A $6 \mathrm{~cm}$ radius circle has an inscribed regular hexagon and a square, with one side of each parallel. Calculate the area of the part of the circle that lies between the parallel sides of the hexagon and the square and does not contain the center of the circle.
We denote the center of the circle by $O$, the intersection points of the square's side with the circle by $A$ and $B$, and the endpoints of the hexagon's side parallel to the square's side by $C$ and $D$. If we subtract the area of the circular segment $\widehat{C D}$ from the area of the circular segment $\widehat{A B}$, we obtain the desired area.

![](https://cdn.mathpix.com/cropped/2024_05_02_f7c639d50dd163394b4eg-1.jpg?height=420&width=433&top_left_y=281&top_left_x=846)

The area of a circular segment is given by: $T=\frac{1}{2} r^{2}(\widehat{\alpha}-\sin \alpha)$, where $r$ is the radius of the circle from which the segment is cut, $\alpha$ is the central angle, and $\widehat{\alpha}$ is its measure in radians.

The side of the square is seen from the center at a $90^{\circ}$ angle, which is $\frac{\pi}{2}$ radians, and the side of the regular hexagon is seen at a $60^{\circ}$ angle, which is $\frac{\pi}{3}$ radians.

The difference in the two areas is:

$$
\frac{1}{2} 6^{2}\left[\left(\frac{\pi}{2}-1\right)-\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)\right] \approx 7.01332 \mathrm{~cm}^{2}
$$

Note. There are two congruent such figures in the circle, and the problem can also be interpreted as requiring the combined area of these two figures. In this case, the answer would be twice the value obtained in the solution, i.e., $\approx 14.03 \mathrm{~cm}^{2}$.