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, āABC ā
āDEF

(b) Given:

ZX = RP

RQ = ZY

ā PRQ = ā XZY

So, āPQR = āXYZ

(c) Given: ā MLN = ā FGH

ā NML = ā GFH

ML = FG

So, āLMN ā
āGFH

(d) Given: EB = DB

AE = BC

ā A = ā C = 90Ā°

So, āABE ā
āCDB

Solution:

(a) SSS congruence criterion

(b) SAS congruence criterion

(c) ASA congruence criterion

(d) RHS congruence criterion

Question 2.

You want to show that āART ā
āPEN,

(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 āART ā
āPEN

ā“ A ā P, R ā E and T ā N

(a) (i) AR = PE

(ii) RT = EN

(iii) AT = PN

(b) āµ 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 āAMP ā
āAMQ. In the following proof, supply the missing reasons.

Solution:

Question 4.

In āABC, ā A = 30Ā°, ā B = 40Ā° and ā C = 110Ā°. In āPQR, ā P = 30Ā°, ā Q = 40Ā° and ā R = 110Ā°. A student says that āABC 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 āRAT 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 āRAT ā
āWON

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 āABC = \(\frac { 1 }{ 2 }\) x 4 x 3 = 6 sq. cm

Area of āCDE = \(\frac { 1 }{ 2 }\) x 4 x 3 = 6 sq. cm

Perimeter of āABC = (3 + 4 + 5) cm = 12 cm

Perimeter of āCDE = (3 + 4 + 5) cm = 12 cm

The two triangles are congruent.

[Perimeter of āABC] = [Perimeter of āCDE]

(ii) Area of āPQR = Area of āPRS

Perimeter of āPQR = (3 + 4 + 5) cm = 12 cm

Perimeter of āPRS = (4 + 3.5 + 4) cm = 1.5 cm

The two triangles are not congruent.

[Perimeter of āPQR] ā [Perimeter of āPRS]

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:

āABC and āDEF are not congruent as any two sides and angle included between these two sides of āABC is not equal to the corresponding two sides and included angle of āDEF.

Question 9.

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

Solution:

Here āABC ā
āPQR

ā“ A ā P, B ā Q and C ā R

Two angles ā B and ā C of āABC are respectively equal to two angles ā Q and ā R of āPQR.

ā“ BC = QR

We use the ASA congruence criterion

Question 10.

Explain, why āABC ā
āFED.

Solution:

āµ 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 āABC are respectively equal to two angles (ā E and ā D) and the included side ED of

ā“ āABC ā
āFED.