10.189. A square is inscribed in a segment whose arc is $60^{\circ}$. Calculate the area of the square if the radius of the circle is $2 \sqrt{3}+\sqrt{17}$.
Solution.

Let $B C=C D=D A=A B=x$ (Fig. 10.156). Consider $\triangle O N C$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-644.jpg?height=446&width=500&top_left_y=93&top_left_x=108)

Fig. 10.156

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-644.jpg?height=330&width=642&top_left_y=211&top_left_x=591)

Fig. 10.157

$\angle O N C=90^{\circ}$. We have $O N^{2}+N C^{2}=O C^{2}$, where $O C=R=2 \sqrt{3}+\sqrt{17}$. $O N \doteq$ $=O M+M N=O M+x, N C=\frac{x}{2}$. In $\triangle O M F \quad \angle O M F=90^{\circ}, \angle M O F=$ $=\frac{\angle E O F}{2}=30^{\circ}$, then $O M=O F \cdot \frac{\sqrt{3}}{2}=R \cdot \frac{\sqrt{3}}{2}$, therefore, $O N=$ $=\frac{R \sqrt{3}}{2}+x$. Thus, we have $\left(\frac{R \sqrt{3}}{2}+x\right)^{2}+\frac{x^{2}}{4}=R^{2}$. Solving the quadratic equation, we get $x=\frac{2 R}{10}(\sqrt{17}-2 \sqrt{3}) ; S_{A B C D}=x^{2}$; $x=\frac{(2 \sqrt{3}+\sqrt{17})(\sqrt{17}-2 \sqrt{3})}{5}=1, S_{A B C D}=1$.

Answer: 1.