11.050. Find the lateral surface area of a regular hexagonal pyramid, the height of which is equal to $h$, and the lateral edge is equal to $l$.
Solution.

$$
S_{60 \mathrm{~K}}=6 \cdot S_{\triangle F S A}=6 \cdot \frac{1}{2} \cdot A F \cdot S K=3 A F \cdot S K \text { (Fig. 11.45); } \triangle S O A
$$

$\angle S O A=90^{\circ}, \quad O A=\sqrt{l^{2}-h^{2}}=A F, \quad S K=\sqrt{l^{2}-\frac{l^{2}-h^{2}}{4}}=\frac{\sqrt{3 l^{2}+h^{2}}}{2} ;$

then $S_{\text {side }}=3 \cdot \sqrt{l^{2}-h^{2}} \cdot \frac{\sqrt{3 l^{2}+h^{2}}}{2}=\frac{3}{2} \sqrt{\left(l^{2}-h^{2}\right)}\left(3 l^{2}+h^{2}\right)$.

Answer: $\frac{3}{2} \sqrt{\left(l^{2}-h^{2}\right)\left(3 l^{2}+h^{2}\right)}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-681.jpg?height=502&width=420&top_left_y=171&top_left_x=174)

Fig. 11.47

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-681.jpg?height=608&width=502&top_left_y=92&top_left_x=749)

Fig. 11.48