NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (EX 7.2) Exercise 7.2

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The solutions to the chapter titled Congruence of Triangles are provided in a simple PDF format on the Extramarks website. The Ch 7 Maths Class 10 Ex 7.2 covers the themes of right-angled triangle congruence and the criteria for triangle congruence. It is advised that Class 7 students complete the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 based on Congruence of Triangles in order to solidify the principles and be ready to answer questions that are frequently asked in the examination. When it comes to examination preparation, using the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 is thought to be the best choice for students. There are numerous exercises in this chapter. On the Extramarks website, in PDF format, the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 are offered. Students can study these solutions straight from the website or mobile application, or they can download them as needed.

Extramarks’ internal subject matter experts, carefully and in accordance with all CBSE regulations, solved the problems and questions from the exercise. Any student in Class 7 who has comprehensive knowledge of all the concepts from the Mathematics textbook and is quite knowledgeable with regard to all the exercises provided in it can easily secure the best scores on the final examination. Students may quickly comprehend the types of questions that may be given in the examination from this chapter and learn the chapter weightage in terms of overall mark distribution with the aid of these NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2. Therefore, students can adequately study for the final examination and score higher marks.

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NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles (EX 7.2) Exercise 7.2

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Access NCERT Solution for Class 7 Mathematics Chapter 7- Congruence of Triangles

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The Class 7 Maths Chapter 7 Exercise 7.2 Solution explains that a triangle has three vertices, three sides, and three angles. Triangles are categorised as equilateral, isosceles, and scalene based on similarities in the side measurements. In this instance, a comparison is made between the same triangle’s sides and angles. In order to compare two distinct triangles, a different set of guidelines is used. Congruent figures are two figures that are alike. These figures are exact replicas of one another. When an object’s counterpart is put over another congruent object, the two appear to be the same figure. Congruent triangles are the same as identical triangles in that they have the same side and angle measurements. Students who face difficulty in understanding the concept can refer to the NCERT Solutions Class 7 Maths Chapter 7 Exercise 7.2 provided by the Extramarks website.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2

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Q.1

Which congruence criterion do you use in the following?(a)Given:AC=DFAB=DEBC=EF


S
o,ΔABCΔDEF

(b)Given:ZX=RP  RQ=ZY    PRQ=XZYSo,ΔPQRΔXYZ

(c)Given:MLN=FGH  NML=GFHML=FG   So,ΔLMNΔGFH

(d)Given:EB=DB  AE=BC  A=C=90°So,ΔABEΔCDB

Ans.

(a) SSS, as the sides of ΔABC are equal to the sides of ΔDEF.(b) SAS, as two sides and the angle included between these sides of ΔPQR are equal to two sides and the angle included between these sides of ΔXYZ.(c) ASA, as two angles and the side included between these angles of ΔLMN are equal to two angles and the side included between these angles of ΔGFH.(d) RHS, as in the given two right-angled triangles, one side and the hypotenuse are respectively equal.

Q.2

You want to show that ΔART@ΔPEN,a If you have to use SSS criterion, then you need to showi AR = ii RT = iii AT =b If it is given that ∠T =N and you are to use SAS criterion, you need to havei RT = andii PN =c If it is given that AT=PN and you are to use ASA criterion, you need to have      i ? ii ?

Ans.

(a)(i) AR=PE (ii) RT=EN (iii) AT = PN(b)(i) RT=EN (ii) PN=AT(c)(i) ATR=PNE (ii) RAT=EPN

Q.3

You have to show that AMPAMQ.In the following

proof, supply the missing reasons.

Steps Reasons
(i) PM = QM (i)…
( ii )PMA =QMA MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGcbaWaaeWaaeaaieqacaWFPbGaa8xAaaGaayjkaiaa wMcaaiabgcIiqlaa=bfacaWFnbGaa8xqaiaa=bcacqGH9aqpcqGHGi c0caWFrbGaa8xtaiaa=feaaaa@4A4C@ (ii)…
(iii) AM = AM (iii)…
( iv )ΔAMPΔAMQ MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGcbaWaaeWaaeaaieqacaWFPbGaa8NDaaGaayjkaiaa wMcaaiaa=r5acaWFbbGaa8xtaiaa=bfacqGHfjcqcaWFuoGaa8xqai aa=1eacaWFrbaaaa@48D9@ (iv)…

Ans.

Steps Reasons
(i) PM = QM (i)…Given
(ii)PMA=QMA (ii)…Given
(iii) AM = AM (iii)…Common
(iv)ΔAMPΔAMQ (iv) SAS,as the two sides and the angle included between these sides of ΔAMP are equal to two sides and the angleincluded between these sides of ΔAMQ

Q.4

In ΔABC,A = 30°,B = 40° and C = 110°InΔPQR,P = 30° ,Q = 40° and R = 110°A student says that ΔABCΔPQR by AAA congruence criterion. Is he justified? Why or why not?

Ans.

No. This property represents that these triangles have their respective angles of equal measure. However, this gives no information about their sides. The sides of these triangles have a ratio somewhat different than 1:1.Therefore, AAA property does not prove the two triangles are congruent.

Q.5

In the figure, the two triangles are congruent. The corresponding parts are marked.

We can write ΔRAT?

Ans.

From the figure, we observe that:RAT = WONART = OWN   AR = OWTherefore, ΔRATΔWON, by ASA criterion.

Q.6

Complete the congruence statement:

Ans.

Given that, BC = BTTA = CABA is common. Therefore, ΔBCAΔBTASimilarly,PQ = RSTQ = QSPT = RQ Therefore, ΔQRSΔTPQ.

Q.7

In a squared sheet, draw two triangles of equal areas such that(i) the triangles are congruent.(ii) the triangles are not congruent.(iii) What can you say about their perimeters?

Ans.

(i)

Here
,ΔABC and ΔPQR have the same area and are congruent to each other also. Also, the perimeter of both the triangles
will be the same
.
(ii)

Here,the two triangles have the same height and base.Thus, their areas are equal.However, these triangles are not congruent to each other.Also, the perimeter of both the triangles will not be the same.

Q.8

Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.

Ans.

A pair of triangles with threee equal angles, and two equalsides are non congruent.


Below is the example:

Q.9

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Ans.

Here,BC = QRSo, ΔABCΔPQR (ASA criterion)

Q.10

Explain, why ΔABC ΔFED

Ans.

Given that, ABC = FED ( 1 ) BAC = EFD (2) The two angles of ΔABC are equal to the two respective angles of ΔFED. Also, the sum of all interior angles of a triangle is 180º. Therefore, third angle of both triangles will also be equal in measure. BCA = EDF( 3 ) Also, given that, BC = ED( 4 ) By using equation ( 1 ), ( 3 ), and ( 4 ),we obtain ΔABC ΔFED (ASA criterion) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGhbGaaeyAaiaabAhacaqGLbGaaeOB aiaabccacaqG0bGaaeiAaiaabggacaqG0bGaaiilaaqaaiabgcIiql aabgeacaqGcbGaae4qaiaabccacqGH9aqpcaqGGaGaeyiiIaTaaeOr aiaabweacaqGebGaaeiiaiablAciljablAciljablAciljablAciln aabmaabaGaaeymaaGaayjkaiaawMcaaaqaaiabgcIiqlaabkeacaqG bbGaae4qaiaabccacqGH9aqpcaqGGaGaeyiiIaTaaeyraiaabAeaca qGebGaeSOjGSKaeSOjGSKaeSOjGSKaeSOjGSKaaeiiaiaacIcacaqG YaGaaiykaaqaaiaabsfacaqGObGaaeyzaiaabccacaqG0bGaae4Dai aab+gacaqGGaGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGZbGa aeiiaiaab+gacaqGMbGaaeiiaiabfs5aejaabgeacaqGcbGaae4qai aabccacaqGHbGaaeOCaiaabwgacaqGGaGaaeyzaiaabghacaqG1bGa aeyyaiaabYgacaqGGaGaaeiDaiaab+gacaqGGaGaaeiDaiaabIgaca qGLbGaaeiiaiaabshacaqG3bGaae4BaiaabccacaqGYbGaaeyzaiaa bohacaqGWbGaaeyzaiaabogacaqG0bGaaeyAaiaabAhacaqGLbaaba Gaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGZbGaaeiiaiaab+ga caqGMbGaaeiiaiabfs5aejaabAeacaqGfbGaaeiraiaac6cacaqGGa GaaeyqaiaabYgacaqGZbGaae4BaiaacYcacaqGGaGaaeiDaiaabIga caqGLbGaaeiiaiaabohacaqG1bGaaeyBaiaabccacaqGVbGaaeOzai aabccacaqGHbGaaeiBaiaabYgacaqGGaGaaeyAaiaab6gacaqG0bGa aeyzaiaabkhacaqGPbGaae4BaiaabkhacaqGGaGaaeyyaiaab6gaca qGNbGaaeiBaiaabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaa bggaaeaacaqG0bGaaeOCaiaabMgacaqGHbGaaeOBaiaabEgacaqGSb GaaeyzaiaabccacaqGPbGaae4CaiaabccacaqGXaGaaeioaiaaicda caGG6cGaaiOlaiaabccacaqGubGaaeiAaiaabwgacaqGYbGaaeyzai aabAgacaqGVbGaaeOCaiaabwgacaGGSaGaaeiiaiaabshacaqGObGa aeyAaiaabkhacaqGKbGaaeiiaiaabggacaqGUbGaae4zaiaabYgaca qGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabkgacaqGVbGaaeiDaiaa bIgacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6gacaqGNbGaae iBaiaabwgacaqGZbGaaeiiaiaabEhacaqGPbGaaeiBaiaabYgaaeaa caqGHbGaaeiBaiaabohacaqGVbGaaeiiaiaabkgacaqGLbGaaeiiai aabwgacaqGXbGaaeyDaiaabggacaqGSbGaaeiiaiaabMgacaqGUbGa aeiiaiaab2gacaqGLbGaaeyyaiaabohacaqG1bGaaeOCaiaabwgaca GGUaaabaGaeyiiIaTaaeOqaiaaboeacaqGbbGaaeiiaiabg2da9iaa bccacqGHGic0caqGfbGaaeiraiaabAeacqWIMaYscqWIMaYscqWIMa YscqWIMaYsdaqadaqaaiaabodaaiaawIcacaGLPaaaaeaacaqGbbGa aeiBaiaabohacaqGVbGaaiilaiaabccacaqGNbGaaeyAaiaabAhaca qGLbGaaeOBaiaabccacaqG0bGaaeiAaiaabggacaqG0bGaaiilaaqa aiaabccacaqGcbGaae4qaiaabccacqGH9aqpcaqGGaGaaeyraiaabs eacqWIMaYscqWIMaYscqWIMaYscqWIMaYscqWIMaYscqWIMaYsdaqa daqaaiaabsdaaiaawIcacaGLPaaaaeaacaqGcbGaaeyEaiaabccaca qG1bGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqGLbGaaeyCaiaa bwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccadaqadaqaai aabgdaaiaawIcacaGLPaaacaGGSaGaaeiiamaabmaabaGaae4maaGa ayjkaiaawMcaaiaacYcacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiam aabmaabaGaaeinaaGaayjkaiaawMcaaiaacYcacaqG3bGaaeyzaiaa bccacaqGVbGaaeOyaiaabshacaqGHbGaaeyAaiaab6gaaeaadaqjEa qaaiabfs5aejaabgeacaqGcbGaae4qaiaabccacqGHfjcqcaqGGaGa euiLdqKaaeOraiaabweacaqGebaaaiaabccacaGGOaGaaeyqaiaabo facaqGbbGaaeiiaiaabogacaqGYbGaaeyAaiaabshacaqGLbGaaeOC aiaabMgacaqGVbGaaeOBaiaacMcaaaaa@8308@

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