Question

In a circle of radius 12 cm., a chord subtends an angle of ${120}^o$ at the centre. Find the arc of the corresponding minor segment of the circle.

Answer 1

We have, $r=12cm$ and $\theta ={120}^o$

Given segment is $APB$

Now, area of the corresponding segment of circle $=$ Area of minor segment

$=\frac{\theta }{360}\times \pi r^2-\frac{1}{2}{r}^2\sin \theta$

$=\frac{120}{360}\times 3.14\times {12}^2-\frac{1}{2}\times (12{)}^2\times \sin {120}^o$

$.........[\because sin{120}^o=\sin ({180}^o-{60}^o)=\sin {60}^o=\frac{\sqrt{3}}{2}]$

$=\frac{1}{3}\times 3.14\times 144-\frac{1}{2}\times 144\frac{\sqrt{3}}{2}$

$=144[\frac{3.14}{3}-\frac{\sqrt{3}}{4}]=144\left[\frac{12.56-3\sqrt{3}}{12}\right]$

$=12(12.56-3\times 1.73)=12(12.56-5.19)=12\times 7.37=88.44c{m}^2$