10.375 An equilateral triangle $ABC$, inscribed in a circle of radius $R$, is rotated around the center of the circle by $90^{\circ}$ to the position $A_{1} B_{1} C_{1}$. Calculate the area of the hexagon $A A_{1} B B_{1} C C_{1}$.
Solution. Let $O$ be the center of the circumcircle of the given equilateral $\triangle A B C$. Then the hexagon $A A_{1} B B_{1} C C_{1}$ will be composed of three equal triangles $\triangle O A A_{1}, \triangle O B B_{1}, \triangle O C C_{1}$ and three equal triangles

$\triangle O A_{1} B, \triangle O B_{1} C, \triangle O C_{1} A$. Since $\angle A O B=120^{\circ}$ and the rotation is performed around $O$ by $90^{\circ}$, then $\angle A O A_{1}=90^{\circ}, \angle A_{1} O B=30^{\circ} \Rightarrow$

$\Rightarrow S_{A A_{1} B B_{1} C C_{1}}=3\left(S_{O A A_{1}}+S_{O A_{1} B}\right)=3\left(\frac{R^{2} \sin 90^{\circ}}{2}+\frac{R^{2} \sin 30^{\circ}}{2}\right)=\frac{9 R^{2}}{4}$.

Answer: $\frac{9 R^{2}}{4}$