Question

Find the area of the segment of circle, given that the angle of the sector is ${120}^o$ and the radius and chord of the circle is $21cm$.(Take $\pi =\frac{22}{7}$)

Answer 1

We know that, area of segment $=$ area of sector $-$ area of $\Delta$

We know that, the area of a sector that subtends an angle $\theta$ at the centre of the circle is $\frac{\theta }{{360}^0}\times \pi r^2$

where, $\theta$ is in degrees.

Also, the $\Delta$ is equilateral.

We know that, area of an equilateral triangle is $\frac{\sqrt{3}}{4}{a}^2$

Hence, area of the segment $=\frac{\theta }{360}\pi r^2-\frac{\sqrt{3}}{4}(21{)}^2$

$=(\frac{120}{360}\times \frac{22}{7}\times (21{)}^2)-\left(\frac{\sqrt{3}}{4}\right(21{)}^2)$

$=\frac{16632}{36}-\frac{441}{4}\sqrt{3}$

$=462-\frac{763.83}{4}$

$=462-190.96$

$=271.04c{m}^2$

Hence, the area of the segment is $271.04c{m}^2$.