If a wire is bent into the shape of a square, then the area of the square is 81 cm2 . When wire is bent into a semi-circular shape, then the area of the semi-circle will be
(a) $22 \mathrm{~cm}^{2}$
(b) $44 \mathrm{~cm}^{2}$
(c) $77 \mathrm{~cm}^{2}$
(d) $154 \mathrm{~cm}^{2}$
We have given that a wire is bent in the form of square of side a cm such that the area of the square is . If we bent the same wire in the form of a semicircle with radius r cm, the perimeter of the wire will not change.
$\therefore$ perimeter of the square $=$ perimeter of semi circle
$4 a=\frac{1}{2}(2 \pi r)+2 r$......(1)
We know that area of the square $=81 \mathrm{~cm}^{2}$.
$\therefore a^{2}=81$
$\therefore a=9$
Now we will substitute the value of a in the equation (1),
$4 \times 9=\frac{1}{2}(2 \pi r)+2 r$
$\therefore 36=\frac{1}{2}(2 \pi r)+2 r$
$\therefore 36=\frac{1}{2}(2 \pi r)+2 r$
$\therefore 36=(\pi r)+2 r$
$\therefore 36=r(\pi+2)$
Now we will substitute $\pi=\frac{22}{7}$.
$\therefore 36=r\left(\frac{22}{7}+2\right)$
$\therefore 36=r\left(\frac{22+14}{7}\right)$
$\therefore 36=r\left(\frac{36}{7}\right)$
Multiplying both sides of the equation by 7 we get, $36 \times 7=r \times 36$
Now we will divide both sides of the equation by 36 we get, r = 7
Therefore, radius of the semi circle is 7cm.
Now we will find the area of the semicircle.
Area of the semicircle $=\frac{1}{2} \times \pi r^{2}$
$=\frac{1}{2} \times \pi \times 7^{2}$
$=11 \times 7$
$=77$
Therefore, area of the semicircle is $77 \mathrm{~cm}^{2}$.
Hence the correct answer is option (c).
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