# A chord of a circle of radius 15 cm subtends an angle of 60º at the centre. Find the areas of the corresponding minor and major segment of the circle. (Use $$\pi \ = \ 3.14$$and$$\sqrt{3} \ = \ 1.73$$)

Here $$r \ = \ 15$$ cm and $$\theta \ = \ 60°$$

$$\therefore$$ Area of the minor segment

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

$$= \ (15)^2[ \frac{3.14 \ × \ 60}{360} \ - \ \frac{1}{2} \ × \ sin60]$$

$$= \ 225[ \frac{3.14}{6} \ - \ \frac{1}{2} \ × \ \frac{ \sqrt{3}}{2}] \$$
$$= \ 225( \frac{3.14 \ × \ 2 \ - \ 3 \sqrt{3}}{12})$$

$$= \ 225( \frac{6.28 \ - \ 3 \ × \ 1.73}{12} ) \$$
$$= \ \frac{225 \ × \ (6.28 \ - \ 5.19)}{12}$$

$$= \ \frac{225 \ × \ 1.09}{12} \$$
$$= \ 20.4375$$cm2

Area of the major segment

= Area of the circle - Area of the minor segment

$$= \ 3.14 \ × \ 225 \ - \ 20.4375 \$$
$$= \ 706.5 \ - \ 20.4375$$

$$= \ 686.0625$$ cm2