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

Given: Radius of the circle $r = 6 \, cm$ and the central angle $\theta = 60^{\circ}$.
Area of the minor sector $= \frac{\pi r^2 \theta}{360^{\circ}}$
$= \frac{22}{7} \times 6 \times 6 \times \frac{60}{360}$
$= \frac{22}{7} \times 36 \times \frac{1}{6}$
$= \frac{22 \times 6}{7} = \frac{132}{7} \, cm^2 \approx 18.86 \, cm^2$.
Area of the major sector $= \text{Area of the circle} - \text{Area of the minor sector}$
$= \pi r^2 - \frac{132}{7}$
$= \frac{22}{7} \times 6 \times 6 - \frac{132}{7}$
$= \frac{792}{7} - \frac{132}{7} = \frac{660}{7} \, cm^2 \approx 94.29 \, cm^2$.
Thus,the area of the minor sector is $\frac{132}{7} \, cm^2$ and the area of the major sector is $\frac{660}{7} \, cm^2$.