A circle and a square of equal area have their centers coinciding. What percentage of the square's area is covered by the circle? (Round the result to the nearest whole percentage.)





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Since the areas of the two figures are equal, the circle cannot be entirely within the square, so there will be "protruding" parts. From the equality of the areas, it also follows that the sum of the areas of the protruding and uncovered parts is equal.

![](https://cdn.mathpix.com/cropped/2024_05_02_aed5ab963194a012fa53g-1.jpg?height=439&width=423&top_left_y=260&top_left_x=840)

Let's try to determine the sum of the areas of these protruding parts. Denote the side of the square by \(a\), the radius of the circle by \(r\), and the center of the circle (which is also the center of the square) by \(O\).

The area of the circular segment can be obtained by subtracting the area of the triangle \(OAB\) from the area of the sector \(OAB\). For both, it would be useful to know the central angle \(2\alpha\) corresponding to the arc \(AB\). From the triangle \(OFB\), \(\cos \alpha = \frac{a}{2r}\); since \(a^2 = r^2 \pi\), we have \(a = r \sqrt{\pi}\) and thus \(\cos \alpha = \frac{\sqrt{\pi}}{2} \approx 0.8862\), from which \(2\alpha \approx 55.194^\circ\).

Using this,

\[
\begin{aligned}
& T_{\text{segment}} = T_{\text{sector}} - T_{\text{triangle}} = \frac{r^2 \pi 2\alpha}{360^\circ} - \frac{r^2 \sin 2\alpha}{2} = \\
& \quad = \frac{r^2}{2} \left( \frac{\pi \cdot 55.194^\circ}{180^\circ} - \sin 55.194^\circ \right) = r^2 \cdot 0.0711
\end{aligned}
\]

The sum of the areas of the four circular segments gives the uncovered part, and subtracting this from the area of the circle gives the area of the covered part:

\[
T_{\text{covered}} = r^2 (\pi - 4 \cdot 0.0711) = r^2 \cdot 2.8572
\]

This is 91% of the area of the square.