10.301. Find the ratio of the area of a square inscribed in a segment with an arc of $180^{\circ}$ to the area of a square inscribed in a segment of the same circle with an arc of $90^{\circ}$.
## Solution.

Let $A B C D$ be a square inscribed in a segment with an arc of $180^{\circ}$ (Fig. 10.91.1), and $A_{1} B_{1} C_{1} D_{1}$ be a square inscribed in a segment of the same circle with an arc of $90^{\circ}$ (Fig. 10.91.2). Let $R$ be the radius of the circle, $A B=x, A_{1} B_{1}=y$. Consider the square $A B C D$. $O$ is the center of the circle. Then $O C=R, O D=C D / 2=x / 2$. From $\triangle C D O\left(\angle C D O=90^{\circ}\right): O D^{2}+C D^{2}=O C^{2} ; \frac{x^{2}}{4}+x^{2}=R^{2} ; x=\frac{2 R}{\sqrt{5}}$. Consider the square $A_{1} B_{1} C_{1} D_{1}$. Draw $O K \perp B_{1} C_{1}$. Then $K$ is the midpoint of $B_{1} C_{1}$, and the point $P$ of intersection of $A_{1} D_{1}$ and $O K$ is the midpoint of $A_{1} D_{1}, K P=A_{1} B_{1}=y$, $K C_{1}=y / 2$. Since $\angle M O N=90^{\circ}$, then $O P=\frac{R \sqrt{2}}{2}$ and $O K=O P+P K=$ $=\frac{R \sqrt{2}}{2}+y$. From $\triangle O K C_{1} \quad\left(\angle O K C_{1}=90^{\circ}\right): O C_{1}^{2}=O K^{2}+K C_{1}^{2} ;$ $R^{2}=\left(\frac{R \sqrt{2}}{2}+y\right)^{2}+\frac{y^{2}}{4} ; 5 y^{2}+4 R y \sqrt{2}-2 R^{2}=0 ; y=\frac{R \sqrt{2}}{5} \cdot$ Therefore, $\frac{S_{A B C D}}{S_{A_{1} B_{1} C_{1} D_{1}}}=\left(\frac{x}{y}\right)^{2}=\left(\frac{2 R \cdot 5}{\sqrt{5} \cdot R \sqrt{2}}\right)^{2}=10$.

Answer: 10:1.