7. Two vertices of a square lie on a circle with a radius of $5 \mathrm{~cm}$, while the other two lie on a tangent to this circle. Find the area of the square.
Solution: $R=5, A B C D$-square, $S_{A B C D}=$ ?

$A B=x, E F=2 R-x, E O=R-E F=x-R$

$\Delta E O C: E C^{2}=O C^{2}-E O^{2}=R^{2}-(x-R)^{2} \Rightarrow\left(\frac{x}{2}\right)^{2}=R^{2}-x^{2}+2 R x-R^{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_8eeeb8d6ace591217257g-23.jpg?height=400&width=468&top_left_y=671&top_left_x=1568)

$\frac{x^{2}}{4}=2 R x-x^{2}, \frac{5}{4} x=2 R, x=\frac{2 R \cdot 4}{5}=8$. Answer: $S_{\triangle A B C D}=64$.