Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

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