A chord of a circle of radius 15 cm subtends an angle of $60^o$ at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt3 = 1.73$)

Given:

A chord of a circle of radius 15 cm subtends an angle of $60^o$ at the centre.

To do: 

We have to find the areas of the corresponding minor and major segments of the circle.

Solution:

Radius of the circle $r = 15\ cm$

Angle at the centre $\theta = 60^o$

Area of the sector $= \frac{\pi r^2 \theta}{360^o}$

$= \frac{3.14 \times 15\times15\times60^o}{360^o}$

$= 117.75\ cm^2$

Area of the triangle formed by radii and chord $= \frac{1}{2}r^2 \theta$

$= \frac{1}{2}(15)^2\ sin\ 60^o$

$= \frac{1}{2} \times 225 \times \frac{\sqrt3}{2}$

$= 97.31\ cm^2$

Area of the minor segment $=$ Area of the sector $-$ Area of the triangle formed by radii and chord

$= 117.75 - 97.31$

$= 20.44\ cm^2$

Area of the circle $= \pi r^2$

$= 3.14 \times 15^2$

$= 706.5\ cm^2$

Area of the major segment $=$ Area of the circle $-$ Area of the minor segment

$= 706.5 - 20.44$

$= 686.06\ cm^2$