NCERT Solutions for Class 9 Mathematics Chapter 7- Triangles

Mathematics is the study of numbers and their applications. It is mainly associated with accounting and various operations of Mathematics. It is broadly categorised into two sections; Algebra and Geometry. Hence, to master Mathematics, one should focus on both the sections in detail.

The main topics covered in the Chapter 7 of Class 9 Mathematics Triangles are congruent triangles, SSS, SAS, ASA and AAS criteria for congruence, why are SSA and AAA rules not valid, properties of isosceles triangles and inequalities in triangles.

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You will find all the properties of the congruence of triangles listed in a structured way in our NCERT solutions for Class 9 Mathematics Chapter 7. It has all the sections and subsections properly divided and highlighted. All the key points in each section are being written in the format required in CBSE examinations. Thus, it proves to be a great guide for students to step up  their preparation to get excellent results.

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Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 7

You have read about the basics of triangles in earlier classes. You are already familiar with   different types of triangles,  their properties and how to refer to a triangle in a figure. 

NCERT Class 9 Mathematics Chapter 7 briefly talks about their congruency and all the properties associated with it. You will know about the rules associated with congruence and its application in geometry. . You can look up NCERT Solutions for Class 9 Mathematics Chapter 7 on our Extramarks’ website where you can find the chapter being covered in detail which clears their concepts and helps them to think analytically to solve a problem.. 

We at Extramarks believe in going beyond the call of our duty to facilitate  students with great learning experience  ; be it note making, guiding the students, designing the mock tests or analysing the results. Hence,  parents, teachers and students have faith and trust in Extramarks NCERT Solutions for reliable and authentic notes which has further strengthened over the years.  

This chapter is mostly based on theorems, formulas  and properties of triangles . Detailed study of this chapter will help students to do the exercises and find solutions effectively and easily.

Introduction

This section helps you recall all the basics of triangles covered in the lower classes. Further, it gives you   detailed  insights of the topics you will study in this chapter.  The Extramarks subject matter experts have meticulously  covered each and every exercise and solution  of the chapter in our NCERT Solutions for Class 9 Mathematics Chapter 7. Students can practice with NCERT books and cross check their answers and if required they may even go ahead and solve other sample papers to strengthen their base.

Congruence of Triangles

The figures which are the same in shape and size or are equal in all aspects are known as  congruent. When two triangles are of the same shape  and of the same size  they are called  congruent triangles.

In this section, you will learn how to interpret congruence of triangles by analysing their sides and angles. All your doubts regarding congruence will be cleared with the help of the illustrations given in the chapter.

Criteria for Congruence of Triangles

There are certain  criterias one needs to note while looking for congruence in two triangles. They are as follows:

Side-Angle-Side or SAS congruence rule –

Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

Angle-Side-Angle or ASA congruence rule- 

Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.

Some properties of a Triangle

In this section, you will learn how to apply the criterion you have read about in the last section when two sides are given equal. You will also get to know about some of the theorems and their uses in various Geometrical concepts with the help of some examples given in the NCERT Solutions for Class 9 Mathematics Chapter 7.

Some More Criteria for Congruence of Triangles

You will read about some more criteria for congruence of triangles in this section. They are listed below:

Side-Side-Side or SSS congruence rule-

If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Right angle-Hypotenuse-Side or RHS congruence rule- 

If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Inequalities in a triangle

Whenever there is a relation between the unequal sides or angles of a triangle it is called as Inequalities of a triangle. There are various activities and examples included from this section in our NCERT Solutions for Class 9 Mathematics Chapter 7 r in such a way that they can get to the point answers without wasting much time on a single topic. It gives a better understanding of all the concepts associated and helps them to solve any difficult or advanced level questions with ease.

NCERT Solutions for Class 9 Mathematics Chapter 7: Exercise &  Solutions

 Testing is the cornerstone of any preparation.; since it shows what you have learnt and how much you can retain. You may gauge  your performance by solving the questions.  However,  the essential prerequisite is  that you must have the right solutions and answers for you to rectify your mistakes and shortcomings. 

 Therefore, NCERT Solutions for Class 9 Mathematics Chapter 7 have  been compiled in a systematic and organised  manner to revise and cross-check your answers. You can access  it from Extramarks’ website as per your convenience to  ace the examinations with flying colours.

The following links has exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 7:

  •  Chapter 7: Exercise 7.1 Question and answers    
  •  Chapter 7: Exercise 7.2 Question and answers
  • Chapter 7: Exercise 7.3 Question and answers
  • Chapter 7: Exercise 7.4 Question and answers
  • Chapter 7: Exercise 7.5Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 7, students can explore NCERT Solutions on our Extramarks’ website for all primary as well as  secondary classes.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5 
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT solutions Class 10
  • NCERT solutions Class 11
  • NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics 

NCERT Exemplar is the collection of NCERT related- questions. It helps in laying the foundation to all the basic as well as advanced level concepts in the manner required for different competitive examinations. It is especially designed by subject matter experts. As a result, teachers and mentors suggest students to include NCERT Exemplar books in their study material. Studying NCERT textbooks alone won’t help you to crack the tests. We therefore recommend students to start using it early to beat the competition and stay ahead of the pack.

The book gives insights of all the topics from the NCERT Class 9 Mathematics textbook. You can find questions ranging from basic to advanced level; thereby making students capable of solving questions of varying degrees to strengthen their base.

Students can be rest assured  that not a single topic has been left untouched in the chapter and hence are definitely going to nail the exams by    referring to NCERT Solutions and NCERT Exemplar. You can get NCERT Exemplar for Class 9 Mathematics easily from the Extramarks website.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 7

In order to score well, you have to learn well and to learn well, you have to read well. Hence, NCERT Solutions for Class 9 Mathematics Chapter 7 has complete academic notes to study from. . The following are the key features: : 

  • It has well-versed short notes to read and learn and revise quickly and easily.
  • You can also find a detailed overview of which  topics to revise the most in our NCERT Solutions.
  • Gradually,  it will boost your confidence and hence be able to leverage your performance.
  • After completing this chapter, you will have a complete idea of the congruency of the triangles and the rules needed to be applied where necessary. 
  • The notes are prepared by highly qualified and experienced faculty who meticulously follow the NCERT textbooks and CBSE guidelines to provide authentic and reliable study material.

 

Q.1

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.2

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.3 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:InΔBOCΔAOD     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.4

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 pq.To prove: ΔABCΔCDAProof: In ΔABCandΔCDABAC=DCA      [Alternate interior angles]     AC=CA          [Common]BCA=DAC      [Alternate interior angles]ΔABCΔCDA      [By​ ASA]

Q.5

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.6 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.7

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.8

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.9 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.10 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.11 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.12

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.13 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.14 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°BCD is right triangle.

Q.15 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.16 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.17

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.18 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.19

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.20 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:InΔABC, BEAC and CFAB and BE = CFTo​ prove: ΔABC is isosceles triangle.        Proof:(i) InΔBCE 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.21 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.22 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.23 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.24

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

Ans.

Given:InΔOAB, 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.]InΔOCD, 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.25 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:Inquadrilateral 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.26 In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Ans.

 Given:InΔPQR, 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.27 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.28 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.29 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.30 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.31 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=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@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.

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