The unit-sided square $ABCD$ is rotated $90^{\circ}$ around vertex $C$. What area does side $AB$ sweep?
Solution. Since the side of the square is one unit, point $B$ traces the points of a quarter-circle arc with center $C$ and a unit radius during the rotation. Point $A$ also moves along a quarter-circle arc with center $C$, but with a radius of $\sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_1298f4632a8affcd5ecfg-1.jpg?height=309&width=458&top_left_y=274&top_left_x=823)

The area swept by segment $AB$ is the sum of the area of the circular segment corresponding to chord $AA'$ and the area of the region $BA'B'$. The area $T_1$ of the circular segment is the difference between the area of a quarter-circle with radius $\sqrt{2}$ and the area of triangle $ACA'$:

$$
T_1 = \frac{(\sqrt{2})^2 \pi}{4} - \frac{2 \cdot 1}{2} = \frac{\pi}{2} - 1
$$

The area $T_2$ of the region $BA'B'$ is the difference between the area of the unit square $BA'B'C$ and the area of a quarter-circle with unit radius:

$$
T_2 = 1 - \frac{\pi}{4}
$$

The desired area $T$ is:

$$
T = \frac{\pi}{2} - 1 + 1 - \frac{\pi}{4} = \frac{\pi}{4}
$$

Remark. It is worth noting that the area in question is exactly the area of a quarter-circle with unit radius: this is the area swept by a unit segment whose endpoint is rotated $90^\circ$ around its other endpoint. Our result seems to suggest that the size of the "swept" area does not depend on the center of rotation.