Solution. Let the radius of the circle be denoted by $r$, and the side of the square by $a$. From the equality of the circumferences: $2 r \pi=4 a$, it follows that $r=\frac{2 a}{\pi}$. Let one vertex of the square be denoted by $A$, and the common center by $O$, then $\overline{A O}=\frac{a \sqrt{2}}{2}$. We claim that

$$
\frac{a \sqrt{2}}{2}>r
$$

thus the square has points outside the circle. Substituting the value of $r$ obtained earlier (1) into this:

$$
\frac{a \sqrt{2}}{2}>\frac{2 a}{\pi}
$$

simplifying by $a$ and rearranging, we get $\pi>2 \sqrt{2}(\approx 2.8284)$. The inequality indeed holds. The square-shaped tablecloth does not cover 4 congruent circular segments of the circle.

Let the circle intersect one side of the square at points $M_{1}, M_{2}$ (see the figure), and let the foot of the perpendicular from $O$ to $M_{1} M_{2}$ be $P$, and the angle $\angle P O M_{1}$ be $\frac{\alpha}{2}$. The area of the circular segment $O M_{1} M_{2}$ is obtained by subtracting the area of the triangle $O M_{1} M_{2}$ from the area of the sector $O M_{1} M_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c17e634ebb7292dba3b4g-1.jpg?height=426&width=426&top_left_y=874&top_left_x=836)

To do this, we first determine the angle $\frac{\alpha}{2}$. From the triangle $P O M_{1}$, $P O=\frac{a}{2}, O M_{1}=r=\frac{2 a}{\pi}$, and

$$
\cos \frac{\alpha}{2}=\frac{O P}{O M_{1}}=\frac{\frac{a}{2}}{\frac{2 a}{\pi}}=\frac{\pi}{4}
$$

from which $\frac{\alpha}{2} \approx 38.24^{\circ}, \alpha \approx 76.48^{\circ}$, and $\sin \alpha \approx 0.9723$.

The area of the sector $O M_{1} M_{2}$: $t_{1}=\frac{\pi \alpha r^{2}}{360^{\circ}}$, the area of the triangle $O M_{1} M_{2}$: $t_{2}=\frac{r^{2} \sin \alpha}{2}$. The area of the circular segment:

$$
t_{1}-t_{2} \approx r^{2}(0.6674-0.4862)=r^{2} \cdot 0.1812
$$

The total area of the four circular segments: $0.7248 r^{2}$. The area covered by the square-shaped tablecloth:

$$
r^{2} \pi-0.7248 r^{2} \approx 2.4168 r^{2}
$$

The area of the circular table is equal to the area of the circle: $r^{2} \pi$. The ratio of the two areas is $\approx \frac{2.4168}{\pi} \approx 0.7693$. The tablecloth covers approximately $77\%$ of the table's area.
A circular table is covered with a square tablecloth such that their centers coincide. The circumference of the circle and the perimeter of the square are equal. What percentage of the table's surface area is covered by the tablecloth?