A chord of a circle of radius 15 cm subtends an angle of 60° at the centre.

Solution:

Here, radius (r) = 15 cm and

Sector angle $(\theta)=60^{\circ}$

$\therefore \quad$ Area of the sector

$=\frac{\theta}{\mathbf{3 6 0}^{\circ}} \times \pi \mathbf{r}^{\mathbf{2}}=\frac{\mathbf{6 0}^{\circ}}{\mathbf{3 B 0}^{\circ}} \times \frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 15 \times 15 \mathrm{~cm}^{2}$

$=\frac{\mathbf{1 1 7 7 5}}{\mathbf{1 0 0}} \mathrm{cm}^{2}=117.75 \mathrm{~cm}^{2}$

Since $\angle \mathrm{O}=60^{\circ}$ and $\mathrm{OA}=\mathrm{OB}=15 \mathrm{~cm}$

$\therefore \quad$ AOB is an equilateral triangle.

$\Rightarrow \mathrm{AB}=15 \mathrm{~cm}$ and $\angle \mathrm{A}=60^{\circ}$

Draw $\mathrm{OM} \perp \mathrm{AB}$, in $\triangle \mathrm{AMO}$

$\therefore \quad \frac{\mathbf{O M}}{\mathbf{O A}}=\sin 60^{\circ}=\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$

$\Rightarrow \mathrm{OM}=\mathrm{OA} \times \frac{\sqrt{\mathbf{3}}}{\mathbf{2}}=\frac{\mathbf{1 5} \sqrt{\mathbf{3}}}{\mathbf{2}} \mathrm{cm}$

Now, $\operatorname{ar}(\Delta \mathrm{AOB})=\frac{\mathbf{1}}{\mathbf{2}} \times \mathrm{AB} \times \mathrm{OM}$

$=\frac{1}{2} \times 15 \times 15 \frac{\sqrt{3}}{2} \mathrm{~cm}^{2}=\frac{225 \sqrt{3}}{4} \mathrm{~cm}^{2}$

$=\frac{225 \times 1.73}{4} \mathrm{~cm}^{2}=97.3125$

$=\frac{225 \times 1.73}{4} \mathrm{~cm}^{2}=97.3125$

Now area of the minor segment

$=($ Area of minor sector $)-(\operatorname{ar} \Delta \mathrm{AOB})$

$=(117.75-97.3125) \mathrm{cm}^{2}=20.4375 \mathrm{~cm}^{2}$

Area of the major segment

= [Area of the circle] – [Area of the minor segment]

$=\pi \mathrm{r}^{2}-20.4375 \mathrm{~cm}^{2}=\left[\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times \mathbf{1 5}^{\mathbf{2}}\right]-20.4375 \mathrm{~cm}^{2}=706.5-20.4375 \mathrm{~cm}^{2}=686.0625 \mathrm{~cm}^{2}$