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Answer :
Let ABC be an equilateral triangle and let O be the circumcentre of the circumcircle of radius 32 cm.
Area of the circle \( = \ \pi r^2 \)
\( = \ \frac{22}{7} \ × \ (32)^2 \ = \ \frac{22528}{7} \) cm2
Area of \( ∆ \ ABC \) \(= \ 3 × \) Area of \(∆ \ BOC \)
\( = \ 3 \ × \ \frac{1}{2} \ × \ OB \ × \ OC \ × \ sinBOC \) \(= \ \frac{3}{2} \ × \ 32 \ × \ 32 \ × \ sin120° \)
\( = \ 3 \ × \ 16 \ × \ 32 \ × \ \frac{ \sqrt{3}}{2} \ = \ 768 \sqrt{3} \)cm2
\(\therefore \) Area of the design
(i.e., shaded region)
= Area of the circle - Area of ∆ ABC
\(= \ ( \frac{22528}{7} \ - \ 768 \sqrt{3}\)) cm2