Students can Download Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Question 1.

Which congruence criterion do you use in the following ?

a) Given : AC = D

AB = DE

BC = EF

So, ∆ ABC = ∆ DEF

SSS congruence criterion

b) Given : ZX = RP

RQ = ZY

∠PRQ = ∠XZY

So, ∆ PQR ≅ ∆ XYZ

SAS congruence criterion

c) Given: ∠MLN = ∠FGH

∠NML = ∠GFH

ML = FG

So, ∆LMN ≅ ∆GFH

ASA congruence criterion.

d) Given : EB = DB

AE = BC

∠A = ∠C = 90°

So, ∆ ABC ≅ ∆ CDB

RHS congruence criterion

Question 2.

You want to show that ∆ ART ≅ ∆ PEN,

a) If you have to use SSS criterion, then you need to show

i) AR =

AR = PE,

ii) RT =

RT = EN,

iii) AT =

AT = PN

b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have

i) RT =

RT = EN and

ii) PN =

PN = AT

c) If it is given that AT = PN and you to use ASA criterion, you need to have

i) ?

∠ATR = ∠PNE and

ii) ?

∠TAR = NPE.

Question 3.

You have to show that ∆ AMP ≅ ∆ AMQ.

In the following proof, supply the missing reasons

Solution:

Question 4.

In ∆ ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°

In ∆ PQR, A student says that ∆ ABC ≅ ∆ PQR by AAA congruence criterion. Is he justified? Why or why not?

Solution:

No, he is not justified because AAA is not a criterion for the congruence of triangles.

Question 5.

In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆ RAT ≅ ?

Solution:

Question 6.

Complete the congruence statement:

Solution:

∆ BCA ≅ ? ∆ QRS ≅ ?

∆ BCA ≅ ? ∆ BTA ≅ ?

∆ QRS ≅ ? ∆ TPQ

Question 7.

In a squared sheet, draw two triangles of equal areas such that

i) The triangles are congruent.

Solution:

Consider the ∆^{S} PQS and SQR

In ∆ PQS and ∆ SQR

PS = QR = 6 cms

∠SPQ = ∠QRS = 90°

QS = QS = common

By RHS congruence criterion

∆PQS ≅ ∆SQR.

Perimeter of the ∆ PQS = PQ + QS + PS

Perimeter of the ∆ SQR = SR + QS + QR

∴ Perimeter of the ∆ PQS = Perimeter of the ∆ SQR (∵ PQ = SR & PS = QR)

ii) the triangles are not congruent. What can you say about their perimeters?

Solution:

∴ area of ∆ PQS = Area of ∆ PQM.

By seeing the figure the ∆ PQS = PQ + PS + SQ = 8 + 6 + 10 = 24 cms.

Perimeter of the ∆ PQM = PQ + PM + QM

= 8 + 7.2 + 7.2

= 22.4 cms.

∴ Their perimeters are not equal.

PM = QM

(PM)^{2} = PN^{2} + MN^{2}

= 4^{2} + 6^{2}

= 16 + 36 = 52

PM = QM = = 7.2

Question 8.

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

Solution:

Considering the two triangles PQR and XYZ

In ∆ PQR and XYZ

PQ = XZ = 7 cms

PR = YZ = 6 cms

RQ = XY = 5 cms

∠PRQ = ∠XYZ

∠PQR = ∠XZY

In the above 2 triangles 5 pairs of the congruent present. Still, they are not congruent.

Question 9.

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

Solution:

The additional corresponding part is BC = RQ by ASA congruence rules

Question 10.

Explain, why ∆ ABC ≅ ∆ FED

Solution:

∠ABC = ∠DEF = 90°

BC = DE ∠ACD = ∠EDF

( ∵ The sum of the measures of three angles of a triangle is 180°.)

∴ ∆ ABC ≅ ∆ DEF

(By ASA congruence criterion.)