11.014. The apothem of a regular hexagonal pyramid is $h$, and the dihedral angle at the base is $60^{\circ}$. Find the total surface area of the pyramid.
Solution.

Since $\angle S K O=60^{\circ}$ (Fig. 11.12), then $O K=\frac{1}{2} S K=\frac{h}{2}$. The base of the pyramid is a regular hexagon, so $\angle K O D=30^{\circ}$ and $K D=\frac{1}{2} O D$. Then $O K^{2}=O D^{2}-K D^{2}=4 K D^{2}-K D^{2}=3 K D^{2}$, i.e., $K D=\frac{h \sqrt{3}}{6}, D E=2 K D=\frac{h \sqrt{3}}{3}$. We obtain: $S_{\text {base }}=\frac{6 D E^{2} \sqrt{3}}{4}=\frac{h^{2} \sqrt{3}}{2}$, $S_{\text {lat }}=3 D E \cdot h=h^{2} \sqrt{3}, S_{\text {tot }}=S_{\text {base }}+S_{\text {lat }}=\frac{3 h^{2} \sqrt{3}}{2}$.

Answer: $\frac{3 h^{2} \sqrt{3}}{2}$.