SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13C

 Exercise 13 C 

Question 1 

(a) What will be the other angles of a right angles isosceles triangle? 

Sol: $x+x+90=180^{\circ} \Rightarrow 2x=90 \Rightarrow x=45^{\circ} \quad 45^{\circ}$

(b) Can you draw an obtuse angles isosceles triangle ? 

Ans: Yes

Question 2

(a) The vertical angle of an isosceles triangle is 110 degree. What must be the size of its base angles? 

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Sol: $\Rightarrow \quad x+x+110^{\circ}=180^{\circ}$

$\Rightarrow \quad 2 x=180^{\circ}-110^{\circ} \Rightarrow \quad 2 x=70^{\circ}$

x = $35^{\circ}$  Ans

(b) What is the size of each exterior angle of an equilateral triangle ? 

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Sol: $180^{\circ}-60^{\circ}=x$

$x=120^{\circ}$

Question 3

$\triangle A B C$ is isosceles with AB= AC, if $\angle A=80^{\circ}$ what is the measure of angle b?

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Sol: Let $\angle B=\angle C=x$

$\Rightarrow x+x+80^{\circ}=180^{\circ}$

$\Rightarrow 2 x=180^{\circ}-80^{\circ}$

$\Rightarrow 2 x=100^{\circ}$

$\Rightarrow  x=50^{\circ}$

$\angle B=\angle L=50^{\circ}$

Question 4

In $\triangle B B C, \angle A=\angle B=50^{\circ}$. Name the pair of sides which are equal . 

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Side AC = BC 

Question 5

In the figure given below AN =AC, $\angle B A C=52^{\circ}$ $\angle A C K=84^{\circ}$ and BCK is  a straight line . prove that NB =NC

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Sol: $\triangle A B C$

$\angle A+\angle B=84^{\circ}$

$52^{\circ}+\angle B=84^{\circ}$

$\begin{aligned} \angle B &=84^{\circ}-52^{\circ} \\ \Rightarrow \angle B &=32^{\circ} \end{aligned}$

$\triangle A N C$

$\Rightarrow \quad 52^{\circ}+x+x=180^{\circ}$

$\Rightarrow \quad 2 x=180^{\circ}-52^{\circ}$ $2x=123^{\circ}$

$\Rightarrow x=\frac{128}{2} \quad \Rightarrow$

x = $64^{\circ}$

∵ BCK Straight line 

$\angle B C N+\angle A C N+\angle A C K=180^{\circ}$

$\angle B C N+64^{\circ}+84^{\circ}=180^{\circ}$

$\angle B C N=180^{\circ}-148$

$\angle B C N=32^{\circ}$

$\because \angle B C N=\angle B$

BN = NC Hence proved 

Question 6

In the figure AB = AC. Prove that BD = BC

Sol: $\Rightarrow x+x+40^{\circ}=180$

$\Rightarrow 2 x=180^{\circ}-40^{\circ}$

$\Rightarrow 2 x=140^{\circ}$

$\Rightarrow x=\frac{140^{\circ}}{2}$

$\Rightarrow x=70^{\circ}$

By ext. angle prop

$\Rightarrow 30^{\circ}+40^{\circ}=y$

$\Rightarrow y=70^{\circ}$

$\because x=y=70^{\circ}$

$\therefore B D=B C$ Hence proved 

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13B

 Exercise 13 B

Question 1

AB and CD bisect each other at K. Prove that AC = BD .

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Sol: AK= BK

$\angle A K C=\angle B K D$ (Vertically opposite angle )

CK = DK 

By SAS $\triangle A K C, \cong \triangle B K D$

C.P.C.T $\quad A C=B D$ Hence proved 

Question 2

The sides BA and CA have been produced such that BA = AD and CA = AE prove that DE||BC 

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Sol: $A B=A D$

$\angle B A C=\angle E A D$  (Vertically opposite angle )

by $S A S \quad \triangle A B C \cong \triangle A E D$

by C.P.CT $\angle B=\angle D \quad \because B C \| D E$  Hence proved .

Question 3

$O A=O B, O C=O D$, $\angle A O B=\angle C O D$ prove that AC = BD 

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Sol: 

$\begin{aligned} O A &=O B \\ \angle A O B &=\angle C O D \\ \angle A O C+\angle C O B=\angle C O B+\angle B O D \\ \angle A O C &=\angle B O D \\ O C &=O D \end{aligned}$

By SAS $\triangle A O C \cong \triangle B O D$

by C.P.CT $A C=B D$ hence proved 

Question 4

If AB and MN bisect each other at O and AM and BN are $\perp$ on xy. Prove that $\triangle S$, OAM and OBN are congruent and hence prove that AM= BN 

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Sol: $\angle A M O=\angle B N O=90^{\circ}$

$O B=O A$

$O M=O N$

by RHS $\triangle A M O \cong \triangle B N O$ 

by $C \cdot P \cdot c \cdot T \quad A M=B N$

Question 5

$\angle XYZ$ is bisected by YP . L is any point on YP and MLN is Perpendicular to YP. Prove that LM = LN 

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Sol: $\angle YLM=\angle YL N=90^{\circ}$

YL = YL (Common)

$\angle M YL=\angle N YL$ (Bisect by YP )

by ASA $\triangle MY L \cong \triangle N Y L$

By C.P.C.T LM = LN Hence proved 

Question 6

In the fig, PM =PN , PM $\perp A B$ AND $P N \perp A C$ Prove $\triangle A M P \cong \triangle A N P$

Sol: (image to be added)

$\angle A N P=\angle A M P=90^{\circ}$ 

$A P=A P \quad($ Comimon $)$

$M P=N P \quad$ (given)

By R.H.S $\triangle A M P \cong \triangle A N P$ Hence proved 

Question 7

In the figure , AD = BC and AD ||BC. Prove that AB =DC 

Sol: (IMAGE TO BE ADDED)

AD=BC 

$\angle C A D=\angle A C B$ [Alt. angles]

AC = AC 

By ASA $\triangle A C D \cong \triangle A B C$

by C.P.CT. $A B=D C$ Hence proved 

Question 8

In the given figure, triangles ABC and DCB are right angles at A and D respectively and AC = DB . Prove that $\triangle A B C \cong \triangle D C B$

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Sol: $\angle B A C=\angle B D C=90^{\circ}$

$B C=B C=$ Common

$A C=D B$

By RHS $\triangle A B C \cong \triangle D C B$ Hence proved

SChand Composite Mathematics Class 7 Chapter 13 Congruence of Triangle Exercise 13A

 Exercise 13 A

Question 1 

State whether or not the following pairs of triangles are congruent. If they are, give reasons.

(i) (Image to be added)

Sol: $5 \frac{3}{8} \mathrm{~cm}$ = $\frac{43}{8}$ = 5.375 cm 

1 cm = 10mm

$5.375 \times 10$ = 53.75mm

 $4 \frac{5}{8} \mathrm{~cm}$ = 46.25mm

$6 \frac{1}{8} \mathrm{~cm}$ = 61.25mm

(BY SSS Triangle are congruent each other)

(ii) (Image to be added)

Sol: 62mm converted into Cm = 6.2cm

Triangle are not congruent

(iii)   (Image to be added)

Sol: 40 mm = 4cm 

70 mm = 7cm 

Acc SAS triangle are congruent 

(iv)   (Image to be added)

Sol: not congruent 

(v) (Image to be added)

Sol: (Acc ASA Triangle are equal)

(v) (Image to be added)

Sol: By RHS Triangle are Congruent 

Question 2

Show by comparing angles and sides, which of the triangles given here are congruent to each other .

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Question 3

In the following figure, state the condition you would use to show that $\triangle A B C$ and $\triangle C D E$ are congruent . 

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Sol: A C=C E

B C=C D

$\angle A C B=\angle DCE$  (Vertically opposite angle)

 By SAS  $\triangle A B C \cong \triangle CDE$

Question 4

 In the figure, $A B C D$ is a rhombus. Are $\triangle A D C$ and $\triangle A B C$ congruent? What can you say about $\triangle A B D$ and $\triangle B C D$ ?

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Sol: $A D=A B$

$C D=B C$

$A C=A C$

Hence By SSS $\triangle A C D \cong \triangle A B C$

$A D=C D$

$A B=B C$

$B D=B D$

$\triangle A B D \cong \triangle  C B D$ (BY SSS)

Question 5

In this figure, $T S \| P Q$ and $T S=P Q$. Prove that the triangles $P Q R$ and $S T R$ are congruent.

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Sol: $\angle R S T=\angle R P Q$

TS= PQ = 5cm

$\angle R T S=\angle R Q P $ 

(By ASA)

$STR\cong \triangle P Q R$

Question 6

$\triangle P R Q \cong \triangle L M N$. If $P Q=6 \mathrm{~cm}, P R=5 \mathrm{~cm}$ and $\angle P=50^{\circ}$, find $N L$ and $\angle L$ if $L M=5 \mathrm{~cm}$ and $Q R=M N$.

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Sol: $\triangle P R Q \cong \triangle L M N$ (Given )

PR = LM 

PQ = LN =NL = 6cm 

Hence corresponding side and angle of congruent angle are equal 

Question 7

In the diagrams of the following figures. Name the triangle which is congruent to $\triangle A B C$ Keeping the letters in the right order. State the congruence condition also. 

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Question 8

In each of the following name the congruent triangles and state the congruence condition.

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