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

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2.

Ex 7.2 Class 7 Maths Question 1.

Which congruence criterion do you use in the following?
(a) Given:
AC = DF
AB = DE
BC = EF
So, ∆ABC ≅ ∆DEF

(b) Given:
ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆PQR ≅ ∆XYZ

(d) Given:
EB = DB
AE = BC
∠A = ∠C = 90°
So, ∆ABE ≅ ∆CDB

Solution:
(a) We have, ∆ABC ≅ ∆DEF   (BY SSS rule)
(b) We have, ∆PQR ≅ ∆XYZ   (BY SAS rule)
(c) We have, ∆LMN ≅ ∆GFH (BY ASA rule)
(d) We have, ∆ABE ≅ ∆CDB   (BY RHS rule)

Ex 7.2 Class 7 Maths Question 2.

You want to show that ∆ART = ∆PEN,
(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 =
(ii) PN =
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ?
(ii) ?

Solution:
(a) For SSS criterion, we need to show
(i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) For SAS criterion, we need to have
(i) RT = EN and
(ii) PN = AT
(c) For ASA criterion, we need to have
(i) ∠A = ∠P
(ii) ∠T = ∠N

Ex 7.2 Class 7 Maths Question 3.

You have to show that ∆AMP ≅ ∆AMQ. In the fallowing proof, supply the missing reasons.

Steps	Reasons
(i) PM = QM	(i) …
(ii) ∠PMA = ∠QMA	(ii) …
(iii) AM = AM	(iii) …
(iv) ∆AMP ≅ ∆AMQ	(iv) …

Solution:

Steps	Reasons
(i) PM = QM	(i) Given
(ii) ∠PMA = ∠QMA	(ii) Given
(iii) AM = AM	(iii) Common
(iv) ∆AMP = ∆AMQ	(iv) SAS rule

Ex 7.2 Class 7 Maths Question 4.

In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°.
A student says that ∆ABC ≅ ∆PQR by AAA congruence criterion. Is he justified? Why or why not?

Solution:
The student is not justified because there is not criterion for AAA congruence rule.

Here, in ∆ABC and ∆PQR, we have

∠A = 30°, ∠B = 40°, ∠C = 110°,
∠P = 30°, ∠Q = 40°, ∠R = 110°
But ∆ABC is not congruent to ∆PQR.

Ex 7.2 Class 7 Maths Question 5.

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

Solution:
In ∆RAT and ∆WON,
AT = ON    (Given)
AR = OW   (Given)
∠A = ∠O     (Given)
∴ ∆RAT ≅ ∆WON      (By SAS rule)

Ex 7.2 Class 7 Maths Question 6.

Complete the congruence statement:

Solution:
Refer to fig. (i).
In ∆BCA and ∆BTA,
∠C = ∠T       (Given)
BC = BT       (Given)
∠CBA = ∠TBA   (Given)
∴ ∆BCA ≅ ∆BTA   (By ASA rule)
Refer to fig. (ii).
In ∆QRS and ∆TPQ,
RS = PQ      (Given)
QS = TQ      (Given)
∠RSQ = ∠PQT     (Given)
∴ ∆QRS ≅ ∆TPQ   (By SAS rule)

Ex 7.2 Class 7 Maths Question 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) On the given square sheet, we have drawn two congruent triangles, i.e.,
∆ABC ≅ ∆DEF such that

AB = DE, BC = EF and AC = DF
On adding, we get
AB + BC + AC = DE + EF + DF
i.e., perimeter of ∆ABC = perimeter of ∆DEF

(ii) On the other square sheet, we have drawn two triangles ABC and PQR which are not congruent.
Such that

Adding both sides, we get
AB + BC + AC ≠ PQ + QR + PR
i.e., perimeter of ∆ABC ≠ the perimeter of ∆PQR.

Ex 7.2 Class 7 Maths 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:
We have ∆PQR and ∆TSU
PQ = SU      (Given)
PR = ST       (Given)
∠Q = ∠S      (Given)
∠P = ∠T       (Given)
∠R = ∠U      (Given)
Since, none of the criteria of congruence is relevant here.
∴ ∆PQR and ∆TSU are not congruent.

Ex 7.2 Class 7 Maths Question 9.

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

Solution:
In ∆ABC and ∆PQR,
∠B = ∠Q (Given)
∠C = ∠R (Given)
For ∆ABC ≅ ∆PQR
BC must equal to QR. Criterion that we used is ASA rule.
Hence, the additional pair of corresponding part is BC = QR.

Ex 7.2 Class 7 Maths Question 10.

Explain, why ∆ABC ≅ ∆FED.

Solution:
In ∆ABC and ∆FED,
∠B = ∠E = 90°       (Given)
∠A = ∠F                 (Given)
∴ ∠A + ∠B + ∠C = ∠E + ∠F + ∠D
[Angle sum property of a triangle]
∴ ∠C = ∠D
BC = ED      (Given)
∴ ∆ABC ≅ ∆FED       (By ASA rule)