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NCERT Solutions Class 10 maths Chapter 6

NCERT Solutions for Class 10 Mathematics Chapter 6 Triangles

Mathematics is one of the most important subjects for Class 10 students as it is scoring in nature and many of the concepts learnt here will be further developed in Class  11 and 12. Mathematics is a vast field as there are several mathematical rules and theorems to cover. A good study resource can really help students to prepare and score higher in this subject. The NCERT class 10 Mathematics Chapter 6 Solutions by Extramarks are created by subject matter experts to assist students in preparing for their examinations. These solutions will come handy not only to prepare for exams but also to complete assignments and for last-minute revision. .

Refer to NCERT Solutions for Class 10 Mathematics Chapter 6 Triangles

Extramarks provides detailed NCERT Solutions for Class 10 Mathematics. Since the only way to be good at Mathematics is to solve problems and practice regularly, these solutions can be very useful for students. Extramarks provides detailed solutions to the questions given in  NCERT Mathematics textbooks of class 10.  All chapters are listed below:

  • Chapter 1 – Real Numbers
  • Chapter 2 – Polynomials
  • Chapter 3 – Pair of Linear Equations in Two Variables
  • Chapter 4 – Quadratic Equations
  • Chapter 5 – Arithmetic Progressions
  • Chapter 6 – Triangles
  • Chapter 7 – Coordinate Geometry
  • Chapter 8 – Introduction to Trigonometry
  • Chapter 9 – Some Applications of Trigonometry
  • Chapter 10 – Circles
  • Chapter 11 – Constructions
  • Chapter 12 – Areas Related to Circles
  • Chapter 13 – Surface Areas and Volumes
  • Chapter 14 – Statistics
  • Chapter 15 – Probability

NCERT Solutions for Class 10 Mathematics Chapter 6 Triangles Details

Chapter 6 of the Class 10 Mathematics textbook covers the following main topics:

Chapter 6 Triangles: 6.1 Introduction

Students in Class 9 were introduced to the concept of triangles and investigated properties such as triangle congruence. The introduction to the chapter essentially acts as a window for students to view what they will learn new under the topic of Triangles.

Chapter 6 Triangles: 6.2 Similar Figures

Students are taught the foundation of resemblance in forms like squares or equilateral triangles with the same side lengths and circles with the same radius. Students will learn that identical figures might have the same shape but not necessarily the same size as they proceed through this lesson. Students are typically asked to prove similarities between figures using theorems in this topic. .

Chapter 6 Triangles: 6.3 Similarity of Triangles

After the students have a basic understanding of the notion of similarity, they get introduced to the criteria for determining if two or more triangles are similar using the Basic Proportionality Theorem.

Chapter 6 Triangles: 6.4 Criteria for Similarity of Triangles

The criterion for triangle similarity gets outlined and explained in this section. The triangles are  considered to be comparable when their respective angles are equal, and their corresponding sides have the same ratio. As theorems are  illustrated using relevant examples, students will be able to visualise them.

Chapter 6 Triangles: 6.5 Areas of Similar Triangles

This section explains the formula and demonstrates how to calculate the surface area of related triangles. Students can use the different theorems in Mathematics NCERT Class 10 Chapter 6 to calculate the area of similar triangles.

Chapter 6 Triangles: 6.6 Pythagoras Theorem

The Pythagoras theorem is applied to similar triangles in NCERT Solutions for Class 10 Mathematics chapter 6 Triangles. In Class 9, students learned the Pythagoras theorem and how to prove it. Students will learn how to apply this theorem using similarity of triangles in this section.

Chapter 6 Triangles: 6.7 Summary

The summary covers all of the concepts and going through the summary will help you recall everything you learned in the chapter. 

List of Exercises in CBSE Class 10 Mathematics Chapter 6

The following is a collection of exercises from Mathematics Class 10 Chapter 6:

  • Ex 6.1 – 3 Questions  (3 Short Answer Questions).
  • Ex 6.2 – 10 Questions  (9 Short Answer Questions, 1 Long Answer Question).
  • Ex 6.3 – 16 Questions  (1 main Question with 6 sub-Questions, 12 Short Answer Questions, 3 Long Answer Questions).
  • Ex 6.4 – 9 Questions  (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions).
  • Ex 6.5 -17 Questions  (15 Short Answer Questions, 2 Long Answer Questions).
  • Ex 6.6 – 10 Questions  (5 Short Answer Questions, 5 Long Answer Questions).Optional*- This exercise is not from the examination point of view.

Benefits of Chapter 6 Mathematics Class 10 NCERT Solutions

There are many advantages of using the  NCERT solutions of class 10 Mathematics chapter 6:

  • The NCERT solutions can be accessed online from Extramark’s official website and used offline as well.
  • The solutions are prepared by experienced faculty  and subject matter experts keeping in mind the latest CBSE updates regarding the examination pattern who ensure they are reliable, accurate  and error-free.
  • Extramarks leaves no stone unturned when it comes to providing the best learning material with unmatchable speed and accuracy for students irrespective of the class and subject.  We have all the answers to your queries. This encourages the students to master the topic and increases their confidence in achieving a high grade

Conclusion

NCERT Class 10 Mathematics Chapter 6 answers are a dependable, reliable, and authentic source of information for the students. They will definitely benefit, if they start using NCERT solutions on a regular basis and especially when they are stuck on a question, theorem, or example to cross-check their answer and clarify their doubts. 

Related Question/Answer

Question: State whether the following quadrilaterals are similar or not:

Answer: :

From the given two figures, we can see their corresponding angles are different or unequal. Therefore, they are not similar.

Q.1 Fill in the blanks using the correct word given in brackets:
(i) All circles are ­­­­­­­­_______. (congruent, similar)
(ii) All squares are______. (similar, congruent)
(iii) All _______triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are _______and
(b) their corresponding sides are_______. (equal, proportional)

Ans.

(i) All circles are­­­­­­ similar.
(ii) All squares are ­­­­­­similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.

Q.2 Give two different examples of pair of
(i) similar figures. (ii) non-similar figures.

Ans.

(i)
(a) Any two circles are similar.
(b) Any two squares are similar.

(ii)
(a) A trapezium and a parallelogram are not similar.
(b) An acute angle triangle and an obtuse angle triangle are not similar.

Q.3 State whether the following quadrilaterals are similar or not:

Ans.

Quadrilaterals PQRS and ABCD are not similar as their corresponding angles are not equal.

Q.4 In the following figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Ans.

(i)

It is given thatDEBC.By using Basic Proportionality Theorem, we get          ADDB=AEEC      1.53=1EC      EC=3×11.5=2EC=2cm

(ii)

It is given thatDE  ||  BC.By using Basic Proportionality Theorem, we get          ADDB=AEEC      AD7.2=1.85.4=13      AD=7.23=2.4AD=2.4cm

Q.5

E and F are points on the sides PQ and PR respectively of aΔPQR. For each of the followingcases, state whether EFQR:(i)PE=3.9cm,EQ=3cm,PF=3.6cm andFR=2.4cm(ii)PE=4cm,QE=4.5cm,PF=8cm andRF=9cm(iii)PQ=1.28cm,PR=2.56cm,PE=0.18cm andPF=0.36cm

Ans.

(i)
As per given information, we have the following triangle.

In the above triangle,PEEQ=3.93=1.3andPFFR=3.62.4=32=1.5Hence,PEEQPFFRi.e., the lineEFdoes not dividePQandPRinthe same ratio and soEFis not parallel toQR.

(ii)
As per given information, we have the following triangle.

In the above triangle,PEEQ=44.5=4045=89andPFFR=89Hence,PEEQ=PFFRi.e., the lineEFdividesPQandPRinthe same ratio and soEFis parallel toQR.

(iii)
As per given information, we have the following triangle.

(iv)In the above triangle,PEEQ=0.181.1=18110=955andPFFR=0.362.2=36220=955Hence,PEEQ=PFFRi.e., the lineEFdividesPQandPRinthe same ratio and soEFis parallel toQR.

Q.6

In the following figure, ifLMCBandLNCD,prove thatAMAB=ANAD.

Ans.

In the given figure,LMCBSo, using basic proportionality theorem inΔABC, we getAMAB=ALAC...(1)Also, in the given figure,LNCDSo, using basic proportionality theorem inΔACD, we getANAD=ALAC...(2)From (1) and (2), we getAMAB=ANAD.

Q.7

In the following figure,DEACandDFAE.Prove thatBFFE=BEEC.

Ans.

InΔABC,DEACSo, using basic proportionality theorem, we get          BDDA=BEEC...(1)InΔBAE,DFAESo, using basic proportionality theorem, we get          BDDA=BFFE...(2)From (1) and (2), we getBEEC=BFFE

Q.8

In the following figure,DEOQandDFOR. Show thatEFQR.

Ans.

InΔPOQ,DEOQSo, using basic proportionality theorem, we get          PEEQ=PDDO...(1)InΔPOR,DFORSo, using basic proportionality theorem, we get          PFFR=PDDO...(2)From (1) and (2), we getPEEQ=PFFRHence, by converse of basic proportionality theorem,EFQR.

Q.9 In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR.Show that BC ∥ QR.

Ans.

In ΔPOQ, AB ll PQ (Given)Therefore, by Basic Proportionality Theorem,APAO=OBBQ...(i)In ΔPOR, AC ll PR (Given)Therefore, by Basic Proportionality Theorem,APAO=OCCR...(ii)From (i) and (ii),OBBQ=OCCRHence, by using Converse of Basic Proportionality Theoremin ΔOQR, we find that BC ll QR.

Q.10 Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Ans.

We consider ΔABC drawn below in which D is the mid-point of side AB and DE is the line segment drawn parallel to BC.

By using Basic Proportionality Theorem, we getADDB = AEEC...(1)D is the mid-point of AB. Therefore, AD=DBFrom (1), we have       ADDB = AEECADAD = AEEC1 = AEECEC = AETherefore, E is the mid-point of AC.

Q.11 Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
(Recall that you have done it in Class IX)

Ans.

We consider ΔABC drawn below in which DE is the line segment joining the mid-points D and E of sides AB and AC respectively.

We haveAD=DBADDB=1andAE=ECAEEC=1Therefore,ADDB=AEECHence, by converse of Basic Proportionality Theorem,DEBC.

Q.12

ABCDis a trapezium in whichABDCand its diagonals intersect each other at the pointO.Show thatAOBO=CODO.

Ans.

Given:ABCD is a trapezium in which AB is parallel to CD and diagonals intersect each other at O.

To prove:AOBO=CODOConstruction:DrawOFABwhich meets AD in F.InΔABD,OFABSo, by converse of Basic  Proportionality Theorem,          DOOB=DFFA...(i)Now inΔADC,OFCD           [SinceABCD]              DFFA=COOA     ...(ii)Comparing equation (i) and (ii), we get                 DOOB=COOAOAOB=CODO

Q.13 The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO = CODO. Show that ABCD is a trapezium.

Ans.

Given:Diagonals of a quadrilateral ABCD intersect each other at point O such thatAOBO=CODO.

To prove:ABCDConstruction:DrawOFABwhich meets AD in F.InΔABD,OFABSo, by converse of Basic  Proportionality Theorem,          DOBO=DFFA...(i)Giventhat      AOBO=CODODOBO=COAODFFA=COAO[From (i)]OFDC [By converse of Basic Proportionality Theorem]Also, OFABTherefore,ABDCHence,ABCD is a trapezium.

Q.14 State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

Ans.

(i)Corresponding angles are equal in the given triangles. i.e.,A=P=60°,B=Q=80°andC=R=40°.Therefore, by AAA similarity criterion,ΔABC~ΔPQR.(ii)Ratios of the corresponding sides of the given pair oftriangles are equal.i.e.,ABQR=BCRP=CAPQ=12Therefore, by SSS similarity criterion,ΔABC~ΔQRP.(iii)The given pair of triangles are not similar as theircorresponding sides are not proportional.(iv)The given pair of triangles are not similar as theircorresponding sides are not proportional.(v)The given pair of triangles are not similar as theircorresponding sides are not proportional.(vi)InΔDEF,      D+E+F=180°70°+80°+F=180°F=180°150°=30°InΔPQR,      P+Q+R=180°P+80°+30°=180°P=180°110°=70°We find out that corresponding angles are equal inΔDEFandΔPQR.i.e.,D=P=70°,E=Q=80°andF=R=30°.Therefore, by AAA similarity criterion,ΔDEF~ΔPQR.

Q.15

Ans.

DOBis a straight line.BOC+DOC=180°DOC=180°BOC=180°125°=55°InΔDOC,      DOC+DCO+ODC=180°DCO=180°DOCODCDCO=180°55°70°=55°It is given thatΔODC~ΔOBA and so corresponding anglesin these triangles are equal.OAB=DCO=55°

Q.16

Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at the point O. Usinga similarity criterion for two triangles, show thatOAOC=OBOD.

Ans.

InΔDOC andΔBOA,CDO=ABO [Alternate interior angles as ABCD]DCO=BAO [Alternate interior angles as ABCD]DOC=BOA [Vertically opposite angles]ΔDOC~ΔBOA [AAA similarity critarion]DOBO=OCOA[Corresponding sides are proportional]OAOC=OBOD

Q.17

In the following figure,QRQS=QTPRand1=2.Show thatΔPQSΔTQR.

Ans.

InΔPQR,       1=2(Given)PQ=PR...(1)It is given thatQRQS=QTPRQRQS=QTQP[From (1), PR=QP]    ...(2)Q is common inΔPQS andΔTQR and the sides includingthis angle in both triangles are proportional as shown inequation (2).Therefore, by SAS critarion,ΔPQS~ΔTQR.

Q.18

S and T are points on sides PR and QR ofΔPQR such thatP=RTS. Show thatΔRPQ~ΔRTS.

Ans.

InΔRPQ andΔRTS,       P=T(Given)andR=RTherefore, by AA criterion,ΔRPQ~ΔRTS.

Q.19

In the following figure, ifΔABEΔACD, show thatΔADE~ΔABC.

Ans.

It is given thatΔABEΔACD.Therefore, by CPCT,         AB=ACand AD=AESo,ADAB=AEACInΔADE andΔABC,       A=AandADAB=AEACTherefore, by SAS criterion,ΔADE~ΔABC.

Q.20

In the following figure, altitudes AD and CE ofΔABC intersect each other at the point P. Show that:(i)ΔAEP~ΔCDP(ii)ΔABD~ΔCBE(iii)ΔAEP~ΔADB(iv)ΔPDC~ΔBEC

Ans.

(i)InΔAEP andΔCDP,       APE=CPD(Vertically opposite angles)andAEP=CDP=90°Therefore, by AA similarity critarion,ΔAEP~ΔCDP.(ii)InΔABD andΔCBE,          ADB=CEB=90°andABD=CBETherefore, by AA similarity critarion,ΔABD~ΔCBE.(iii)InΔAEP andΔADB,          AEP=ADB=90°andPAE=BADTherefore, by AA similarity critarion,ΔAEP~ΔADB.(iv)InΔPDC andΔBEC,          PDC=BEC=90°andDCP=ECBTherefore, by AA similarity critarion,ΔPDC~ΔBEC.

Q.21

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CDat F. Show thatΔABE~ΔCFB.

Ans.

In the above figure, ABCD is a parallelogram in which ABDC.Opposite angles in parallelogram are equal. Therefore,EAB=BCFAlso,ABE=CFB [Alternate interior angles]Therefore, by AA similarity critarion,ΔABE~ΔCFB.

Q.22

In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively.Prove that:(i)ΔABC~ΔAMP(ii)CAPA=BCMP

Ans.

(i)InΔABC andΔAMP,ABC=AMP=90°and,CAB=PAMTherefore, by AA similarity critarion,ΔABC~ΔAMP.(ii)Corresponding sides are proportional in similar triangles.Therefore,      ΔABC~ΔAMPCAPA=BCMP

Q.23

CD and GH are respectively the bisectors ofΔACB andΔEGF such that D and H lie on sides AB and FEofΔABC andΔEFG respectively. IfΔABC~ΔFEG, show that:(i)    CDGH=ACFG(ii)   ΔDCB~ΔHGE(iii)  ΔDCA~ΔHGF

Ans.

It is given thatΔABC~ΔFEG.A=F,B=E,BCA=EGFAlso,     12BCA=12EGFACD=FGH andDCB=HGEBy AA similarity critarion,      ΔDCA~ΔHGF andΔDCB~ΔHGECDGH=ACFG

Q.24

In the following figure, E is a point on side CB produced of an isosceles triangle ABCwith AB=AC. If ADBC and EFAC, prove thatΔABDΔECF.

Ans.

InΔABD andΔECF,          BDA=CFE=90°andABD=ECF [AB=ACB=C]Therefore, by AA similarity critarion,ΔABD~ΔECF.

Q.25

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR andmedian PM ofΔPQR (see the following figure). Show thatΔABC~ΔPQR.

Ans.

It is given that AD and PM are medians.Therefore,12BC=BD=DCand12QR=QM=MRIt is given that        ABPQ=BCQR=ADPM    ABPQ=12BC12QR=ADPM    ABPQ=BDQM=ADPMTherefore, by SSS similarity critarion,ΔABD~ΔPQM and soinΔABC andΔPQR,B=Q andABPQ=BCQR.Therefore, by SAS similarity critarion,ΔABC~ΔPQR.

Q.26

D is a point on the side BC of a triangle ABC such thatADC=BAC. Show that CA2=CB·CD.

Ans.

InΔBAC andΔADC,ADC=BAC (Given)ACD=BCA (Common angle)ΔADC~ΔBAC (By AA similarity critarion)We know that corresponding sides of similar trianglesare proportional.   CACD=CBCA  CA2=CBCD

Q.27

Sides AB and AC and median AD of a triangle ABCare respectively proportional to sides PQ and PR andmedian PM of another triangle PQR. Show that ΔABC ∼ ΔPQR

Ans.

It is given that,ABPQ=ACPR=ADPMWe extend AD and PM up to point E and L respectively,such that AD=DE and PM=ML. We join B to E, C to E,Q to L and R to L.

It is given that AD and PM are medians.Therefore,12BC=BD=DC    and    12QR=QM=MRIn quadrilateral ABEC, diagonals AE and BC bisect eachother at point D. Therefore, quadrilateral ABEC is aparallelogram.AC=BE and AB=ECSimilarly, we can prove that quadrilateral PQLR isa parallelogram and PR=QL, PQ=LR.It is given that        ABPQ=ACPR=ADPM    ABPQ=BEQL=2AD2PM    ABPQ=BEQL=AEPLTherefore, by SSS similarity critarion,ΔABE~ΔPQL and soinΔABE andΔPQL,BAE=QPL...(1)Similarly, we can show thatΔAEC~ΔPLR andCAE=RPL...(2)Adding (1) and (2), we getBAE+CAE=QPL+RPLCAB=RPQAlso, we haveABPQ=ACPRTherefore, by SAS similarity critarion,ΔABC~ΔPQR.

Q.28

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a towercasts a shadow 28 m long. Find the height of the tower.

Ans.

Let CD is a pole and DF is its shadow.Let AB is a tower and BE is its shadow.At the same time in a day, the sun rays will fall on poleand tower at the same angle.Therefore,DCF=BAEAlso,CDF=ABE=90°ΔABE~ΔCDF [By AA similarity critarion]Corresponding sides are proportional in similar triangles.Therefore,ABCD=BEDFAB6=284=7AB=42Therefore, the height of the tower is 42 m.

Q.29

If AD and PM are medians of triangles ABC and PQR, respectively whereΔABC~ΔPQR, prove thatABPQ=ADPM.

Ans.

It is given that AD and PM are medians.Therefore,12BC=BD=DC and12QR=QM=MRAlso,ΔABC~ΔPQRTherefore,        ABPQ=BCQR=ACPRandA=P,B=Q,  C=RInΔABD andΔPQM,ABPQ=12BC12QR=BDQMandB=QΔABD~ΔPQM [By SAS similarity critarion]ABPQ=ADPM

Q.30

LetΔABC~ΔDEF and their areas be, respectively, 64 cm2and 121 cm2. If EF=15.4 cm, find BC.

Ans.

It is given thatΔABC~ΔDEF.ar(ΔABC)ar(ΔDEF)=BC2EF2GiventhatEF=15.4  cmar(ΔABC)=64cm2ar(ΔDEF)=121cm2    ar(ΔABC)ar(ΔDEF)=BC2EF264121=BC2(15.4)2811=BC15.4BC=8×15.411BC=11.2

Q.31 Diagonals of a trapezium ABCD with AB ║ DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Ans.

InΔCOD andΔAOB,ODC=OBA [Alternate interior angles as ABCD]DCO=BAO [Alternate interior angles as ABCD]COD=AOB [Vertically opposite angles]ΔCOD~ΔAOB [AAA similarity critarion]ar(ΔCOD)ar(ΔAOB)=CD2AB2ar(ΔCOD)ar(ΔAOB)=CD2(2CD)2=14[Given AB=2CD]ar(ΔAOB):ar(ΔCOD)=4:1

Q.32 In the following figure, ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O, show that ar(ΔABC) (ΔDBC) = AO DO.

Ans.

We draw perpendiculars AP and DM on line BC.

We know that area of a triangle=12×Base×Heightar(ΔABC)ar(ΔDBC)=12BC×AP12BC×DM=APDMInΔAPO andΔDMO,APO=DMO=90°AOP=DOM [Vertically opposite angles]ΔAPO~ΔDMO [By AA similarity critarion]APDM=AODOar(ΔABC)ar(ΔDBC)=AODO

Q.33 If the areas of two similar triangles are equal, prove that they are congruent.

Ans.

LetΔABC~ΔPQR,then we have      ar(ΔABC)ar(ΔPQR)=(ABPQ)2=(BCQR)2=(ACPR)2ar(ΔABC)ar(ΔABC)=(ABPQ)2=(BCQR)2=(ACPR)2[Givenar(ΔABC)=ar(ΔPQR)]1=(ABPQ)2=(BCQR)2=(ACPR)21=ABPQ=BCQR=ACPRAB=PQ,BC=QRandAC=PRΔABCΔPQR[By SSS congruence criterion]

Q.34

D, E and F are respectively the mid-points of sides AB, BC and CA ofΔABC. Find the ratio of the areasofΔDEF andΔABC.

Ans.

D and E are mid-points ofΔABC.DEACand DE=12ACInΔBED andΔBCA,BED=BCA [Corresponding angles]BDE=BAC [Corresponding angles]EBD=CBA [Common angles]ΔBED~ΔBCA [AAA similarity criterion]ar(ΔBED)ar(ΔBCA)=(DEAC)2=(12ACAC)2=14ar(ΔBED)=14ar(ΔBCA)Similarly,ar(ΔCFE)=14ar(ΔCBA)andar(ΔADF)=14ar(ΔABC)Also,ar(ΔDEF)=ar(ΔABC)[ar(ΔBED)+ar(ΔCFE)+ar(ΔADF)]ar(ΔDEF)=ar(ΔABC)34ar(ΔABC)=14ar(ΔABC)ar(ΔDEF)ar(ΔABC)=14

Q.35 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Ans.

Letustaketwo triangles ABC and PQR such thatΔABC~ΔPQR.LetAD and PS be the medians ofΔABC andΔPQRrespectively.Now,ΔABC~ΔPQR    ABPQ=BCQR=ACPRandA=P,B=Q,C=R.SinceAD and PS are medians, so we haveBD=DC=BC2and QS=SR=QR2InΔABD andΔPQS,B=QandABPQ=BCQR=BDQSΔABD~ΔPQS[SAS similarity criterion]ABPQ=BDQS=ADPSNow,        ΔABC~ΔPQRar(ΔABC)ar(ΔPQR)=(ABPQ)2ar(ΔABC)ar(ΔPQR)=(ADPS)2

Q.36

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of theequilateral triangle described on one of its diagonals.

Ans.

LetABCD is a square and length of its one side isaunit.Therefore, length of its diagonal=2aEquilateral triangles as per question are formed here asΔABE andΔDBF.Length of a sideofequilateralΔABE=aandlength of a sideofequilateralΔDBF=2aWeknow that equilateral triangles are similar to each other.Therefore, ratio of their areas is equal to ratio of squares oftheir sides.i.e.,ar(ΔABE)ar(ΔDBF)=(a2a)2=12

Q.37

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of trianglesABC and BDE is(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

Ans.

It is given thatΔABC andΔBDE are equilateral and D is themid point of BC. Therefore, BD=BC2Weknow that equilateral triangles are similar to each otherand so ratio of their areas is equal to square of the ratio oftheir sides.ar(ΔABC)ar(ΔBDE)=(ABBD)2=(BCBD)2[ΔABC is equilateral]=(BCBC2)2=41ar(ΔABC):ar(ΔBDE)=4:1Hence, the correct answer is (C).

Q.38

Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio(A)2:3(B)4:9(C)81:16(D)16:81

Ans.

Itis given that sides of two similar triangles are in theratio 4:9.Therefore,Ratio of areas of these triangles=4292=1681Hence, the correct answer is (D).

Q.39 Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Ans.

(i) Given sides of the triangle are 7 cm, 24 cm and 25 cm.
Squares of the given sides of the triangle are 49 cm2, 576 cm2 and 625 cm2.
Now,
49 cm2 + 576 cm2 = 625 cm2
Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.
Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 25 cm.

(ii)

Given sides of the triangle are 3 cm,8 cm and 6 cm.Squares of the given sides of the triangle are 9 cm2,64 cm2and 36 cm2.Now,9+3664We find that sum of the squares of two sides is not equal tothe square of third side.Therefore, triangle of the given sides is not a right triangle.(iii)Given sides of the triangle are 50 cm,80 cm and 100 cm.Squares of the given sides of the triangle are 2500 cm2,6400 cm2and 10000 cm2.Now,2500+640010000We find that sum of the squares of two sides is not equal tothe square of third side.Therefore, triangle of the given sides is not a right triangle.

(iv) Given sides of the triangle are 13 cm, 12 cm and 5 cm.
Squares of the given sides of the triangle are 169 cm2, 144 cm2 and 25 cm2.
Now, 144 c m2 + 25 cm2 = 169 cm2
Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.
Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 13 cm.

Q.40 PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that ( PM )2 = QM.MR.

Ans.

LetMPR=θInΔMPR,MRP=180°90°θ=90°θSimilarly, inΔMPQ,MPQ=90°MPR=90°θMQP=180°90°(90°θ)=θNow, inΔMPR andΔMQP, we haveMQP=MPR,MPQ=MRP andPMQ=PMR.Therefore, by AAA similarity critarion,ΔMPR~ΔMQP.Therefore, we have     QMPM=MPMRPM2=QM×MR

Q.41 In the following figure, ABD is a triangle right angled at A and AC BD. Show thati AB2 = BC . BDii AC2 = BC . DCiii AD2 = BD . CD

Ans.

(i)InΔADB andΔCAB, we have            DAB=ACB=90°and    ABD=CBA [Common angle]Therefore, by AA similarity critarion,ΔADB~ΔCAB.Therefore, we have     ABCB=BDABAB2=BCBD(ii)LetCAB=θInΔCBA, we haveCBA=180°90°θ=90°θInΔCAD, we haveCAD=90°CAB=90°θCDA=180°90°(90°θ)=θInΔCBA andΔCAD, we haveCBA=CAD,CAB=CDA andACB=DCA=90°Therefore, by AAA similarity critarion,ΔCBA~ΔCAD.Therefore, we have     ACDC=BCACAC2=BCDC(iii)InΔDCA andΔDAB,DCA=DAB=90°,CDA=ADBTherefore, by AA similarity critarion,ΔDCA~ΔDAB.Therefore, we have     DCDA=DADBAD2=BDCD

Q.42 ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.

Ans. Given that ABC is an isosceles triangle right angled at C.
Therefore, AC = BC

Using Pythagoras theorem in the given triangle,
we have
AB2 = AC2 + BC2 = AC2 + AC2 = 2AC2

Q.43 ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

Ans.


Given that ABC is an isosceles triangle with AC = BC and AB2 = 2AC2.
Therefore,
AB2 = 2AC2 = AC2 + BC2
Therefore, by converse of Pythagoras theorem, ABC is a right triangle.

Q.44 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Ans.

It is given that ABC is an equilateral triangle of side 2a.Let AD is an altitude.We know that altitude bisects opposite side in anequilateral triangle.Therefore,BD=CD=a.Using Pythagoras theorem inΔADB, we get      AB2=BD2+AD2AD2=AB2BD2=(2a)2a2=4a2a2=3a2AD=a3We know that all the altitudes in an equilateral triangleare of same length. Therefore, length of each altitude is a3.

Q.45 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Ans.

Let ABCD is a rhombus.We know that diagonals of a rhombus bisect each otherat right angles.Using Pythagoras Theorem inΔAOB,ΔBOC,ΔCOD andΔAOD,we getAB2=OA2+OB2,BC2=OB2+OC2,CD2=OC2+OD2,DA2=OA2+OD2Adding all these equations, we getAB2+BC2+CD2+DA2=2(OA2+OB2+OC2+OD2)                                              =2[2(AC2)2+2(BD2)2]                                              =AC2+BD2

Q.46

In the following figure, O is a point in the interior of a triangle ABC,OD  BC, OE  AC and OF  AB. (i) OA2+OB2+OC2 OD2OE2OF2 = AF2+BD2+CE2,(ii)  AF2+BD2+CE2 = AE2+CD2+BF2.

Ans.

We join O to A,B and C.(i)InΔAOF,OA2=OF2+AF2[By Pythagoras theorem]InΔBOD,OB2=OD2+BD2[By Pythagoras theorem]InΔCOE,OC2=OE2+EC2[By Pythagoras theorem]Adding these equations, we getOA2+OB2+OC2=OF2+AF2+OD2+BD2+OE2+EC2OA2+OB2+OC2OD2OE2OF2=AF2+BD2+EC2(ii)We have form above result,AF2+BD2+EC2=OA2+OB2+OC2OD2OE2OF2                                 =(OA2OE2)+(OB2OF2)+(OC2OD2)                                 =AE2+BF2+CD2

Q.47 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Ans.

Let AB is a ladder and AC is wall.A represents the windowand BC is the distance of the foot of the ladder from thebase of the wall.ABC is a right angle triangle. Therefore, using Pythagorastheorem, we get     AB2=BC2+AC2102=BC2+82BC2=10064=36BC=6Therefore, the distance of the foot of the ladder from thebase of the wall is 6 m.

Q.48 A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Ans.

Let AC is a pole and AB is a guy wire with stake at B.Then, ABC is a right angle triangle. Therefore, using Pythagorastheorem, we get     AB2=BC2+AC2242=BC2+182BC2=242182=576324=252BC=252=4×9×7=67Therefore, the distance of the foot of the ladder from thebase of the wall is 67m.

Q.49 An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west ata speed of 1200 km per hour. How far apart will be the two planes after 112  hours?

Ans.

Distance travelled by the plane flying towards north in112hours=1000×112km=1500kmDistance travelled by the plane flying towards west in112hours=1200×112km=1800kmWe represent thsese distances by OA and OB.Also, AB represents the distance between the two planes.Using Pythagoras theorem, we getAB2=OA2+OB2=15002+18002=22,50,000+32,40,000AB2=54,90,000AB=54,90,000=9×6,10,000=30061Therefore, distance between the two planes=30061km

Q.50

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feetof the poles is 12 m, find the distance between their tops.

Ans.

Let AB and CD be two poles of heights 6 m and 11mrespectively.Let their feet are on the ground at points Band D.It is given that distance between feet is 12 m.Therefore,BD=12 m.Also,AB=DE=6mandCE=CDDE=11 m6 m=5 mNow,AEC is a right triangle.Therefore,by Pythagoras theorem,AC=AE2+CE2=122+52=144+25=169=13mTherefore,Distance between tops of the two poles=13m

Q.51

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.Prove thatAE2+BD2=AB2+DE2.

Ans.

InΔACE,      AE2=AC2+CE2                                                                ...(1)InΔDCB,BD2=CD2+BC2                                                              ...(2)On adding (1) and (2), we haveAE2+BD2=AC2+CE2+CD2+BC2                                       ...(3)InΔACB,AB2=AC2+BC2                                                                            ...(4)InΔDCE,DE2=CD2+CE2                                                                             ...(5)On adding (4) and (5), we haveAB2+DE2=AC2+BC2+CD2+CE2                                         ...(6)From (3) and (6), we find thatAE2+BD2=AB2+DE2

Q.52

The perpendicular from A on side BC of aΔABC intersects BC at D such thatDB=3CD(see thefollowing figure). Prove that2AB2=2AC2+BC2.

Ans.

InΔADC,      AC2=CD2+AD2AD2= AC2CD2                                       ...(1)InΔADB,AB2=AD2+DB2AD2=AB2DB2                                       ...(2)From (1) and (2), we haveAC2CD2=AB2DB2                                          ...(3)Now, it is given that DB=3CDAlso,      BC=CD+DBBC=CD+3CD=4CDCD=BC4                                                                 ...(4)Again,      BC=CD+DBBC=DB3+DB=4DB3DB=3BC4                                                                 ...(5)From (3), (4) and (5), we have       AC2CD2=AB2DB2AC2BC216= AB29BC21616AC2BC2=16AB29BC216AB2=16AC2BC2+9BC2=16AC2+8BC22AB2=2AC2+BC2

Q.53

In an equilateral triangle ABC, D is a point on side BC such that BD=13BC. Prove that 9AD2=7AB2.

Ans.

It is given that ABC is an equilateral triangle.Let length of each side ofΔABC be a.It is given that D is a point on side BC such thatBD=13BC=a3.Let AE is an altitude of equilateral triangle ABC. So,BE=EC=12BC=a2.Also, AE=AC2EC2=a2a24=32aAgain, DE=BEBD=a2a3=a6InΔADE,     AD2=AE2+DE2[By Pythagoras Theorem]AD2=(32a)2+a236=34a2+a236=27a2+a236=2836a2=79a2AD2=79a2=79AB2[AB=a]9AD2=7AB2

Q.54 In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Ans.

Let ABC is an equilateral triangle and AE is its altitude.Let length of each side ofΔABC be a.So,BE=EC=12BC=a2.Using Pythagoras theorem inΔACE, we haveAE2=AC2EC2=a2a24=34a23a2=4AE2Therefore, three times the square of one side of anequilateral triangle is equal to four times the squareof one of its altitude.

Q.55

Tick the correct answer and justify: InΔABC, AB=63cm, AC=12 cm and BC=6 cm.The angle B is :(A) 120°   (B) 60°(C) 90° (D) 45°

Ans.

It is given that inΔABC,AB=63cm, AC=12 cm and BC=6 cmAB2=108 cm2, AC2=144 cm2and BC2=36cm2Now,AB2+BC2=108 cm2+36cm2=144cm2=AC2Therefore, by converse of Pythagoras theorem, we find thatinΔABC, angle B is 90°.Therefore, correct answer is (C).

Q.56

In the following figure, PS is the bisector ofQPR ofΔPQR. Prove thatQSSR=PQPR.

Ans.

We draw a line segment RT parallel to SP whichintersects extended line segment QP at point T.It is given that PS is angle bisector ofQPR.Therefore,QPS=SPR (1)Also,SPR=PRT (As PSTR) (2)QPS=QTR (As PSTR) (3)Using these equations we get,PRT=QTRSo, PT=PRNow inΔQTR andΔQPS,QSP=QRT (As PSTR)QPS=QTR (As PSTR)Q is common     ΔQTR~ΔQPSQSSR=QPPTQSSR=QPPR[PT=PR]

Q.57

In the following figure, D is a point on hypotenuse AC ofΔABC, DMBC and DNAB. Prove that:(i)DM2=DN.MC(ii)DN2=DM.AN

Ans.

(i)

We have, DNCB, DMAB andB=90°So,DNMB is a rectangle.DN=MB and DM=NB2+3=90°...(1)InΔCDM,1+2+DMC=180°1+2=90°                                                                     ...(2)InΔDMB3+DMB+4=180°3+4=90°                                                                     ...(3)From equation(1)and(2),1=3From equation(1)and(3),2=4ΔDCM~ΔBDMDMBM=MCDMDM2=BM×MC=DN×MC [BM=DN](ii)InΔDBN,5+7=90°...(4)InΔDAN,6+8=90°...(5)InΔDAB,5+6=90°                           ...(6)From equation(4)and(6),we get6=7From equation(5)and(6),we get5=8ΔBND~ΔDNA [By AA similarity critarion]ANDN=DNNBDN2=AN×NB=AN×DM [As NB=DM]

Q.58 

In the following figure, ABC is a triangle in whichABC > 90° and ADCB produced. Prove thatAC2=AB2+BC2+2BC.BD.

Ans.

Applying Pythagoras theorem in ΔADB, we get
AB2 = AD2 + DB2 ….(1)
Applying Pythagoras theorem in ΔACD, we get
AC2 = AD2 + DC2
⇒AC2 = AD2 + ( BD + BC )2
⇒AC2 = AD2 + DB2 + BC2 + 2BD x BC
Now using equation ( 1 ), we get
AC2 = AB2 + BC2 + 2BD × BC

Q.59

Ans.

ApplyingPythagorastheoreminΔADB,we getAD2+BD2=AB2AD2=AB2BD2                       ...(1)ApplyingPythagorastheoreminΔADC,we getAD2+DC2=AC2Nowusingequation(1),we getAB2BD2+DC2=AC2AB2BD2+(BCBD)2=AC2AC2=AB2BD2+BC2+BD22BC×BDAC2=AB2+BC22BC×BD

Q.60

In the following figure, AD is a median of a triangle ABC and AMBC. Prove that:(i) AC2=AD2+BC.DM+(BC2)2(ii) AB2=AD2BC . DM+(BC2)2(iii) AC2+AB2=2AD2+12BC2

Ans.

We have the following figure.

(i)Applying Pythagoras theorem inΔAMD,we getAM2+DM2=AD2...(1)Applying Pythagoras theorem inΔAMC,we get      AM2+MC2=AC2                              AM2+(DM+DC)2=AC2(AM2+DM2)+DC2+2DM.DC=AC2Usingequation(1),wegetAD2+DC2+2DM.DC=AC2AD2+(BC2)2+2DM.BC2=AC2AD2+(BC2)2+DMBC=AC2(ii)Applying Pythagoras theorem inΔABM, we getAB2=AM2+MB2=(AD2DM2)+MB2=(AD2DM2)+(BDDM)2=AD2DM2+BD2+DM22BD×DM=AD2+BD22BD×DM=AD2+(BC2)22BC2×DM=AD2+(BC2)2BC×DM(iii)Applying Pythagoras theorem inΔAMB, we getAM2+MB2=AB2...(1)Applying Pythagoras theorem inΔAMC, we getAM2+MC2=AC2...(2)Adding equations (1) and (2), we get2AM2+MB2+MC2=AB2+AC22AM2+(BDDM)2+(MD+DC)2=AB2+AC22AM2+BD2+DM22BD.DM+MD2+DC2+2DM.DC=AB2+AC22AM2+2DM2+BD2+DC2+2DM(BD+DC)=AB2+AC22(AM2+DM2)+(BC2)2+(BC2)2+2DM(BC2+BC2)=AB2+AC22AD2+BC22=AB2+AC2

Q.61 Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Ans.

Let ABCD be a parallelogram. We draw perpendiculars AF on CD and DE on extended side BA.

Applying Pythagoras theorem inΔDEA, we getDE2+EA2=DA2...(i)Applying Pythagoras theorem inΔDEB,we getDE2+EB2=DB2DE2+(EA+AB)2=DB2(DE2+EA2)+AB2+2EA×AB=DB2DA2+AB2+2EA×AB=DB2...(ii)Applying Pythagoras theorem inΔADF,we getAD2=AF2+FD2Applying Pythagoras theorem inΔAFC,we getAC2=AF2+FC2         =AF2+(DCFD)2         =AF2+DC2+FD22DC×FD         =(AF2+FD2)+DC22DC×FDAC2=AD2+DC22DC×FD...(iii)Since ABCD is a parallelogramAB=CD...(iv)and,BC=AD...(v)InΔDEA andΔADF,DEA=AFD=90°,EAD=ADF(EADF)Therefore,FAD=EDAAD is common in both triangles.Therefore,by ASA congruence critarion, we getΔEADΔFDAEA=DF...(vi)Adding equation (ii) and (iii), we getDA2+AB2+2EA×AB+AD2+DC22DC×FD=DB2+AC2DA2+AB2+AD2+DC2+2EA×AB2DC×FD=DB2+AC2BC2+AB2+AD2+DC2+2EA×AB2AB×EA=DB2+AC2(Using equations (iv), (v) and (vi))AB2+BC2+CD2+DA2=AC2+BD2Hence,thesumofthesquaresofthediagonalsofaparallelogramisequaltothesumofthesquaresofitssides.

Q.62

In the following figure, two chords AB and CD intersect each other at the point P. Prove that :(i)ΔAPC~ΔDPB (ii) AP.PB=CP.DP

Ans.

Let us join CB.(i) InΔAPC andΔDPB,APC=DPB [Vertically opposite angles]CAP=BDP [Angles in same segment for chord CB]ΔAPC~ΔDPB [By AA similarity criterion](ii)We know that corresponding sides of similar trianglesare proportional.Therefore,ΔAPC~ΔDPBAPDP=PCPBAPPB=PCDP

Q.63

In the following figure, two chords AB and CD of a circle intersect each other at the point P(when produced) outside the circle. Prove that(i)ΔPAC~ΔPDB (ii)PAPB=PCPD

Ans.

(i)InΔPAC andΔPDB,P=P (common)PAC=PDB(Exterior angle of a cyclic quadrilateral isequal to opposite interior angle)ΔPAC~ΔPDB(ii)We know that corresponding sides of similar trianglesare proportional.ΔPAC~ΔPDBPAPD=PCPBPAPB=PCPD

Q.64 In the following figure, D is a point on side BC of ΔABC such that BD CD = AB AC . Prove that AD is the bisector of BAC.

Ans.

Let us extend BA to P such that AP=AC. Join PC.It is given that     BDCD=ABACBDCD=ABAPTherefore, by converse of basic proportinality theorem, ADPC.Therefore,BAD=APC...(1)DAC=ACP...(2)Also,   AP=ACAPC=ACPAPC=ACP=DAC [From (2)]BAD=DAC                                [From (1)]Therefore, AD is the bisector of angle BAC.

Q.65 Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see the following figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Ans.

Let AB be the height of the tip of fishing rod from watersurface.Let BC be the horizontal distance of fly from the tip offishing rod.AC is the length of string.On applying Pythagoras theorem inΔABC, we getAC2=AB2+BC2AB2=(1.8)2+(2.4)2AB2=3.24+5.76AB2=9.00AB=9=3Therefore, length of the string is 3 m.It is given that string is pulled at the rate of 5 cm per second.So, string pulled in 12 seconds=12×5=60cm=0.6m

Let after 12 second Fly be at point D.Length of string out after 12 second is ADAD = AC – string pulled by Nazima in 12 seconds    = 3.00 – 0.6    = 2.4In ΔADB,AB2 + BD2 = AD2(1.8)2 + BD2 = (2.4)2BD2 = 5.76 

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FAQs (Frequently Asked Questions)

1. What are the essential theorems in Chapter 6 Triangles in Class 10?

 The following are the most important theorems in Chapter 6 Triangles in Class 10:

  • Pythagoras Theorem
  • Midpoint Theorem
  • Remainder Theorem
  • Angle Bisector Theorem
  • Inscribed Angle Theorem

2. How much weightage is given to the Chapter 6 Triangles in Class 10 Board exams?

Chapter 6 Triangles is a part of the ‘Geometry’ unit in Class 10. The unit has a weightage of a total of 15 marks in the exam. Therefore, Chapter 6 is most likely to carry questions of 5-6 marks in the examination paper.