2.7. 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.
2.7. Since $\angle S K O=60^{\circ}$ (Fig. 2.7), 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 $K D=\frac{1}{2} O D$. Therefore, $O K^{2}=O D^{2}-K D^{2}=4 K D^{2}-K D^{2}=3 K D^{2}$,

Fig. 2.7 i.e., $K D=\frac{h \sqrt{3}}{6}, D E=2 K D=\frac{h \sqrt{3}}{3}$. Thus,

$$
S_{\mathrm{OCH}}=\frac{6 D E^{2} \sqrt{3}}{4}=\frac{h^{2} \sqrt{3}}{2}, S_{6 \text { side }}=3 D E \cdot h=h^{2} \sqrt{3}
$$

Finally, we find $S_{\text {total }}=S_{\text {base }}+S_{\text {side }}=\frac{3 h^{2} \sqrt{3}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-459.jpg?height=412&width=387&top_left_y=1465&top_left_x=45)

Fig. 2.8

Answer: $1.5 h^{2} \sqrt{3}$.