11.095. A sphere is circumscribed around a regular triangular prism, the height of which is twice the side of the base. How does its volume relate to the volume of the prism?
## Solution.

The volume of the sphere is: $V_{s}=\frac{4}{3} \pi R^{3}$ (Fig. 11.82). The volume of the prism is: $V_{p}=S \cdot H$. The area of the base of the prism is: $S=\frac{a^{2} \sqrt{3}}{4}$, and the height of the prism $H=2 a$. Therefore, $V_{p}=\frac{a^{3} \sqrt{3}}{2}$. Consider $\triangle A O B: O B=\frac{1}{2} H=a$;
$A B=r=\frac{a \sqrt{3}}{3}$ (the radius of the circumscribed circle for an equilateral triangle).

Then $R=\sqrt{A B^{2}+O B^{2}}=\sqrt{a^{2}+\frac{3 a^{2}}{9}}=\frac{2 a \sqrt{3}}{3}=\frac{2 a}{\sqrt{3}}$. Therefore, $V_{s}=\frac{4}{3} \pi \frac{8 a^{3}}{3 \sqrt{3}}$. The desired ratio:

$\frac{V_{s}}{V_{p}}=\frac{4 \cdot 8 a^{3} \pi \cdot 2}{3 \cdot 3 \sqrt{3} \cdot a^{3} \sqrt{3}}=\frac{64 \pi}{27}$.

Answer: $\frac{64 \pi}{27}$.