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

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).