To decorate the exterior of a building, a large square was painted on the wall. They drew the circumscribed circle of the square and also semicircles extending outward from the sides of the square. The arcs of the circles define four crescent-shaped fields. What is the side length of the square if the area of one such field is $1 \mathrm{~m}^{2}$?
Solution. Let the side of the square be denoted by $a$, its center by $O$, and the endpoints of one of its sides by $A$ and $B$ as shown in the figure. The radius of the circumcircle of the square is: $\frac{a \sqrt{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3ae763952045d43b236fg-1.jpg?height=537&width=426&top_left_y=306&top_left_x=839)

The area of the circular segment over $AB$ is obtained by subtracting the area of triangle $ABO$ from the area of a quarter circle:

$$
\frac{1}{4}\left(\frac{a \sqrt{2}}{2}\right)^{2} \cdot \pi-\frac{a^{2}}{4}=\frac{a^{2}}{4}\left(\frac{\pi}{2}-1\right)
$$

The area of the crescent is the difference between the area of the semicircle over $AB$ and the area of the circular segment, and we know that this is equal to 1. That is,

$$
1=\frac{1}{2}\left(\frac{a}{2}\right)^{2} \pi-\frac{a^{2}}{4}\left(\frac{\pi}{2}-1\right)=\frac{a^{2}}{4}\left(\frac{\pi}{2}-\frac{\pi}{2}+1\right)=\frac{a^{2}}{4}
$$

from which $a=2$, so the side of the square is $2 \mathrm{~m}$.