# GSEB Solutions Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Question 1.
Which congruence criterion do you use in the following?
(a) Given:
AC = DF
AB = DE
BC = EF
So, āABC ā āDEF

(b) Given:
ZX = RP
RQ = ZY
ā PRQ = ā XZY
So, āPQR = āXYZ

(c) Given: ā MLN = ā FGH
ā NML = ā GFH
ML = FG
So, āLMN ā āGFH

(d) Given: EB = DB
AE = BC
ā A = ā C = 90Ā°
So, āABE ā āCDB

Solution:
(a) SSS congruence criterion
(b) SAS congruence criterion
(c) ASA congruence criterion
(d) RHS congruence criterion

Question 2.
You want to show that āART ā ā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 ā T = ā N and you are to use SAS criterion, you need to have

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have P

(i) ā ATR =
(ii) ā TAR =
Solution:
Here āART ā āPEN
ā“ A ā P, R ā E and T ā N

(a) (i) AR = PE
(ii) RT = EN
(iii) AT = PN

(b) āµ ā T = ā N
(i) RT = EN
(ii) PN = AT

(c) (i) ā ATR = ā PNE
(ii) ā TAR = ā NPE

Question 3.
You have to show that āAMP ā āAMQ. In the following proof, supply the missing reasons.

Solution:

Question 4.
In āABC, ā A = 30Ā°, ā B = 40Ā° and ā C = 110Ā°. In āPQR, ā P = 30Ā°, ā Q = 40Ā° and ā R = 110Ā°. A student says that āABC ā APQR by AAA congruence criterion. Is he justified? Why or why not?
Solution:
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 āRAT ā?

Solution:
We have $$\left.\begin{array}{l} \mathrm{O} \leftrightarrow \mathrm{A} \\ \mathrm{N} \leftrightarrow \mathrm{T} \\ \mathrm{W} \leftrightarrow \mathrm{R} \end{array}\right\}$$ ā āRAT ā āWON

Question 6.
Complete the congruence statement:

Solution:
(i) We have:
$$\left.\begin{array}{l} A \leftrightarrow A \\ B \leftrightarrow B \\ T \leftrightarrow C \end{array}\right\} \Rightarrow \Delta B C A \cong \Delta B T A$$

(ii) $$\left.\begin{array}{l} R \leftrightarrow P \\ Q \leftrightarrow T \\ S \leftrightarrow Q \end{array}\right\} \Rightarrow \Delta Q R S \cong \Delta T P Q$$

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) Area of āABC = $$\frac { 1 }{ 2 }$$ x 4 x 3 = 6 sq. cm
Area of āCDE = $$\frac { 1 }{ 2 }$$ x 4 x 3 = 6 sq. cm
Perimeter of āABC = (3 + 4 + 5) cm = 12 cm
Perimeter of āCDE = (3 + 4 + 5) cm = 12 cm
The two triangles are congruent.
[Perimeter of āABC] = [Perimeter of āCDE]

(ii) Area of āPQR = Area of āPRS
Perimeter of āPQR = (3 + 4 + 5) cm = 12 cm
Perimeter of āPRS = (4 + 3.5 + 4) cm = 1.5 cm
The two triangles are not congruent.
[Perimeter of āPQR] ā  [Perimeter of āPRS]

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:
A pair of triangle with 3 equal angles and two equal sides are non-congruent are as follows:

āABC and āDEF are not congruent as any two sides and angle included between these two sides of āABC is not equal to the corresponding two sides and included angle of āDEF.

Question 9.
If āABC and āPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:
Here āABC ā āPQR
ā“ A ā P, B ā Q and C ā R
Two angles ā B and ā C of āABC are respectively equal to two angles ā Q and ā R of āPQR.
ā“ BC = QR
We use the ASA congruence criterion

Question 10.
Explain, why āABC ā āFED.

Solution:
āµ ā A = ā F (Given)
ā“ ā C = ā D (Third angles are equal)
Also, BC = ED (Given)
Two angles (ā B and ā C) and included side BC of āABC are respectively equal to two angles (ā E and ā D) and the included side ED of
ā“ āABC ā āFED.