Gujarat Board 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, ∆ABC ≅ ∆DEF

(b) Given:

ZX = RP

RQ = ZY

∠PRQ = ∠XZY

So, ∆PQR = ∆XYZ

(c) Given: ∠MLN = ∠FGH

∠NML = ∠GFH

ML = FG

So, ∆LMN ≅ ∆GFH

(d) Given: EB = DB

AE = BC

∠A = ∠C = 90°

So, ∆ABE ≅ ∆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 ∆ART ≅ ∆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 ∠T = ∠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) ∠ATR =

(ii) ∠TAR =

Solution:

Here ∆ART ≅ ∆PEN

∴ A ↔ P, R ↔ E and T ↔ N

(a) (i) AR = PE

(ii) RT = EN

(iii) AT = PN

(b) ∵ ∠T = ∠N

(i) RT = EN

(ii) PN = AT

(c) (i) ∠ATR = ∠PNE

(ii) ∠TAR = ∠NPE

Question 3.

You have to show that ∆AMP ≅ ∆AMQ. In the following proof, supply the missing reasons.

Solution:

Question 4.

In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. In ∆PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°. A student says that ∆ABC ≅ 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 ∆RAT ≅?

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\}\) ⇒ ∆RAT ≅ ∆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 ∆ABC = \(\frac { 1 }{ 2 }\) x 4 x 3 = 6 sq. cm

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

Perimeter of ∆ABC = (3 + 4 + 5) cm = 12 cm

Perimeter of ∆CDE = (3 + 4 + 5) cm = 12 cm

The two triangles are congruent.

[Perimeter of ∆ABC] = [Perimeter of ∆CDE]

(ii) Area of ∆PQR = Area of ∆PRS

Perimeter of ∆PQR = (3 + 4 + 5) cm = 12 cm

Perimeter of ∆PRS = (4 + 3.5 + 4) cm = 1.5 cm

The two triangles are not congruent.

[Perimeter of ∆PQR] ≠ [Perimeter of ∆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:

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

Question 9.

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

Solution:

Here ∆ABC ≅ ∆PQR

∴ A ↔ P, B ↔ Q and C ↔ R

Two angles ∠B and ∠C of ∆ABC are respectively equal to two angles ∠Q and ∠R of ∆PQR.

∴ BC = QR

We use the ASA congruence criterion

Question 10.

Explain, why ∆ABC ≅ ∆FED.

Solution:

∵ ∠A = ∠F (Given)

∴ ∠C = ∠D (Third angles are equal)

Also, BC = ED (Given)

Two angles (∠B and ∠C) and included side BC of ∆ABC are respectively equal to two angles (∠E and ∠D) and the included side ED of

∴ ∆ABC ≅ ∆FED.