10.284. A square and an equilateral triangle are described around a circle of radius $R$, with one side of the square lying on a side of the triangle. Calculate the area of the common part of the triangle and the square.
Solution.

$\triangle A B C$ and square $E D D_{1} E_{1}$ (Fig. 10.78) are the ones mentioned in the problem, $B N$ is the altitude of $\triangle A B C$, $O$ is the center of the given circle, $M$ and $F$ are the points of intersection of $A B, M_{1}$ and $F_{1}$ are the points of intersection of $B C$ with the sides of the square, $N_{1}$ is the point of tangency of $D D_{1}$ with the circle and the intersection of $D D_{1}$ with $B N$. The hexagon $E M F F_{1} M_{1} E_{1}$ is the figure of the desired area $S$. $S=S_{E D D_{1} E_{1}}-\left(S_{\triangle M D F}+S_{\triangle M_{1} D_{1} F_{1}}\right)=S_{E D D_{1} E_{1}}-2 S_{\triangle M D F} . B N=3 R ; N N_{1}=2 R$; $B N_{1}=R ; D N_{1}=R$.

In $\triangle B N_{1} F \quad\left(\angle B N_{1} F=90^{\circ}\right): \quad F N_{1}=B N_{1} \operatorname{tg} \angle F B N_{1}=R \operatorname{tg} 30^{\circ}=\frac{R \sqrt{3}}{3}$,

$$
\begin{aligned}
& D F=D N_{1}-F N_{1}=R-\frac{R \sqrt{3}}{3}=\frac{R(3-\sqrt{3})}{3} \cdot \angle D F M=\angle B F F_{1}=60^{\circ} . \text { In } \triangle M D F \\
& \left(\angle M D F=90^{\circ}\right): M D=D F \operatorname{tg} \angle D F M=\frac{R(3-\sqrt{3})}{3} \cdot \sqrt{3}=R(\sqrt{3}-1) \\
& \quad S_{\triangle M D F}=\frac{1}{2} D F \cdot D M=\frac{1}{2} \cdot \frac{R \sqrt{3}(\sqrt{3}-1)}{3} \cdot R(\sqrt{3}-1)= \\
& \quad=\frac{R^{2} \sqrt{3}(\sqrt{3}-1)^{2}}{6}=\frac{R^{2} \sqrt{3}(2-\sqrt{3})}{3} ; S_{E D D_{1} E_{1}}=4 R^{2} \\
& \text { Then } S=4 R^{2}-\frac{2 R^{2} \sqrt{3}(2-\sqrt{3})}{3}=R^{2} \cdot \frac{12-4 \sqrt{3}+6}{3}=\frac{R^{2} \sqrt{3}(6 \sqrt{3}-4)}{3} \\
& \text { Answer: } \frac{R^{2} \sqrt{3}(6 \sqrt{3}-4)}{3}
\end{aligned}
$$