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


Answer :


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

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