10.349. A circle of radius $R$ is divided into 4 large and 4 small parts, which alternate with each other. The large part is twice as long as the small one. Determine the area of the octagon whose vertices are the points of division of the circle.
## Solution.

Let the small arc contain $x$ radians. Then $4 x+8 x=2 \pi, x=\frac{\pi}{6}$. Thus, the octagon contains four triangles with a central angle of $\frac{\pi}{3}$ (their total area is $4 \cdot \frac{R^{2} \sqrt{3}}{4}$) and four triangles with a central angle of $\frac{\pi}{6}$ (their total area is $4 \cdot \frac{R^{2}}{4}$). The desired area is $S=R^{2}(\sqrt{3}+1)$.

Answer: $S=R^{2}(\sqrt{3}+1)$.