12.283. The side of the base of a regular quadrilateral pyramid is equal to $a$. The angle between adjacent lateral faces is $\alpha$. Find the lateral surface area of the pyramid.
Solution.

Let $SO$ be the height of the regular pyramid $SABCD$ (Fig. 12.146), $AD = \alpha$.

In the plane of the face $SDC$, draw the perpendicular $DE$ to $SC$.

$SO \perp BD, AC \perp BD$, therefore the line $BD$ is perpendicular to the plane $ASC$. Hence, $BD \perp SC$.

$SC \perp BD, SC \perp DE$. We have that the line $SC$ is perpendicular to the plane $BED$. Then $\angle BED$ is the angle between the adjacent lateral faces $SCD$ and $SCB$, $\angle BED = \alpha$, and $OE \perp SC$.

Right triangles $DEC$ and $BEC \left(\angle DEC = \angle BEC = 90^\circ\right)$ are equal by the leg and hypotenuse. Then $BE = ED$.

$O$ is the midpoint of $BD \Rightarrow OE \perp BD, \angle OED = \frac{1}{2} \angle BED = \frac{\alpha}{2}$.

If $\angle SOC = \beta$, then from $\triangle EOD \left(\angle EOD = 90^\circ\right)$:

$$
DE = \frac{OD}{\sin \angle OED} = \frac{a \sqrt{2}}{2 \sin \frac{\alpha}{2}}; \quad EO = OD \operatorname{ctg} \angle OED = \frac{a \sqrt{2}}{2} \operatorname{ctg} \frac{\alpha}{2}
$$

In $\triangle OEC \left(\angle OEC = 90^\circ\right)$:

$EO = OC \sin \angle SCO = \frac{a \sqrt{2}}{2} \sin \beta \Rightarrow \sin \beta = \operatorname{ctg} \frac{\alpha}{2}$.

In $\triangle SOC \left(\angle SOC = 90^\circ\right)$:

$$
\begin{aligned}
& SC = \frac{OC}{\cos \angle SCO} = \frac{a \sqrt{2}}{2 \cos \beta} = \frac{a \sqrt{2}}{2 \sqrt{1 - \sin^2 \beta}} = \frac{a \sqrt{2}}{2 \sqrt{1 - \operatorname{ctg}^2 \frac{\alpha}{2}}} = \frac{a \sqrt{2} \sin \frac{\alpha}{2}}{2 \sqrt{-\cos \alpha}} \\
& S_{\triangle DSC} = \frac{1}{2} SC \cdot DE = \frac{1}{2} \cdot \frac{a \sqrt{2} \sin \frac{\alpha}{2}}{2 \sqrt{-\cos \alpha}} \cdot \frac{a \sqrt{2}}{2 \sin \frac{\alpha}{2}} = \frac{a^2}{4 \sqrt{-\cos \alpha}}
\end{aligned}
$$

The lateral surface area of the pyramid

$S_6 = 4 S_{\triangle DSC} = \frac{a^2}{\sqrt{-\cos \alpha}}$.

Answer: $\frac{a^2}{\sqrt{-\cos \alpha}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1039.jpg?height=662&width=565&top_left_y=79&top_left_x=81)

Fig. 12.147