If a wire is bent into the shape of a square, then the area enclosed by the square is $81c{m}^2$ .  When the same wie is bent into a semi-circular shape, then the area enclosed by the semicircle will be

Area of square $=81c{m}^2$

$a^2=81$

$a=\sqrt{81}=9$

Side $=9cm$

Perimeter $=49$

$=4\times 9$

$=36cm$

Perimeter of semi circle $=$  Perimeter of square

$\pi r+2r=36$

$r\left[\pi +2\right]=36$

$r=\frac{36}{\pi +2}cm$

Area of semi circle $=\frac{1}{2}\pi r^2$

$=\frac{1}{2}\pi {\left(\frac{36}{\pi +2}\right)}^2$

$=\frac{1}{2}\times \frac{22}{7}\times \frac{36}{\pi +2}\times \frac{36}{\pi +2}$

$=\frac{1}{2}\times \frac{22}{7}\times \frac{36}{\frac{36}{7}}\times \frac{36}{\frac{36}{7}}=\frac{1}{2}\times \frac{22}{7}\times \frac{7\times 36}{36}\times \frac{7\times 36}{36}=77{\text{cm}}^2$

$=\frac{11}{7}\times 7\times 7=77c{m}^2$ .