance between Salma and Mandeep (BC) is equal to 6 m eachTo Find: Distance between Reshma and Mandeep i.e., AC.Proof:In ΔODC, ODC=90°        So, by Pythagoras Theorem,            OC2=OD2 + DC2            DC2=OC2 OD2            DC2=52x2     ...(i)           In ΔBDC, BDC=90°       So, by Pythagoras Theorem,            BC2=BD2 + DC2            DC2=BC2 BD2            DC2=62(5x)2   ...(ii)From equation(i) and equation(ii), we get              52x2=62(5x)2        25x2=3625+10xx2        2511=10x             x=1410=1.4Substituting value of x in equation(i), we get            DC2=52(1.4)2                      =251.96                  =23.04              DC=23.04                  =4.8mThus, AC=2DC              =2×4.8m              =9.6mTherefore, the distance between Reshma and Mandeep is 9.6 m.

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

Let A, B and C be the positions of Ankur, Syed and David respectively.Radius of circle is 20 m.AD is median of triangle. O is centre of circle as well as intersection point of medians too. Let each side of ΔABCbe 2x.Since, cetroid divides a median in 2:1, thenAOOD=2120OD=21OD=202=10mTherefore, AD=20+10=30ar(ΔABC)=12×BC×AD =12×2x×30=30x m2In ΔBOD, ODB=90°So, by Pythagoras theorem,OB2=OD2+BD2(20)2=(10)2+x2x2=400100 =300x=300=103So,BC=2x =2×103=203Thus, the length of the string of each phone is 203 m.

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

Given:In circle, with centre O, BOC=30° and AOB=60°.To find:ADCSol:∵AOC=60°+30° =90°Since, the angle subtended by an arc at the centre is double theangle subtended by it at any point on the remaining part of the circle.AOC=2ADC90°=2ADCADC=90°2 =45°Thus, the required angle is 45°.

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

Given:Let O is the centre of circle and chord AB is equal to the radius of circle.To find: ACB and ADB.Proof: In ΔAOB, OA=AB=OB. So, AOB=60°       [Angle of equilateral triangle.]             ACB=12AOB  [By theorem]                       =12×60°                       =30°Since, ​ADBC is a cyclic quadrilateral. So, ACB+ADB=180°     30°+ADB=180°          ADB=180°30°                      =150°

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

Ans

Given:In circle with centre O, PQR=100°.To find:OPRSolution: Since, The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. ReflexPOR=2 PQR                    =2×100°                    =200°Now,  POR+ReflexPOR=360°                   POR+200°=360°                         POR=360°200°                                      =160°In ΔOPR,        OP=OR                 ORP=OPR   [Angles opposite to equal sides are equal.]ORP+OPR+POR=180°   [By angle sum property.]   OPR+OPR+160°=180°                               2OPR=180°160°                               2OPR=20°                    OPR=10°Thus, the OPR is 10°.

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

Ans

Given: In circle, ABC = 69°,ACB=31°.To find:BDCSolution: In ΔABC, BAC+ABC+ACB=180°                        BAC+69°+31°=180°                                          BAC=180°100°                                               =80°Since, angles in same segment of a circle are equal.                                       BDC=BAC                                         =80°

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

Given:In a circle, chords AC and BD intersect at E. BEC=130°           and ECD=20°.To find: BAC.Solution:   CEB+CED=180°         [Linear pair of angles.]                130°+CED=180°                        CED=180°130°                               =50°In ΔDEC,   ECD+CED+CDE=180°               20°+50°+CDE=180°                       CDE=180°70°                            =130°                          BDC=130°Since, angles in same segment of a circle are equal.                         BAC=BDC                             =130°Thus, the required angle BAC is 130°.

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

Given:ABCD is a cyclic quadrilateral. DBC=70°,BAC=30°.To find:BCD,ECD(if AB=BC)Since, angles in the same segments of the circle are equal. So, DAC=DBC=70°DAB=DAC+BAC =70°+30° =100°Since, sum of opposite angles in a cyclic quadrilateral is 180°.Then, DAB+DCB=180°100°+DCB=180°DCB=180°100°or BCD=80°If AB=BC

Then,                BAC=BCA    [Opposite angles of equal sides areequal.]                             30°=BCASince,​                  BCD=80°Then,  BCA+ACD=80°                30°+ECD=80°               ECD=80°30°                           =50°

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

Given:ABCD is a cyclic quadrilateral.AC and BD are the diameters of the circle.To prove: ABCD is a rectangle.Proof:Since, angle is semicircle is right angle. So, ABC=90°[AC is diameter.] and ADC=90°[AC is diameter.] Now, BAD=90°[BD is diameter.] and BCD=90°[BD is diameter.]Since, each angles of ABCD is 90°, so ABCD is a rectangle.

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

Ans

Given:Let ABCD be trapezium in which AD=BC. To prove: ABCD is a cyclic quadrilateral.Construction: Draw ECAD and produce AB upto E.           Proof: Since, AEDC and ADEC                   So, AECD is a prallelogram. Then, AD=EC but AD=BC                       EC=BC                     And CEA=CDA      [Opposite angles of a parallelogram are equal.]In​ ΔCEB,              CBE=CEB       [Opposite angles of equal sidesare equal.]            CBE+CBA=180°         [Linear pair of angles]            CEB+CBA=180°         [∵CBE=CEB]      CEA+CBA=180°      CDA+CBA=180°        [∵CEA=CDA]∵CDA+CBA+ABC+ADC=360°  [Angle sum property in quadrilateral.]                180°+ABC+ADC=360°                          ABC+ADC=360°180°                             ABC+ADC=180°Since, sum of opposite angles is 180°, so ABCD is a cyclic quadrilateral.

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

Given:Intersection points of two circles are B and C. Two line segments ABD and PBQ are drawn to intersect the circles at A, D and P,Q respectivelyTo prove:ACP=QCDProof:Since, angles in the same segments of a circle are equal. ACP=ABP(i)  and QCD=QBD(ii)But  ABP=QBD(iii)[Vertical opposite angles]ACP=QCD[From equation(i) and equation(ii) and equation (iii)]    Hence proved.

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

Given:ABC is a triangle. Two circles are drawn with          diameter AB and AC respectively.To prove: Point of intersection(D) of circles lie on BC.

Proof:  ADC=90°       [Angle in semi circle is 90°.] and ADB=90°       [Angle in semi circle is 90°.]           ADC+ADC=90°+90°                             BDC=180°BDC is a straight line.D lies on BC i.e., third side of triangle.Thus, point of intersection of both circles lie on third side of triangle.

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

Ans

Given:ΔABC and ΔDAC have equal hypotenous AC.To prove:CAD=CBDProof:In quadrilateral ABCD, B+D=180° So, quadrilateral ABCD is cyclic.Then, CAD=CBD[Angles in same segment of circle are equal.] Hence Proved.

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

Ans

Given:ABCD is a cyclic parallelogram.To prove:ABCD is a rectangle.Proof:A+C=180°[ABCD is a cyclic quadrilateral.]A=C[Opposite​ angles of a parallelogram.]Then,A=C=90°Since, a parallelogram will be a rectangle if any angle is right angle. Thus, ABCD is a rectangle.

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

Ans

Given:Two circles with centres P and Q are intersecting at A and B.To prove:PAQ=PBQProof:In ΔAPQ and ΔBPCAP=BP[Radii of same circle.]AQ=BQ[Radii of same circle.]PQ=PQ[Common]ΔAPCΔBPC[By S.S.S.]PAQ=PBQ[By C.P.C.T.]Thus, centres P and Q subtend equal angle at A and B.

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

Let OP=x m and OQ=( 6x )m. In right Δ AOP, by Pythagoras Theorem, OA 2 = AP 2 + OP 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@82E8@

OA2=(AB2)2+(x)2     [Since, perpendicular from centrebisect the chord.] OA2=(52)2+x2 OA2=254+x2      ...(i)In ΔCQO, Q=90°So, by Pythagoras theorem, we have OC2= CQ2+ OQ2       =(CD2)2+(6x)2       =(112)2+(6x)2 OC2=1214+3612x+x2...(ii)Since,  OA2=OC2     [Radius of same circle.]    254+x2=1214+3612x+x2              254=1214+3612x            12x=1214+36254               =964+36               =24+36            x=6012=5Substituting value of x in equation(i), we get OA2=254+52            =254+25        OA=1254            =552            =5×2.242            =5.6m  (approx)Thus, the radius of circle is 5.6 m (approx).

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.

In right Δ AOP, by Pythagoras Theorem, OA2= AP2+ OP2 OA2=(AB2)2+(4)2[Since, perpendicular from centrebisect the chord.] OA2=(62)2+42 OA2=9+16OA=25OA=5 cm.In ΔCQO, Q=90°So, by Pythagoras theorem, we have OC2= CQ2+ OQ2 (5)2=(CD2)2+x2[∵OC=OA=5cm] 25=(82)2+x2 x2=2516x=9=3Thus, the distance of chord CD from centre O is 3 cm.

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

Given: AD=CE, ABC is in exterior of circle with centre O.To prove:ABC=12(DOEAOC)Construction: Join AC, OA, OD, OC and OE.Proof:  In ΔAOD and ΔCOE,             OA=OC        [Radius of the same circle]             OD=OE        [Radius of the same circle]             AD=CE        [Given]           ΔAODΔCOE   [By S.S.S.]    OAD=OCE  [By C.P.C.T.]andODA=OEC  [By C.P.C.T.]ButOAD=ODA   [Opposite angles of equal sides of atriangle are equal.]Then,OAD=ODA=OCE=OEC=x  (let)In ΔOAC, OA=OC         OCA=OAC       [Angle opposite to equal sides are equalin a triangle.]                 =y(let)and     AOC=180°2y   [Angle sum property in a triangle]In ΔODE, OD=OE         OED=ODE      [Angle opposite to equal sides are equalin a triangle.]                =z (let)and     DOE=180°2z  [Angle sum property in a triangle]Now,       DOEAOC=(180°2z)(180°2y)                            =2(yz)12(DOEAOC)=yz    ...(i)                        BAC=180°(x+y)     [Linear pair of angles]and                 BCA=180°(x+y)     [Linear pair of angles]In ΔBAC,ABC+BAC+BCA=180°                              ABC=180°BACBCA                                      =180°{180°(x+y)}{180°(x+y)}                                 B=2(x+y)180°or              2(x+y)B=180°    ...(ii)In ΔBDE,BDE+BED+DBE=180°                              DBE=180°BDEBED                                    B=2(x+y)B(x+z)(x+z)                                                [From equation(ii)]                      2B=2(x+y)2(x+z)                        B=(yz)    ...(iii)From equation(i) and equation(iii), we have           ABC=12(DOEAOC)  Hence proved.

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

Given:Let ABCD be a rhombus and circle is drawn with diameter AB.To prove:Circle passes through point O.Proof: In circle with diameter AB and angle in semicircle is right angle.So, AOB=90°Since, diagonals of a rhombus bisect at 90°. So,AOB=90°Point O, lies on circumference. Thus, circle passes throughthe point of intersection of diagonals of rhombus.

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

Given:Let ABCD be a parallelogram and circle passing through vertex A, B and C intersects at E on side CD.To prove:AD=AEProof:Since, ABCD is a parallelogram. So, B=D     ...(i)       [Opposite angles of parallelogram areequal.]Since,ABCE is a cyclic quadrilateral, so      B+AEC=180° (ii)     [Sum of opposite angles in a cylicquadrilateral is 180°.]   AED+AEC=180°...(iii)       [Linear pair of angles]From equation (ii) and equation(iii), we have     B+AEC=AED+AEC                 B=AED   ...(iv)From equation(i) and equation(iv), we have            D=AED       ADE=AED   [∵D=ADE]So,              AE=AD       [Opposite sides of equal angles are equalin ΔADE.]                              Hence proved.

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

Given:In circle with centre O, AC and BD are two chords which bisect each other.To prove:(i) AC and BD are diameters, (ii)ABCD is a rectangle.Proof:In ΔAOB and ΔCOD AO=OC[Given]AOC=BOD[Vertical opposite angles] BO=OD[Given]ΔAOBΔCOD[By SAS]AB=DC[By C.P.C.T.]In ΔAOD and ΔBOC AO=OC[Given]AOD=BOC[Vertical opposite angles] DO=OB[Given]ΔAODΔBOC[By SAS]AD=BC[By C.P.C.T.]Since, opposite sides of quadrilateral ABCD are equal,so ABCD is a prallelogram.Then, A=C(i)[Opposite angles of a parallelogram are equal.]But ABCD is a cyclic quadrilateral.So,A+C=180°(ii)From equation(i) and (ii), we haveA=90°If any angle in a parallelogram is right angle, then it is a rectangle. Here, in parallelogram ABCD, A=90°, so ABCD is a rectangle  Hence proved.

Q.124

Bisectors of angles A, B and C of a triangle ABC intersect itscircumcircle at D ,E and F respectively.Prove that theangles of the triangle DEF are 90°12A, 90°12Band 90°12C.

Ans

Given:ABC is a triangle.AD, BE and CF are bisector of A, B and C respectively.To prove: D=90°12A, E=90°12B,F=90°12C

Proof: BEF=BCF        [Angles in same segment of circleare equal.]                  =C2            BED=BAD         [Angles in same segment of circleare equal.]                   =A2So,       E=BEF+BED                 =C2+A2                 =12(C+A)                 =12(180°B)                =90°12BSimilarly,            F=90°12C and D=90°12A.Thus, D=90°12A,  E=90°12B and F=90°12C.

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

Given: Two congruent circles intersect at A and B. PAQ intersects circles at P and Q respectively.To prove: BP=BQProof:Since, AB is common chord in two congruent triangles. So, the angles subtended by common arc at any point on the circumference of both circles will be equal. BPA=BQABPQ=BQPIn ΔBPQ,BPQ=BQPBQ=BP[Opposite sides of equal anglesare equal.]BP=BQ

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

Given: AD is bisector of A in ΔABC.To prove: AD and pependicular of BC intersect at point D on             circumcircle.Proof:In ΔBOM and ΔCOM            OB=OC           [radius of circle]           OMB=OMC    [Each 90°]            MB=CM           [Given]          ΔBOMΔCOM     [By S.A.S.]          BOM=COM     [By C.P.C.T.]           BOC=2BOM   ...(i)The angle subtended by an arc at the centre is double the anglesubtended by it at any point on the remaining part of the circle.                BOC=2BAC     ...(ii)Similarly , for arc BD,                 BOD=2BAD    ...(iii)Since, AD is bisector of A, then            BAD=12BAC                 =12×12BOC    [From equation(ii)]           BAD=12BOM         [From equation(i)]       12BOD=12BOM          [From equation(iii)]    BOD=BOMperpendicular bisector of BC and bisector of A intersect      circumcircle at the point D.

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

Ans

Given:Two congruent circles with centre O and O’ i.e., OA=O’P, OB=O’Q and AOB=POQ.To prove: chord AB=chord PQProof: In ΔAOB and ΔPOQ OA=OP  [Radii of congruent circles]AOB=POQ[Given]             OB=OQ        [Radii of congruent circles]      ΔAOBΔPOQ    [By S.A.S.]Therefore, AB=PQ[By C.P.C.T.]Thus,chord AB=chord PQ   Hence proved.

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

Number of families = 1500 (i) P(2 girls)=Number of families having two girlsTotal families               =4751500               =1960 (ii) P(1 girl)=Number of families having one girlTotal families               = 8141500               =407750 (iii) P(No girl)=Number of families having one girlTotal families                  = 2111500 The sum of probabilities                 =4751500 + 8141500+2111500                 =15001500                = 1

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

           Number of student in class IX= 40Number of students born in August=6P(A student born in August)=Number of students born in AugustNumber of student in class IX                                  =640                                  =320Thus, the probability of a student born in August is 320.

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

Outcome 3 heads 2 heads 1 head No head
Frequency 23 72 77 28

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

Ans

Total  number of tossing of two coins=200    Frequency of coming up two heads=72P(2 heads coming up)=Frequency of coming up two headsTotal  number of tossing of two coins                               =72200                               =925                               =0.36Thus, the probability of coming up two heads is 0.36.

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

Total number of families=2400(i)​ Number of families earning Rs 10000 13000 per month and owning exactly 2 vehicles=29                   P(exactly two vehicles)=292400(ii) Number of families earning Rs 16000 or more per month and owning exactly 1 vehicle=579                 P(exactly two vehicles)=5792400(iii) Number of families earning less than Rs 7000 per month and does not own any vehicle=10      P(does not own any vehicle)=102400                                      =1240(iv) Number of families earning Rs 1300016000 per month          and more than 2 vehicles=25                P(more than 2 vehicles)=252400                                       =196(v) Number of families owning not more than 1 vehicle                                      =10+1+2+1+160+305+535                                            +469+579                                      =2062P(owning not more than 1 vehicle)                                     =20622400                                     =10311200

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

Total number of students in a class=7+10+10+20+20+15+8                                              =90(i) Number of students obtained marks less than 20%=7                   P(a student obtained less than 20% marks)=790(ii) Number of students obtained marks 60 or above=15+8                                                              =23            P(a student obtained marks 60 or above)=2390

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

                  Total number of students=135+65                   =200(i)   Number of students who likes statistics=135                     P(who likes statistics)=135200                      =2740(ii)Number of students who dislike statistics=65            P(who likes statistics)=65200                 =1340

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

Total number of engineers=40(i)Number of engineers living in less than 7 km from her place of work=9    P(engineers living in less than 7 km from her place)   =940(iii)Number of engineers living in more than or equal to 7 km from her place of work=31     P(engineers living in more than or equal to 7 km from her place)   =3140(ii)Number of engineers living within12  km from her place of work=0    Pengineers living within  12  km from her place   =040   =0

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

Total number of vehicles=100 Number of two-wheelers=40                 P(Twowheeler)=Number of two-wheelersTotal number of vehicles                                   =40100                                   =25

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

The 3digits numbers which are divisible by 3 are:342, 504, 405, 519, 516, 612, 615,  528Total numbers divisible by 3 =8       P(Number divisible by 3)=Total numbers divisible by 3Total written Numbers                                  =830                                  =415Thus, the probability of a selected number to be divisible by 3 is 415.

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

Number of flour bags = 11Number of flour bags contains more than 5 kg of flour                          =7P(bag contains more than 5 kg flour)                          = bags containing more than 5 kg flourTotal number of flour                          =711Thus, the required probability of a selected bag having more than 5 kg flour is (7/11).

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

    Total number of days=30Number of days having the concentration of sulphur dioxide in the interval 0.120.16=2P(day having 0.120.16 concentration of SO2)                             =230                             =115

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

Total number of students=30Number of students having blood group AB                               =3              P(Blood group AB)=Number of students having blood group ABTotal number of students                               =330                               =110                               =0.1Thus, the required probability is 0.1.

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