## Task B-2.1.

A circle and two semicircles are inscribed in a square as shown in the figure. What is the area of the square if the area of the shaded part is equal to $48 \pi - 36 \sqrt{3}$?

![](https://cdn.mathpix.com/cropped/2024_05_30_ea3ef29009412d1e0d2bg-06.jpg?height=423&width=439&top_left_y=765&top_left_x=800)
## Solution.

Notice that the shaded area is equal to the sum of the areas of two congruent figures. Each figure is formed by the intersection of a circle inscribed in a square and a semicircle constructed over the side of the square.

Let the length of the side of the square be $x$, then $r=\frac{x}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_ea3ef29009412d1e0d2bg-06.jpg?height=411&width=414&top_left_y=1639&top_left_x=244)

Observe that each of the congruent figures consists of two equilateral triangles with side length $r$ and 4 circular segments of a circle with radius $r$ corresponding to an angle of $60^{\circ}$.

The area of one of the triangles is $P_{\triangle}=\frac{r^{2} \sqrt{3}}{4}=\frac{x^{2} \sqrt{3}}{16}$.

The area of the circle inscribed in the square is $P_{k}=\frac{x^{2}}{4} \pi$.

The area of the circular sector is $\frac{1}{6}$ of the area of the circle, i.e., $P_{i}=\frac{1}{6} P_{k}=\frac{1}{6} \cdot \frac{x^{2}}{4} \pi=\frac{x^{2} \pi}{24}$.
$P_{\text {segment }}=P_{i}-P_{\triangle}=\frac{x^{2} \pi}{24}-\frac{x^{2} \sqrt{3}}{16}$.

The area of the shaded part of the square is

$$
P=8 \cdot P_{i}+4 \cdot P_{\triangle}=8\left(\frac{x^{2} \pi}{24}-\frac{x^{2} \sqrt{3}}{16}\right)+\frac{x^{2} \sqrt{3}}{4}=\frac{x^{2} \pi}{3}-\frac{x^{2} \sqrt{3}}{4}
$$

Since the area of the shaded part of the square is $48 \pi-36 \sqrt{3}$, solving the equation $48 \pi-36 \sqrt{3}=\frac{x^{2} \pi}{3}-\frac{x^{2} \sqrt{3}}{4}$ gives:

$$
\begin{aligned}
& 48 \pi-36 \sqrt{3}=\frac{4 x^{2} \pi-3 \sqrt{3} \cdot x^{2}}{12} \\
& 12(4 \pi-3 \sqrt{3})=\frac{x^{2}(4 \pi-3 \sqrt{3})}{12} \\
& x^{2}=144
\end{aligned}
$$

Therefore, the area of the square is 144.