3.75. Find the lateral surface area and volume of a right parallelepiped if its height is $h$, the diagonals form angles $\alpha$ and $\beta$ with the base, and the base is a rhombus.
3.75. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a right parallelepiped, $A B C D$ is a rhombus, $A A_{1}=h, \angle A C A_{1}=\alpha, \angle D B D_{1}=\beta$ (Fig. 3.76). From $\triangle A_{1} A C$ and $\triangle D_{1} D B$ we find $A C=h \operatorname{ctg} \alpha, B D=\boldsymbol{h} \operatorname{ctg} \beta$; then

$$
S_{A B C D}=\frac{1}{2} A C \cdot B D=\frac{1}{2} h^{2} \operatorname{ctg} \alpha \operatorname{ctg} \beta, V=\frac{1}{2} h^{3} \operatorname{ctg} \alpha \operatorname{ctg} \beta
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-547.jpg?height=411&width=650&top_left_y=1425&top_left_x=258)

Fig. 3.76

Finally, from $\triangle A O B$ we get

$$
A B=\sqrt{\frac{1}{4} h^{2}\left(\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta\right)}=\frac{1}{2} h \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}
$$

therefore $S_{\text {side }}=2 h^{2} \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$.