Find the area of the minor segment of a circle of radius $ 14 \mathrm{~cm} $, when the angle of the corresponding sector is $ 60^{\circ} . $

Given:

Radius of the circle $=14\ cm$.

The angle of the corresponding sector $=60^{\circ}$.

To do:

We have to find the area of the minor segment of the circle.

Solution:

Let $AB$ be the chord of the circle and $O$ the centre of the circle.

In $\Delta \mathrm{AOB}$,

$\mathrm{OA}=\mathrm{OB}=r$

This implies,

$\Delta \mathrm{AOB}$ is an isosceles triangle.

Let $\angle \mathrm{OAB}=\angle \mathrm{OBA}=\theta$

In $\Delta \mathrm{OAB}$, 

$\angle \mathrm{AOB}+\angle \mathrm{OAB}+\angle \mathrm{OBA}=180^{\circ}$

$\Rightarrow 60^{\circ}+\theta+\theta=180^{\circ}$

$\Rightarrow 2 \theta=120^{\circ}$

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

Therefore,

$\angle \mathrm{OAB}=\angle \mathrm{OBA}=60^{\circ}$

This implies,

$\triangle \mathrm{AOB}$ is an equilateral triangle.

$\mathrm{OA}=\mathrm{OB}=\mathrm{AB}=14 \mathrm{~cm}$

Area of $\Delta O A B=\frac{\sqrt{3}}{4}(14)^{2}$

$=\frac{\sqrt{3}}{4} \times 196$

$=49 \sqrt{3} \mathrm{~cm}^{2}$

Area of the sector $\mathrm{OBAO}=\frac{\theta}{360^{\circ}} \times \pi r^{2} $

$=\frac{22}{7} \times \frac{60^{\circ}}{360} \times 196$

$=\frac{22 \times 2 \times 14}{6}$

$=\frac{22 \times 14}{3}$

$=\frac{308}{3} \mathrm{~cm}^{2}$

Area of the minor segment $=$ Area of sector $OBAO -$ Area of $\Delta OAB$

$=(\frac{308}{3}-49 \sqrt{3}) \mathrm{cm}^{2}$

The area of the minor segment is $(\frac{308}{3}-49 \sqrt{3}) \mathrm{cm}^{2}$.