Question

A chord AB of a circle, of radius $14$ cm makes an angle of ${60}^{\circ }$ at the centre of the circle. Find the area of the minor segments of the circle.

Answer 1

$\Rightarrow$ Radius $\left(r\right)=14cm=OA=OB$

$\Rightarrow$ $\theta ={60}^o$

In $△AOB,\angle A=\angle B$ [ Angle opposite to the equal sides ]

$\therefore$ $\angle A=\angle B=\angle O={60}^o$

$\therefore$ $△ABO$ is equilateral triangle.

$\Rightarrow$ Area of equilateral $△ABO=\frac{\sqrt{3}}{4}\times (OA{)}^2=\frac{\sqrt{3}}{4}\times (14{)}^2=84.87c{m}^2$

$\Rightarrow$ Area of sector $OABO=\frac{\theta }{{360}^o}\left(\pi r^2\right)$

$\Rightarrow$ Area of sector $OABO=\frac{{60}^o}{{360}^o}\times \frac{22}{7}\times 14\times 14$

$\therefore$ Area of sector $OABO=102.66c{m}^2$

$\Rightarrow$ Area of minor segment = Area of sector $OABO$ - Area of equilateral $△ABO$

$\therefore$ Area of minor segment $=(102.66-84.87)c{m}^2=17.79c{m}^2$