Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
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 \( \theta \ = \ 270° \)
\(= \ \frac{270}{360} \ × \ 3.14 \ × \ 10^2 \)
\( = \ (\frac{3}{4} \ × \ 3.14 \ × \ 100) \ = \ 235.5 \ cm^2 \)