1.63. The entire arc of a circle with radius $R$ is divided into four large and four small segments, alternating with each other. The large segment is twice as long as the small one. Determine the area of the octagon whose vertices are the points of division of the circle's arc.
1.63. Let the small arc contain $x$ degrees. Then $4 x+8 x=2 \pi$, from which $x=\frac{\pi}{6}$. Therefore, 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 $4 \cdot\left(\frac{R^{2}}{4}\right)$. The desired area is $S=R^{2}(\sqrt{3}+1)$.

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