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)
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
$A=\left\{\frac{3.14 \times 60^{\circ}}{360^{\circ}}-\sin \frac{60^{\circ}}{2} \cos \frac{60^{\circ}}{2}\right\} \times 14 \times 14$
$=\left\{\frac{3.14}{6}-\sin 30^{\circ} \cos 30^{\circ}\right\} \times 196$
$=\frac{3.14 \times 196}{6}-\frac{1}{2} \times \frac{\sqrt{3}}{2} \times 196$
$=102.573-84.868$
$A=17.70 \mathrm{~cm}^{2}$
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