If a wire is bent into the shape of a square, | vedclass

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(C) The area of the square is $81 \, cm^2$. Since the area of a square is $side^2$,the side length is $\sqrt{81} = 9 \, cm$. The perimeter (length) of

Vedclass · Accepted Answer

$77$

Vedclass · Answer

(C) The area of the square is $81 \, cm^2$. Since the area of a square is $side^2$,the side length is $\sqrt{81} = 9 \, cm$. The perimeter (length) of the wire is $4 	imes 9 = 36 \, cm$. When this wire is bent into a semi-circular shape,its perimeter is given by the formula $P = \pi r + 2r = r(\pi + 2)$. Setting the perimeter equal to $36 \, cm$: $r(\frac{22}{7} + 2) = 36$ $r(\frac{22 + 14}{7}) = 36$ $r(\frac{36}{7}) = 36$ $r = 7 \, cm$. The area of the semi-circle is $\frac{1}{2} \pi r^2$: Area $= \frac{1}{2} 	imes \frac{22}{7} 	imes 7^2 = \frac{1}{2} 	imes 22 	imes 7 = 77 \, cm^2$.

If a wire is bent into the shape of a square,then the area of the square is $81 \, cm^2$. When the wire is bent into a semi-circular shape,the area of the semi-circle will be (in $cm^2$):

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