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

NCERT Solutions Class 7 Mathematics
Chapter – 7 (Congruence of Triangles)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 7 Congruence of Triangles Exercise 7.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 7 Congruence of Triangles

• NCERT Solution Class 7 Maths Exercise – 7.1

Exercise – 7.2

1. Which congruence criterion do you use in the following?
(a) Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF

Solution –
By SSS congruence property –  Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.
ΔABC ≅ ΔDEF

(b) Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ

Solution –
By SAS congruence property – Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.
ΔACB ≅ ΔDEF

(c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
∠ML = ∠FG
So, ΔLMN ≅ ΔGFH

Solution –
By ASA congruence property – Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
ΔLMN ≅ ΔGFH

(d) Given: EB = DB
AE = BC
∠A = ∠C = 90^o
So, ΔABE ≅ ΔACD

Solution –
By RHS congruence property – Two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second.
ΔABE ≅ ΔACD

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) ZA  (ii) ZT

Solution –

(a) For SSS criterion, we need
(i) AR = PE
(ii) RT = EN
(iii) AT = PN

(b) For SAS criterion, we need
(i) RT = EN and
(ii) PN = AT

(c) For ASA criterion, we need
(i) ∠A = ∠P
(ii) ∠T = ∠N

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

Solution –

4. In ΔABC, ∠A = 30^o, ∠B = 40^o and ∠C = 110^o
In ΔPQR, ∠P = 30^o, ∠Q = 40^o and ∠R = 110^o
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or Why not?

Solution –

No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be enlarged copy of the other.

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

Solution –
From the given figure,
We may observe that,
∠TRA = ∠OWN
∠TAR = ∠NOW
∠ATR = ∠ONW
Hence, ΔRAT ≅ ΔWON

6. Complete the congruence statement:

ΔBCA  ≅  ?                                                                 ΔQRS ≅ ?

Solution –
First consider the ΔBCA and ΔBTA
From the figure, it is given that,
BT = BC
Then,
BA is common side for the ΔBCA and ΔBTA
Hence, ΔBCA ≅ ΔBTA

Consider the ΔQRS and ΔTPQ
From the figure, it is given that
PT = QR
TQ = QS
PQ = RS
Hence, 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) On the given square sheet, we have draw two congruent triangles i.e.
∆ABC = ∆DEF

such that

On adding, we get
=
i.e. perimeters 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
, and

Adding both sides, we get
≠
i.e., perimeter of ∆ABC ≠ the perimeter of ∆PQR.

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 –
Let us draw triangles LMN and FGH.

In the above figure, all angles of two triangles are equal. But, out of three sides only two sides are equal.
Hence, ΔLMN is not congruent to ΔFGH.

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

Solution –
By observing the given figure, we can say that
∠ABC = ∠PQR
∠BCA = ∠PRQ
The other additional pair of corresponding part is BC = QR
∴ ΔABC ≅ ΔPQR

10. Explain, why ΔABC ≅ ΔFED

Solution –
From the figure, it is given that,
∠ABC = ∠DEF = 90^o
∠BAC = ∠DFE
BC = DE
By ASA congruence property, two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.
ΔABC ≅ ΔFED