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

These NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 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
∠A = ∠C = 90°
AE = BC
So, ΔABE ≅ ΔCDB

Answer:
(a) Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
By SSS Congruence Criterion

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

(c) Given:
∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
ΔLMN ≅ ΔGFH
By ASA Congruence Criterion

(d) Given: EB = DB
AE = BC
∠A = ∠C = 90°
So, ΔABE ≅ ΔCDB
By RHS Congruence Criterion.

Question 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) ∠RAT =
(ii) ∠ATR =
Answer:
ΔART ≅ ΔPEN
(a) (i) AR = PE
(ii) RT = EN
(iii) AT = PN

(b) Given: ∠T = ∠N
(i) RT = EN
(ii) PN = AT

(c) (i) ∠RAT = ∠EPN
(ii) ∠ATR = ∠PNE

Question 3.
You have to show that ΔAMP = AMQ. In the following proof, provide the missing reasons.

Answer:

Question 4.
In ΔABC, ∠A =30°, ∠B = 40° and ∠C= 110°
In APQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A Student says that ΔABC = ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
Answer:
No, he is not justified because AAA is not a congruence criterion.

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

Answer:
We have N ↔T
O ↔ A
W ↔ R
∴ ΔRAT ≅ ΔWON

Question 6.
Complete the congruence statement:
ΔBCA ≅ ?

(i) We have
A ↔ A
B ↔ B
T ↔ C
∴ ΔBCA ≅ ΔBTA

(ii) We have
R ↔ P
Q ↔ T
S ↔ Q
∴ ΔQRS ≅ ΔTPQ

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?
Answer:

(i) Area of ΔABC = 1/2 × 4 × 3 sq cm = 6 sq cm
Area of ΔCDE =  = 6 sq cm
Perimeter of ΔABC = (3 + 4 + 5) cm = 12 cm
Perimeter of ΔCDE = (3 + 4 + 5) cm = 12 cm
The two triangles are congruent.
(∵ Perimeter of ΔABC = Perimeter of ΔCDE)

Perimeter of ΔPQR = (3 + 4 + 5) cm = 12cm
Perimeter of ΔPRS = (4 + 3.5 + 4) cm = 11.5 cm
∴ The two triangles are not congruent.
(∵ Perimeter of ΔPRS ≠ Perimeter of ΔPQR)

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.
Answer:
In ΔABC and ΔDEF

AB = 2 cm DF = 2cm
∴ AB = DF
BC = 4 cm, ED = 4 cm
∴ BC = ED
AC = 3 cm, EF = 3 cm
∴ AC = EF
∠BAC = ∠EDF
∠ABC = ∠DEF
But ΔABC is not congruent to ΔDEF.

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

Answer:
Given ΔABC = ΔPQR
∴ A ↔ P; B ↔ Q and C ↔ R
Two angles ∠B and ∠C of ΔABC are respectively equal to two angles ∠Q and
∠R of ΔPQR
If BC = QR then ΔABC ≅ ΔPQR (using ASA congruence criterion)
We use ASA congruence criterion.

Question 10.
Explain why AABC = AFED

Answer:
ZB = ZE (each 90°)
ZA = ZF (Given)
ZC = ZD
(3^rd angle are equal) BC = ED (Given)
Two angles ZB and ZC and included side BC of AABC are respectively equal to the angle ZE and ZD and the included side ED of ADEF.
∴ ΔABC ≅ ΔFED (ASA)