12.209. 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.
Solution.

Let the rhombus $ABCD$ be the base of a right prism

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0956.jpg?height=458&width=677&top_left_y=1370&top_left_x=582)

Fig. 12.76
$ABCD A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.76), $AA_{1}=BB_{1}=h, O$ - the point of intersection of the diagonals $AC$ and $BD$ of the rhombus, $\angle A_{1}CA=\alpha, \angle B_{1}DB=\beta$.

In $\triangle B_{1}BD\left(\angle B_{1}BD=90^{\circ}\right): BD=BB_{1} \operatorname{ctg} \angle B_{1}DB=h \operatorname{ctg} \beta$.

From $\triangle A_{1}AC\left(\angle A_{1}AC=90^{\circ}\right): AC=AA_{1} \operatorname{ctg} \angle A_{1}CA=h \operatorname{ctg} \alpha$.

The area of the base of the prism $S=\frac{1}{2} AC \cdot BD=\frac{1}{2} h^{2} \operatorname{ctg} \alpha \operatorname{ctg} \beta \Rightarrow$ the volume of the prism $V=S h=\frac{1}{2} h^{3} \operatorname{ctg} \alpha \operatorname{ctg} \beta$.

$$
OA=\frac{1}{2} AC=\frac{1}{2} h \operatorname{ctg} \alpha, OD=\frac{1}{2} BD=\frac{1}{2} h \operatorname{ctg} \beta
$$

In $\triangle AOD\left(\angle AOD=90^{\circ}\right): AD=\sqrt{AO^{2}+OD^{2}}=\frac{1}{2} h \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$.

The lateral surface area of the prism: $S_{\sigma}=4 AD \cdot h=2 h^{2} \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$.

Answer: $2 h^{2} \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta} ; \frac{1}{2} h^{3} \operatorname{ctg} \alpha \operatorname{ctg} \beta$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0957.jpg?height=499&width=668&top_left_y=905&top_left_x=92)

Fig. 12.77