Question

A chord of a circle of radius 14 cm subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle.

Hint:

For finding the area of a circular segment, always subtract the area of the triangle from the area of the sector.

Answer 1

Step 1: Area of sector
The angle at the centre is $120^\circ$.
Area of sector $OAB = \dfrac{\theta}{360^\circ} \times \pi r^2$
$= \dfrac{120}{360} \times \pi \times (14)^2$
$= \dfrac{1}{3} \times \pi \times 196$
$= \dfrac{196\pi}{3}$ cm$^2$

Step 2: Area of triangle $OAB$
Here, $\triangle OAB$ is an isosceles triangle with $OA = OB = 14$ cm and $\angle AOB = 120^\circ$.
Area of $\triangle OAB = \dfrac{1}{2} \times OA \times OB \times \sin(120^\circ)$
$= \dfrac{1}{2} \times 14 \times 14 \times \dfrac{\sqrt{3}}{2}$
$= 49\sqrt{3}$ cm$^2$

Step 3: Area of segment
Area of segment = Area of sector $-$ Area of triangle
$= \dfrac{196\pi}{3} - 49\sqrt{3}$ cm$^2$
\[ \boxed{\dfrac{196\pi}{3} - 49\sqrt{3} \ \text{cm}^2} \]