# A chord of a circle of radius 12 cm subtends an angle of 120º at the centre. Find the area of the corresponding segment of the circle. (Use $$\pi \ = \ 3.14$$ and $$\sqrt{3} \ = \ 1.73$$)

Here$$r \ = \ 12$$ cm and $$\theta \ = \ 120°$$

Area of the corresponding segment of the circle

= Area of the minor segment

$$= \ r^2[ \frac{ \pi \theta}{360} \ - \ \frac{1}{2}sin\theta]$$

$$= \ (12)^2[ \frac{3.14 \ × \ 120}{360} \ - \ \frac{1}{2} \ × \ sin120]$$

$$= \ 144[ \frac{3.14}{3} \ - \ \frac{1}{2} \ × \ \frac{ \sqrt{3}}{2}] \$$
$$= \ 225( \frac{3.14 \ × \ 2 \ - \ 3 \sqrt{3}}{12})$$
[ $$\therefore sin120° \ = \ sin(180° \ - \ 60°) \$$
$$= \ sin60° \ = \ \frac{\sqrt{3}}{2}$$]

$$= \ 144 \ × \ \frac{3.14 \ × \ 4 \ - \ 3 \sqrt{3}}{12} \$$
$$= \ 12 \ × \ (12.56 \ - \ 3 \ × \ 1.73)$$

$$= \ 12 \ × \ (12.56 \ - \ 5.19) \$$
$$= \ 12 \ × \ 7.37 \ = \ 88.44$$cm2