11.086. A sphere is inscribed in a cone, the axial section of which is an equilateral triangle. Find the volume of the cone if the volume of the sphere is $32 \pi / 3 \mathrm{~cm}^{3}$.
## Solution.

The volume of the sphere $V_{\mathrm{w}}=\frac{4}{3} \pi r^{3}=\frac{32}{3} \pi$ (Fig. 11.76). From this, $r=2$ cm. The volume of the desired cone is found as $V_{\mathbf{k}}=\frac{1}{3} \pi R^{2} H$. Since $\triangle A S B$ is equilateral, the radius of the inscribed circle $r=\frac{a \sqrt{3}}{6}$. From this, $a=\frac{12}{\sqrt{3}}=4 \sqrt{3}$ cm. The radius of the base of the cone $R=\frac{a}{2}=2 \sqrt{3}$ cm. From $\triangle S O B$, $SO=H=$ $=\sqrt{SB^{2}-OB^{2}}=\sqrt{(4 \sqrt{3})^{2}-(2 \sqrt{3})^{2}}=6$ cm. Finally, we get: $V_{\mathbf{k}}=$ $=\frac{1}{3} \pi(2 \sqrt{3})^{2} \cdot 6=24 \pi$.

Answer: $24 \pi \mathrm{cm}^{3}$.