12.249. The lateral edge of a regular truncated quadrilateral pyramid forms an angle $\alpha$ with the plane of the base. A rectangular parallelepiped is inscribed in the pyramid such that its upper base coincides with the upper base of the pyramid, and its lower base lies in the plane of the lower base of the pyramid. Find the ratio of the lateral surfaces of the pyramid and the parallelepiped, if the diagonal of the parallelepiped forms an angle $\beta$ with its base.
## Solution.

Let in the right truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.115) $\angle D_{1} D B=\alpha . A_{1} B_{1} C_{1} D_{1} A_{2} B_{2} C_{2} A_{2}$ - the inscribed rectangular parallelepiped $\angle B_{1} D_{2} B_{2}=\beta$.

Assume the lateral edge of the parallelepiped is 1, then from $\Delta B_{1} B_{2} D_{2}\left(\angle B_{1} B_{2} D_{2}=90^{\circ}\right): B_{2} D_{2}=B_{1} B_{2} \operatorname{ctg} \angle B_{1} D_{2} B_{2}=\operatorname{ctg} \beta$,

$C_{2} D_{2}=\frac{\sqrt{2}}{2} \operatorname{ctg} \beta$.

The lateral surface of the parallelepiped

$$
\begin{aligned}
& S_{1}=4 C_{2} D_{2} \cdot D_{1} D_{2}=2 \sqrt{2} \operatorname{ctg} \beta . \\
& \mathrm{B} \triangle D_{1} D_{2} D\left(\angle D_{1} D_{2} D=90^{\circ}\right): D_{2} D=D_{1} D_{2} \operatorname{ctg} \angle D_{1} D D_{2}=\operatorname{ctg} \alpha . \\
& B D=B_{2} D_{2}+2 D_{2} D=\operatorname{ctg} \beta+2 \operatorname{ctg} \alpha \Rightarrow D C=\frac{1}{\sqrt{2}} B D=\frac{\operatorname{ctg} \beta+2 \operatorname{ctg} \alpha}{\sqrt{2}}
\end{aligned}
$$

If $D_{1} E$ is the height of the trapezoid $D D_{1} C_{1} C$, then $D_{2} E \perp C D$.

From the isosceles right $\triangle D_{2} E D$ :

$D_{2} E=\frac{1}{\sqrt{2}} D_{2} D=\frac{1}{\sqrt{2}} \operatorname{ctg} \alpha$.

In $\triangle D_{1} D_{2} E\left(\angle D_{1} D_{2} E=90^{\circ}\right)$ :

$D_{1} E=\sqrt{D_{1} D_{2}^{2}++D_{2} E^{2}}=\sqrt{1+\frac{\operatorname{ctg}^{2} \alpha}{2}}=\frac{\sqrt{2+\operatorname{ctg}^{2} \alpha}}{\sqrt{2}}$.

The lateral surface area of the pyramid

$$
\begin{aligned}
& S_{2}=4 S_{D D_{1} C_{1} C}=4 \cdot \frac{C D \cdot C_{1} D_{1}}{2} \cdot D_{1} E=2\left(\frac{\operatorname{ctg} \beta+2 \operatorname{ctg} \alpha}{\sqrt{2}}+\frac{\operatorname{ctg} \beta}{\sqrt{2}}\right) \cdot \frac{\sqrt{2+\operatorname{ctg}^{2} \alpha}}{\sqrt{2}}= \\
& =2(\operatorname{ctg} \beta+\operatorname{ctg} \alpha) \sqrt{1+\frac{1}{\sin ^{2} \alpha}}=2 \cdot \frac{\sin (\alpha+\beta) \sqrt{1+\sin ^{2} \alpha}}{\sin ^{2} \alpha \sin \beta}
\end{aligned}
$$

Then $\frac{S_{2}}{S_{1}}=\frac{2 \sin (\alpha+\beta) \sqrt{1+\sin ^{2} \alpha}}{\sin ^{2} \alpha \sin \beta}: 2 \sqrt{2} \operatorname{ctg} \beta=\frac{\sin (\alpha+\beta) \sqrt{2\left(1+\sin ^{2} \alpha\right)}}{2 \sin ^{2} \alpha \cos \beta}$.

Answer: $\frac{\sin (\alpha+\beta) \sqrt{2\left(1+\sin ^{2} \alpha\right)}}{2 \sin ^{2} \alpha \cos \beta}$.