A silver wire when bent in the form of a squa | vedclass

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Question

(C) The perimeter of the wire remains constant regardless of the shape.Let the side of the square be $a \, cm$.Given,area of the square $= a^{2} = 484

Vedclass · Accepted Answer

$616$

Vedclass · Answer

(C) The perimeter of the wire remains constant regardless of the shape.Let the side of the square be $a \, cm$.Given,area of the square $= a^{2} = 484 \, cm^{2}$.Therefore,$a = \sqrt{484} = 22 \, cm$.Perimeter of the square $= 4a = 4 	imes 22 = 88 \, cm$.Since the same wire is used to form a circle,the circumference of the circle is equal to the perimeter of the square.Let the radius of the circle be $r \, cm$.Circumference $= 2\pi r = 88 \, cm$.$2 	imes \frac{22}{7} 	imes r = 88$.$r = \frac{88 	imes 7}{44} = 14 \, cm$.Area of the circle $= \pi r^{2} = \frac{22}{7} 	imes 14 	imes 14 = 22 	imes 2 	imes 14 = 616 \, cm^{2}$.

$A$ silver wire when bent in the form of a square,encloses an area of $484 \, cm^{2}$. Now if the same wire is bent to form a circle,the area enclosed by the circle would be? (in $cm^{2}$)

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