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Answer :
Area of the circular portion
= Area of the circle - Area of the sector
\(= \ \pi r^2 \ - \ \frac{60}{360} \ \pi \ r^2 \ = \ \pi r^2(1 \ - \ \frac{1}{6}) \)
\(= \ \frac{5}{6} \pi r^2 \), where r = 6
\(= \ \frac{5}{6} \ × \ \frac{22}{7} \ × \ 36 \ = \ \frac{660}{7} \) cm2
Area of the equilateral \( ∆ \ OAB \)
\( = \ \frac{ \sqrt{3}}{4}(side)^2 \ = \ ( \frac{ \sqrt{3}}{4} \ × \ 144) \)
\( = \ 36 \sqrt{3} \)cm2
\(\therefore \) Area of the shaded region
\(= \ ( \frac{660}{7} \ + \ 36 \sqrt{3}) \) cm2