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A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment

(ii) major sector (Use \( \pi \ = \ 3.14 \))


Answer :


Here, \( r \ = \ 10 \) cm, \( \theta \ = \ 90° \)

(i) Area of the minor segment

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

\( = \ (10)^2[ \frac{3.14 \ × \ 90}{360} \ - \ \frac{1}{2} sin90°] \)

\(= \ 100(0.785 \ - \ 0.5) \ \)
\( = \ 100 \ × \ 0.285 \ = \ 28.5 \ cm^2 \)

(ii) Area of the major sector

\(= \ \frac{ \theta }{360} \ × \ \pi r^2 \),
where \(r \ = \ 10 \) and \( \theta \ = \ 270° \)

\(= \ \frac{270}{360} \ × \ 3.14 \ × \ 10^2 \)
\( = \ (\frac{3}{4} \ × \ 3.14 \ × \ 100) \ = \ 235.5 \ cm^2 \)

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