2.57. 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.
2.57. According to the condition, $\angle O A A_{1}=60^{\circ}$ (Fig. 2.47); 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

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-486.jpg?height=458&width=488&top_left_y=98&top_left_x=333)

Fig. 2.47

$$
S_{\text {lower base}}=6 \frac{R^{2} \sqrt{3}}{4}=\frac{3 R^{2} \sqrt{3}}{2}, S_{\text {upper base}}=\frac{1}{4} S_{\text {lower base}}=\frac{3 R^{2} \sqrt{3}}{8} \text {. }
$$

Finally, we get

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

Answer: $\frac{21 R^{3}}{16}$.