11.173. A regular hexagonal prism is circumscribed around a sphere of radius $R$. Determine its total surface area.
## Solution.

A regular hexagon $A B C D E F$ is the base of a regular prism circumscribed around a sphere with center $O$ (Fig. 11.66). Point $O$ is the midpoint of the axis $L L_{1}$ of the prism, $L L_{1}=2 R$. $M$ is the point of tangency of the sphere with the face $C C_{1} D_{1} D$ of the prism, $K$ is the midpoint of $C D$. Then $O M$ and $L K$ are perpendiculars to the face $C C_{1} D_{1} D$ and, therefore, $O M \| L K$, and since $L O$ is parallel to the face $C C_{1} D_{1} D$, $L K=O M=R \cdot L K$ is the height of the equilateral $\triangle C L D$.

Then $C D=\frac{2 L K \sqrt{3}}{3}=\frac{2 R \sqrt{3}}{3}, S_{\triangle C L D}=\frac{C D^{2} \sqrt{3}}{4}=\frac{4 R^{2}}{3} \cdot \frac{\sqrt{3}}{4}=\frac{R^{2} \sqrt{3}}{3}$.

The area of the base of the prism $S_{\text {base }}=6 S_{\triangle C L D}=2 R^{2} \sqrt{3}$, the perimeter of the base $P_{\text {base }}=6 C D=4 R \sqrt{3}$, the lateral surface area of the prism $S_{6}=P_{\text {base }} \cdot L L_{1}=8 R^{2} \sqrt{3}$.

The total surface area of the prism $S=2 S_{\text {base }}+S_{6}=12 R^{2} \sqrt{3}$.

Answer: $12 R^{2} \sqrt{3}$.