A chord AB of a circle, of radius 14 cm makes an angle
Question:

A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle. (Use π = 22/7)

Solution:

We know that the area of minor segment of angle θ in a circle of radius r is,

$A=\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}$

It is given that,

$r=14 \mathrm{~cm}$

$\theta=60^{\circ}$

Substituting these values in above formula

$A=\left\{\frac{3.14 \times 60^{\circ}}{360^{\circ}}-\sin \frac{60^{\circ}}{2} \cos \frac{60^{\circ}}{2}\right\} \times 14 \times 14$

$=\left\{\frac{3.14}{6}-\sin 30^{\circ} \cos 30^{\circ}\right\} \times 196$

$=\frac{3.14 \times 196}{6}-\frac{1}{2} \times \frac{\sqrt{3}}{2} \times 196$

$=102.573-84.868$

$A=17.70 \mathrm{~cm}^{2}$