12.275. The base of the pyramid is a rectangle. Two lateral faces are perpendicular to the base plane, and the other two form angles $\alpha$ and $\beta$ with it. Find the lateral surface area of the pyramid if the height of the pyramid is $H$.
Solution.

Let the rectangle $ABCD$ be the base of the pyramid $KACBD$ (Fig. 12.139).

The lateral faces $ABK$ and $CBK$ are perpendicular to the base plane of the pyramid, their common edge $KB$ is the height of the pyramid, $KB=H$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1029.jpg?height=739&width=847&top_left_y=30&top_left_x=285)

Fig. 12.139

$BA$ and $BC$ are the projections of the lateral edges $KA$ and $KC$ on the base plane, $BA \perp AD, BC \perp CD$. Therefore, $KA \perp AD, KC \perp CD, \angle KAB$ and $\angle KCB$ are the angles of inclination of the lateral faces $KAD$ and $KCD$ to the base plane, $\angle KAB=\alpha, \angle KCB=\beta$.

In $\triangle ABK\left(\angle ABK=90^{\circ}\right): AB=H \operatorname{ctg} \alpha, AK=\frac{H}{\sin \alpha}$.

In $\triangle CBK\left(\angle CBK=90^{\circ}\right): BC=H \operatorname{ctg} \beta, KC=\frac{H}{\sin \beta}$.

The lateral surface area of the pyramid

$$
\begin{aligned}
& S=S_{\triangle ABK}+S_{\triangle CBK}+S_{\triangle DAK}+S_{\triangle DCK}=\frac{1}{2} AB \cdot KB+\frac{1}{2} BC \cdot KB+\frac{1}{2} AK \cdot AD+ \\
& +\frac{1}{2} KC \cdot CD=\frac{H^2}{2}\left(\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\frac{\operatorname{ctg} \beta}{\sin \alpha}+\frac{\operatorname{ctg} \alpha}{\sin \beta}\right)=\frac{H^2}{2}\left(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}+\right. \\
& \left.+\frac{\cos \beta+\cos \alpha}{\sin \alpha \sin \beta}\right)=\frac{H^2}{2 \sin \alpha \sin \beta}\left(2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}+2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\right)= \\
& =\frac{H^2 \cos \frac{\alpha+\beta}{2}}{\sin \alpha \sin \beta}\left(\sin \frac{\alpha+\beta}{2}+\cos \frac{\alpha-\beta}{2}\right)=\frac{H^2 \cos \frac{\alpha+\beta}{2}}{\sin \alpha \sin \beta}\left(\sin \frac{\alpha+\beta}{2}+\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left.+\sin \left(90^{\circ}-\frac{\alpha-\beta}{2}\right)\right)=\frac{2 H^2 \cos \frac{\alpha+\beta}{2} \cos \left(\frac{\alpha}{2}-45^{\circ}\right) \sin \left(45^{\circ}+\frac{\beta}{2}\right)}{\sin \alpha \sin \beta}= \\
& =\frac{2 H^2 \cos \frac{\alpha+\beta}{2} \sin \left(45^{\circ}+\frac{\alpha}{2}\right) \sin \left(45^{\circ}+\frac{\beta}{2}\right)}{\sin \alpha \sin \beta} \\
& \text { Answer }: \frac{2 H^2 \cos \frac{\alpha+\beta}{2} \sin \left(45^{\circ}+\frac{\alpha}{2}\right) \sin \left(45^{\circ}+\frac{\beta}{2}\right)}{\sin \alpha \sin \beta}
\end{aligned}
$$