11.111. In a regular quadrilateral pyramid, the dihedral angle at the lateral edge is $120^{\circ}$. Find the lateral surface area of the pyramid if the area of its diagonal section is $S$.
## Solution.

$E O$ is the height of the regular pyramid $E A B C D, \angle B K D$ is the linear angle of the dihedral angle at the lateral edge $E C$ (Fig. 11.9), $\angle B K D=120^{\circ}$. Then $O K \perp E C, S_{\triangle A E C}=2 S_{\triangle E O C}=E C \cdot O K, D K \perp E C, S_{\triangle D E C}=\frac{1}{2} E C \cdot D K$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0826.jpg?height=537&width=601&top_left_y=303&top_left_x=93)

Fig. 11.9

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0826.jpg?height=775&width=555&top_left_y=67&top_left_x=708)

Fig. 11.10

From $\triangle D O K\left(\angle D O K=90^{\circ}\right): O K=D K \cos \angle D K O=D K \cos 60^{\circ}=\frac{1}{2} D K$.

Therefore, the lateral surface area of the pyramid $S_{6}=4 \cdot S_{\triangle D E C}=2 E C \cdot D K$. Thus, $\frac{S_{6}}{S_{\triangle A E C}}=\frac{2 E C \cdot D K}{E C \cdot \frac{1}{2} D K}=4 . S_{6}=4 \cdot S_{\triangle A E C}=4 S$.
Answer: $4 S$