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
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 1

(b) Given : ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 2

(c) Given:
∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ΔLMN ≅ ΔGFH
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 3

(d) Given : EB = DB
∠A = ∠C = 90°
AE = BC
So, ΔABE ≅ ΔCDB
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 4

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.

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

Question 2.
You want to show that ΔART ≅ ΔPEN,
(a) If you have to use SSS criterion, then you need to show
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 5
(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.
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 6
Answer:

Steps Reasons
(i) PM = QM Given
(ii) ∠PMA = ∠QMA Given
(iii) AM = AM Common
(iv) ΔAMP ≅ ΔAMQ By SAS Congruence rule

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

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 ≅ ?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 7
Answer:
We have N ↔T
O ↔ A
W ↔ R
∴ ΔRAT ≅ ΔWON

Question 6.
Complete the congruence statement:
ΔBCA ≅ ?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 8
ΔQRS ≅ ?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 9
(i) We have
A ↔ A
B ↔ B
T ↔ C
∴ ΔBCA ≅ ΔBTA

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

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

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:
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 10
(i) Area of ΔABC = \(\frac { 1 }{ 2 }\) × 4 × 3 sq cm = 6 sq cm
Area of ΔCDE = NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 11 = 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)
(ii)
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 12
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
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 13
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.

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

Question 9.
If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 14
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
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 15
Answer:
ZB = ZE (each 90°)
ZA = ZF (Given)
ZC = ZD
(3<sup>rd</sup> 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)

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