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In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region).


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

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