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