# Question #e6f05

Feb 3, 2018

The area of the semi-circle will be =$77 s q . c m .$

#### Explanation:

Given: If a wire is bent into the shape of a square, then the area of the square is 81 sq.cm.

That means each side of the square = $\sqrt{81} = 9 c m$ ---as, area of square =$s i {\mathrm{de}}^{2}$

The length $l$, of the wire will be equal to the perimeter of this square:

$l = 4 \times 9 = 36 c m$

Now, when the wire is bent into a semi-circular shape, the perimeter will remain same = $36 c m$

And perimeter of semicircle will be given as:

$\left(\frac{1}{2} \times 2 \times \pi \times r\right) + \left(2 \times r\right)$------ half the circumference of complete circle + diameter.

SO,
$\left(\frac{1}{2} \times 2 \times \pi \times r\right) + \left(2 \times r\right) = 36$

$\implies \left(\frac{1}{\cancel{2}} \times \cancel{2} \times \pi \times r\right) + \left(2 \times r\right) = 36$

$\implies \pi r + 2 r = 36$ -------assume value of $\pi = \frac{22}{7}$

$\implies \frac{22}{7} \times r + 2 r = 36$

$\implies \frac{22}{7} \times r + 2 r \times \frac{7}{7} = 36$

$\implies \frac{22 r + 14 r}{7} = 36$

$\implies 36 r = 36 \times 7$

$\implies r = 7 c m$

Area of this semicircle 'a' will be :

$a = \frac{1}{2} \pi {r}^{2}$

$\implies a = \frac{1}{2} \times \frac{22}{7} \times {7}^{2}$

$\implies a = \frac{1}{\cancel{2}} ^ 1 \times {\cancel{22}}^{11} / {\cancel{7}}^{1} \times {\cancel{49}}^{7} = 11 \times 7$

$\therefore a = 77 s q . c m .$ will be the area of the semi-circle.