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# 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