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 \( \sqrt{3} \ = \ 1.73 \))


Answer :


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

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