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

## 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
(c) Given: MLN = FGH
NML = GFH
ML = FG
So, ∆LMN = ∆GFH
(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
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

AB + BC + ACPQ + 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)

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