A circle with a radius of  $21 \mathrm{cm}$ has an angle of $60°$at its centre. Find the area of the sector formed.

Given: $\mathrm{r}=21 \mathrm{cm} \mathrm{and} \mathrm{\theta }={60}^{\mathrm{o}}$

Now, if r is the circle's radius, then the area of the sector of the angle $=\frac{\mathrm{\theta }}{{360}^{°}}\times {\mathrm{\pi r}}^2$

$=\frac{60}{360}\times \frac{22}{7}\times 21\times 21$

$=231 {\mathrm{cm}}^2$