# In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.

Question:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chor

Solution:

Here, radius $=21 \mathrm{~cm}$ and $\theta=60^{\circ}$

(i) Circumference of the circle $=2 \pi \mathrm{r}$

$=2 \times \frac{22}{7} \times 21 \mathrm{~cm}=2 \times 22 \times 3 \mathrm{~cm}=132 \mathrm{~cm}$

$\therefore \quad$ Length of $\operatorname{arc}$ APB

$=\frac{\theta}{360^{\circ}} \times \mathbf{2} \pi \mathbf{r}=\frac{\mathbf{6 0}^{\circ}}{\mathbf{3 C D}^{\circ}} \times 132 \mathrm{~cm}$

$=\frac{\mathbf{1}}{\mathbf{6}} \times 132 \mathrm{~cm}=22 \mathrm{~cm}$

(ii) Area of the sector with sector angle 60°

$=\frac{\mathbf{6 0}^{\circ}}{\mathbf{3 B 0}^{\circ}} \times \pi \mathbf{r}^{\mathbf{2}}=\frac{\mathbf{6 0}^{\circ}}{\mathbf{3 6 0}^{\circ}} \times \frac{22}{\mathbf{7}} \times 21 \times 21 \mathrm{~cm}^{2}$

$=11 \times 21 \mathrm{~cm}^{2}=231 \mathrm{~cm}^{2}$

(iii) Area of the segment $\mathrm{APB}=[$ Area of the sector $\mathrm{AOB}]-[$ Area of $\triangle \mathrm{AOB}$ ]..(1)

In $\Delta \mathrm{AOB}, \mathrm{OA}=\mathrm{OB}=21 \mathrm{~cm}$

$\therefore \angle \mathrm{A}=\angle \mathrm{B}=60^{\circ} \quad\left[\because \angle \mathrm{O}=60^{\circ}\right]$

$\Rightarrow \mathrm{AOB}$ is an equilateral $\Delta$.

$\therefore \quad \mathrm{AB}=21 \mathrm{~cm}$

$\therefore \quad$ area of $\Delta \mathrm{AOB}=\frac{\sqrt{3}}{4}(\text { side })^{2}$

$=\frac{\sqrt{3}}{4} \times 21 \times 21 \mathrm{~cm}^{2}=\frac{441 \sqrt{3}}{4} \mathrm{~cm}^{2} ...(2)$

From (1) and (2), we have

Area of segment $=\left[231 \mathrm{~cm}^{2}\right]-\left[\frac{\mathbf{4 4 1} \sqrt{\mathbf{3}}}{4} \mathbf{c m}^{2}\right]=\left(231-\frac{441 \sqrt{3}}{4}\right) \mathrm{cm}^{2}$