ML AGGARWAL CLASS 8 Chapter 15 Circle Exercise 15.1

 Exercise 15.1

Question 1

Sol :











length of card $\overline{A B}=2.5 \mathrm{~cm}$


Question 2

Sol :










Question 3

Sol :







Given 

radius of circle =3cm

$\overline{O B}=5 \mathrm{~cm}$

$\triangle A O B$ is a right angled triangle

$O B^{2}=O A^{2}+A B^{2}$

$5^{2}=3^{2}+A B^{2}$

$25=9+A B^{2}$

$A B^{2}=25-9 \Rightarrow A B^{2}=16 .$

AB=4 cm

Question 4

Sol :







Given

CP=20 cm
PT=16 cm

In $\triangle^{l e}$ PTC

$\begin{aligned} P C^{2} &=P T^{2}+T C^{2} \\ 20^{\prime} &=16^{2}+P r^{2} \\ r^{2} &=20^{2}-16^{2} \\ &=400-256 \\ r^{2} &=64 \\ x &=\sqrt{64} \end{aligned}$

r=8 cm

Question 5

Sol :

(i) $x=90-32=58^{\circ}$ $\left(\because\right.$ Angle in semi circle $\left.=90^{\circ}\right)$

$y=90-50=40^{\circ} \quad\left(\therefore\right.$ Angle in Semicircle $\left.=90^{\circ}\right)$



(ii) $y=90-37^{\circ} \quad(\because$ Angle in rectangle $)$

$y=53^{\circ}$

$x=90-53^{\circ}=37^{\circ}$



(iii) $2 x=90^{\circ} \quad\left(\because\right.$ Angle in semi circle $\left.=90^{\circ}\right)$
$x=45^{\circ}$



(iv) 







$\begin{aligned} 112^{\circ}+\angle A B C &=180^{\circ} \\ \angle A B C &=180-122^{\circ} \\ & \angle A B C=58^{\circ} \end{aligned}$

$x+\angle A B C=90^{\circ}$ $\left(\therefore\right.$ Anglein semi circle $\left.=90^{\circ}\right)$

$x=90-\angle A B C$

$x=90-58$

$x=32^{\circ}$



(v) 








$x+40^{\circ}=90$ (∵Angle in semi circle 90')
x=90-40
$x=50^{\circ}$

In$\Delta$ $A B C$  (∵ Sum of angle is quadrilateral is equal to $360^{\circ}$)

$\begin{aligned} x+y+90+90=& 360 \\ x+y+180 &=360 \\ x+y &=180 \\ 50+y &=180 \\ & y=180-50 \\ & y=130^{\circ} \end{aligned}$



(vi) 











$\operatorname{In} \triangle A B C$

$\begin{aligned} \angle A B C=\angle B A C &=x \\ &(: B C=A C) \end{aligned}$

$\angle A B C+\angle B A C+\angle A C B=180^{\circ}$  $\left(\because \angle A C B=90^{\circ}\right)$

x+x+90=180

2 x= 90

$x=\frac{90}{2}$

$x=45^{\circ}$

$\begin{aligned} \angle A B C+\angle C B D &=180^{\circ} \\ 45+\angle C B D &=180^{\circ} \\ \angle C B D &=180-45 \\ \angle C B D &=135^{\circ} \end{aligned}$

$\angle B C D=\angle B D C=y(\because B D=B C)$

$\begin{aligned} \angle B C D+\angle B D C &+\angle C D B=180^{\circ} \\ y+y+195 &=180 \\ 2 y &=45 \\ y &=22.5^{\circ} \end{aligned}$



(vii) 







$\begin{aligned} x+65 &=90^{\circ} \\ x &=90-65 \\ x &=25^{\circ} \end{aligned}$



(viii) 










$\angle B A C=\angle A B C$  $(\because B C=A C)$

x=y

x+y=90

x+x=90

2 x=90

 $x=45^{\circ}$



(ix) 










$\angle O C A=90^{\circ}$

$\begin{aligned} x+36 &=90^{\circ} \\ x &=90-36^{\circ} \\ x &=54^{\circ} \end{aligned}$

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