Question

An arc of a circle of radius 14 cm subtends an angle of 60${}^o$ at the centre. Find the area of the major sector.

Answer 1

We can draw the above figure by using the information given.

The arc subtends an angle of ${60}^o$ at the centre.

We know that,

Area of a sector of a circle $=\frac{\theta }{{360}^o}\times \pi r^2$

where $\theta$ is the angle subtended by the arc at the centre.

Also, area of a circle $=\pi r^2$

$\therefore$ Area of the given circle $=\pi \times (14{)}^2=\frac{22}{7}\times 14\times 14=616c{m}^2$

and Area of the minor sector $=\frac{{60}^o}{{360}^o}\times \pi \times (14{)}^2$

$\Rightarrow \frac{1}{6}\times \frac{22}{7}\times 14\times 14=102.67c{m}^2$

Area of major sector $=$ Area of circle $-$ Area of minor sector

$\Rightarrow 616-102.67=513.33c{m}^2$

Hence, the area of the major sector is $513.33c{m}^2$.