In the given figure,ABCD is a square. EFGH is | vedclass

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(B) Area of the circle $= 38.5 	ext{ cm}^2$.Let $r$ be the radius of the circle.$\pi r^2 = 38.5 \Rightarrow \frac{22}{7} 	imes r^2 = 38.5 \Rightarro

Vedclass · Accepted Answer

$196$

Vedclass · Answer

(B) Area of the circle $= 38.5 	ext{ cm}^2$.Let $r$ be the radius of the circle.$\pi r^2 = 38.5 \Rightarrow \frac{22}{7} 	imes r^2 = 38.5 \Rightarrow r^2 = \frac{38.5 	imes 7}{22} = 12.25$.$r = \sqrt{12.25} = 3.5 	ext{ cm}$.The side of the square $LMNO$ is equal to the diameter of the inscribed circle.Side of $LMNO = 2r = 2 	imes 3.5 = 7 	ext{ cm}$.Area of square $LMNO = (	ext{side})^2 = 7^2 = 49 	ext{ cm}^2$.When a square is formed by joining the midpoints of another square,its area is exactly half the area of the outer square.Area of square $EFGH = 2 	imes (	ext{Area of square } LMNO) = 2 	imes 49 = 98 	ext{ cm}^2$.Area of square $ABCD = 2 	imes (	ext{Area of square } EFGH) = 2 	imes 98 = 196 	ext{ cm}^2$.

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