Question

A chord of a circle of radius 12 cm subtends an angle of ${120}^{\circ }$ at the center. Find the area of the corresponding segment of the circle.

Answer 1

REF.Image.

Given$\to$

Radius = 12 cm

$\angle AOB={120}^{\circ }$

Area of segement (APB)

= Area of sector APBO

- Area of triangle AOB

$=\frac{\theta }{{360}^{\circ }}\times \pi {\mu }^2-\frac{1}{2}{\mu }^{2}sin\theta$

$=\frac{120}{360}\times \pi (12{)}^2-\frac{1}{2}(12{)}^{2}sin{120}^{\circ }$

$=\frac{144\times 22}{7\times 3}-\frac{1}{2}\times 144\times \frac{\sqrt{3}}{2}$

$=(150.85-62.35)c{m}^2$

$=88.40c{m}^2$

Area of $△AOB=2\times (\frac{1}{2}\times OD\times DB)$

$=2\times \frac{1}{2}\left(rcos\frac{\theta }{2}\right)r\left(sin\frac{\theta }{2}\right)$

$=\frac{1}{2}\times \left(2cos\frac{\theta }{2}sin\frac{\theta }{2}\right)r^2$

$=\frac{1}{2}{r}^{2}sin\theta$

$[\because sin2\theta =2sin\theta cos\theta ]$