Question

A chord PQ of a circle with a radius of 15 cm subtends an angle of ${60}^{\circ }$ with the center of the circle. Find the area of the minor as well as the major segment. $(\pi =3.14,\sqrt{3}=1.73)$

Answer 1

The radius of the circle, $r=15cm$
Let O be the center and PQ be the chord of the circle.

$\angle POQ=\theta ={60}^{\circ }$

Area of the minor segment = Area of the shaded region

$=r^2(\frac{\pi \theta }{360°}-\frac{\sin \theta }{2})$
$={\left(15\right)}^2\times (\frac{3.14\times 60°}{360°}-\frac{\sin 60°}{2})$
$=225\times \frac{3.14}{6}-225\times \frac{\sqrt{3}}{4}$
$=117.75-97.31$
$=20.44{cm}^2$

Now,
Area of the circle =

$[\pi r^2=3.14\times {\left(15\right)}^2=3.14\times 225]=706.5c{m}^2$
∴ Area of the major segment = Area of the circle − Area of the minor segment $=706.5-20.44=686.06c{m}^2$
Thus, the areas of the minor segment and major segment are $20.44c{m}^2$ and $686.06c{m}^2$, respectively.