In a circle with radius $6\, cm ,$ a minor are subtends an angle of measure $60$ at the centre. Find the area of the minor sector and the major sector corresponding to that arc.

Here, the radius of the circle $r=6 cm$ and the measure of the angle subtended at the centre $\theta=60$.

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

$=\frac{22}{7} \times \frac{6 \times 6 \times 60}{360}$

$=\frac{132}{7} cm ^{2}$

Area of a major sector

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

Area of a major sector

= Area of the circle-Area of the minor sector

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

$=\pi r^{2}\left(1-\frac{\theta}{360}\right)$

$=\frac{22}{7} \times 6 \times 6\left(1-\frac{60}{360}\right)$

$=\frac{22}{7} \times 6 \times 6 \times \frac{5}{6}$

$=\frac{660}{7} cm ^{2}$

Thus, the area of the minor sector is $\frac{132}{7} cm ^{2}$

and the area of the major sector is $\frac{660}{7-2} cm ^{2}$.