Exercise 15.1
Question 1
(diagram should be added)
length of card $\overline{A B}=2.5 \mathrm{~cm}$
Question 2
(diagram should be added)
Question 3
(diagram should be added)
Given
radius of circle =3cm
$\overline{O B}=5 \mathrm{~cm}$
$\triangle A O B$ is a rightangled triangle
$O B^{2}=O A^{2}+A B^{2}$
$5^{2}=3^{2}+A B^{2}$
$25=9+A B^{2}$
$A B^{2}=25-9 \Rightarrow A B^{2}=16 .$
$A B=4 \mathrm{Cm}$
Question 4
(diagram should be added)
Given
$C P=20 \mathrm{~cm}$
$P T=16 \mathrm{~cm}$
In $\triangle^{l e}$ PTC
$\begin{aligned} P C^{2} &=P T^{2}+T C^{2} \\ 20^{\prime} &=16^{2}+P r^{2} \\ r^{2} &=20^{2}-16^{2} \\ &=400-256 \\ r^{2} &=64 \\ x &=\sqrt{64} \end{aligned}$
$r=8 \mathrm{~cm}$
Question 5
(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)
(diagram should be added)
$\begin{aligned} 112^{\circ}+\angle A B C &=180^{\circ} \\ \angle A B C &=180-122^{\circ} \\ & \angle A B C=58^{\circ} \end{aligned}$
$x+\angle A B C=90^{\circ}$ $\left(\therefore\right.$ Anglein semi circle $\left.=90^{\circ}\right)$
$x=90-\angle A B C$
$x=90-58$
$x=32^{\circ}$
(v)
(diagram should be added)
$x+40^{\circ}=90$ (∵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) (diagram should be added)
$\operatorname{In} \triangle A B C$
$\begin{aligned} \angle A B C=\angle B A C &=x \\ &(: B C=A C) \end{aligned}$
$\angle A B C+\angle B A C+\angle A C B=180^{\circ}$ $\left(\because \angle A C B=90^{\circ}\right)$
$x+x+90=180$
$2 x= 90$
$x=\frac{90}{2}$
$x=45^{\circ}$
$\begin{aligned} \angle A B C+\angle C B D &=180^{\circ} \\ 45+\angle C B D &=180^{\circ} \\ \angle C B D &=180-45 \\ \angle C B D &=135^{\circ} \end{aligned}$
$\angle B C D=\angle B D C=y(\because B D=B C)$
$\begin{aligned} \angle B C D+\angle B D C &+\angle C D B=180^{\circ} \\ y+y+195 &=180 \\ 2 y &=45 \\ y &=22.5^{\circ} \end{aligned}$
(vii) (daigram should be added)
$\begin{aligned} x+65 &=90^{\circ} \\ x &=90-65 \\ x &=25^{\circ} \end{aligned}$
(viii) (daigram should be added)
$\angle B A C=\angle A B C$ $(\because B C=A C)$
$x=y$
$x+y=90$
$x+x=90$
$2 x=90$
$x=45^{\circ}$
(ix) (daigram should be added)
$\angle O C A=90^{\circ}$
$\begin{aligned} x+36 &=90^{\circ} \\ x &=90-36^{\circ} \\ x &=54^{\circ} \end{aligned}$
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