12.044. In a rectangular parallelepiped, the diagonal of the base is equal to $d$ and forms an angle $\alpha$ with a side of the base. A plane is drawn through this side and the opposite side of the upper base, forming an angle $\beta$ with the plane of the base. Find the lateral surface area of the parallelepiped.
Solution.

Given $A C=d, \angle A C D=\alpha, \angle A_{1} D A=\beta$ (Fig. 12.46).

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-746.jpg?height=64&width=1050&top_left_y=1178&top_left_x=155)

$D C=d \cos \alpha$. From $\triangle A A_{1} D, \angle A A_{1} D=90^{\circ}, A A_{1}=d \sin \alpha \cdot \operatorname{tg} \beta$, then $S_{A A_{1} D_{1} D}=A D \cdot A_{1} A=d^{2} \sin ^{2} \alpha \cdot \operatorname{tg} \beta ; S_{D D_{1} C_{1} C}=d^{2} \cos \alpha \cdot \sin \alpha \cdot \operatorname{tg} \beta$. We have $S_{6}=2 d^{2} \cdot \operatorname{tg} \beta \cdot \sin \alpha(\sin \alpha+\cos \alpha)=2 \sqrt{2} d^{2} \cdot \operatorname{tg} \beta \cdot \sin \alpha\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}+\cos \alpha \cdot \frac{1}{\sqrt{2}}\right)=$ $=2 \sqrt{2} d^{2} \operatorname{tg} \beta \cdot \sin \alpha\left(\sin \alpha \cdot \cos \frac{\pi}{4}+\cos \alpha \cdot \sin \frac{\pi}{4}\right)=$ $=2 \sqrt{2} d^{2} \cdot \sin \alpha \cdot \operatorname{tg} \beta \cdot \sin \left(\alpha+\frac{\pi}{4}\right)$.

Answer: $\quad 2 \sqrt{2} d^{2} \cdot \sin \alpha \cdot \operatorname{tg} \beta \cdot \sin \left(\alpha+\frac{\pi}{4}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-747.jpg?height=478&width=398&top_left_y=161&top_left_x=193)

Fig. 12.47

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-747.jpg?height=544&width=528&top_left_y=94&top_left_x=688)

Fig. 12.48