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If a wire is bent into a 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?

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{154 cm}^{2}

{77 cm}^{2}

{22 cm}^{2}

{44 cm}^{2}

answer is B.

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Detailed Solution

Let the side of the square be a cm
Perimeter is 4a Area of the square is

a^{2}

area of the square is 81

{cm}^{2}

is given

a^{2} = 81

∴

Perim

a = 9

eter is

4 a = 4 * 9 = 36 cm

Now, the wire is bent into a semi-circular shape
Perimeter o f square = perimeter of semi circle

36 = 2 r + πr

36 = r (2 + π)

36 = r (2 + \frac{22}{7})

r = 7 cm

Area of semicircle =

\frac{π r^{2}}{2} = \frac{\frac{22}{7} * {(7)}^{2}}{2} = \frac{22 * 7}{2} = 77 {cm}^{2}

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