 # NCERT Solutions for Class 7 Mathematics: Chapter 7 - Congruence of Triangles

Check NCERT Solutions for Class 7 Mathematics (Chapter 7 - Congruence of Triangles). With this article, you will get solutions to all the questions of Class 7 Maths NCERT Textbook, Chapter 7 - Congruence of Triangles. CBSE Class 7

Check NCERT Solutions for Class 7 Mathematics, Chapter 7 - Congruence of Triangles. This chapter is one of the most important chapters of CBSE Class 7 Mathematics subject. Students of Class 7 are advised to prepare this chapter well, as concepts given in this chapter are required to solve questions of other chapters.

NCERT Solutions for Class 7 Mathematics, Chapter 7 - Congruence of Triangles

Exercise 7.1

1. Complete the following statements:

(a) Two line segments are congruent if ___________.

(b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is ___________.

(c) When we write ∠A = ∠B, we actually mean ___________.

Solutions:

(a) they have the same length

(b) 70°

(c) measure of ∠A = measure of ∠B

2. Give any two real-life examples for congruent shapes.

Solutions:

(i) Pages of a notebook.

(ii) Breads in the same packet

3. If ΔABC ≅ΔFED under the correspondence ABC ↔ FED, write all the corresponding congruent parts of the triangles.

Solutions:

As these triangles are congruent it means the corresponding angles and sides are equal to each other.

∠A ↔ ∠F

∠B ↔ ∠E

∠C ↔ ∠D

CA ↔ DF

AB ↔ FE

BC ↔ ED

4. If ΔDEF ≅ΔBCA, write the part(s) of ΔBCA that correspond to Solutions:

(i) ∠C

(ii) CA

(iii) ∠A

(iv) BA

Exercise 7.2

1. Which congruence criterion do you use in the following?

(a) Given: AC = DF AB = DE

BC = EF

So, ΔABC ≅ ΔDEF

(b) Given: ZX = RP RQ = ZY

∠PRQ = ∠XZY

So, ΔPQR ≅ ΔXYZ

(c) Given: ∠MLN = ∠FGH ∠NML = ∠GFH

ML = FG

So, ΔLMN ≅ ΔGFH

(d) Given: EB = DB AE = BC

∠A = ∠C = 90°

So, ΔABE ≅ ΔCDB

Solution: Criterion: RHS

2. You want to show that ΔART ≅ΔPEN,

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

(i) AR =

(ii) RT =

(iii) AT =

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

(i) RT = and

(ii) PN =

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

(i) ?

(ii) ? (a)

(i) PE = AR

(ii) EN = RT

(iii) PN = AT

(b)

(i) EN = RT

(ii) AT = PN

(c)

(i) ∠PNE = ∠ATR

(ii) ∠EPN = ∠RAT

3. You have to show that ΔAMP ≅ΔAMQ. In the following proof, supply the missing reasons.

 Steps Reasons (i) PM = QM (i) ... (ii) ∠PMA = ∠QMA (ii) ... (iii) AM = AM (iii) ... (iv) ΔAMP ≅ ΔAMQ (iv) ...

Solution:

(i) Given in the question

(ii) Given in the question

(iii) Common

(iv) By SAS congruency ΔAMP ≅ ΔAMQ

4. In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°

In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°

A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?

Solution:

No. Because two traingle with different sides can have all their respective angles equal.

5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ΔRAT ≅ ? Solution:

Answers of the remaining questions will be available here shortly.

As,

∠RAT = ∠WON

∠ART = ∠OWN

AR = OW

So, ΔRAT ≅ ΔWON, by ASA criterion.

6. Complete the congruence statement: Solution:

Given, BC = BT &TA = CA

Also, BA is common.

So, ΔBCA ≅ ΔBTA

Similarly, PQ = RS, TQ = QS and PT = RQ

So, ΔQRS ≅ ΔTPQ

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

(i) The triangles are congruent.

(ii) The triangles are not congruent.

What can you say about their perimeters?

Solution:

(i) These triangles are congruent and have same areas (half of the product of the length of base and height).

(ii) Here measures of angles of these triangles are different but areas (half of the product of the length of base and height) are same.

8. If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use? Solution: It is BC = QR

By ASA criterion, ΔABC ≅ ΔPQR

9. Explain, why ΔABC ≅ΔFED. Solution:

Here, ∠ABC = ∠FED ....(i)

∠BAC = ∠EFD ....(ii)

Sidem BC = ED

So,  ΔABC ≅ ΔFED (by ASA criterion).

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