11.174. A regular truncated hexagonal pyramid is inscribed in a sphere of radius $R$, with the plane of the lower base passing through the center of the sphere, and the lateral edge forming an angle of $60^{\circ}$ with the base plane. Determine the volume of the pyramid.
Solution.

According to the condition, $\angle O A A_{1}=60^{\circ}$ (Fig. 11.68); hence, $\angle O_{1} O A_{1}=30^{\circ}$ and $A_{1} O_{1}=\frac{1}{2} A_{1} O=\frac{R}{2}, O O_{1}=\frac{R \sqrt{3}}{2}$. We find $S_{\text {lower.base }}=6 \frac{R^{2} \sqrt{3}}{4}=\frac{3 R^{2} \sqrt{3}}{2}$, $S_{\text {upper.base }}=\frac{1}{4} S_{\text {lower.base }}=\frac{3 R^{2} \sqrt{3}}{8}$.

We obtain that $V=\frac{1}{3} \cdot \frac{R \sqrt{3}}{2}\left(\frac{3 R^{2} \sqrt{3}}{2}+\frac{3 R^{2} \sqrt{3}}{8}+\sqrt{\frac{9 R^{4} \cdot 3}{16}}\right)=\frac{21 R^{3}}{16}$. Answer: $\frac{21 R^{3}}{16}$.