11.113. In a slanted parallelepiped, the projection of a lateral edge onto the base plane is 5 dm, and the height is 12 dm. A section perpendicular to the lateral edge is a rhombus with an area of 24 dm² and a diagonal of 8 dm. Find the lateral surface area and the volume of the parallelepiped.
Solution.

Let $C O_{1}$ be the height of the oblique parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.11), $C_{1} O=12$ dm, then $C O$ is the projection of $C C_{1}$ onto the base plane, $C O=5$ dm, and the rhombus $M N K L$ is the perpendicular cross-section of the parallelepiped.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0828.jpg?height=420&width=600&top_left_y=307&top_left_x=661)

Fig. 11.11 From $\triangle C_{1} O C\left(\angle C_{1} O C=90^{\circ}\right): C C_{1}=\sqrt{C_{1} O^{2}+C O^{2}}=13$ dm.

Therefore, the volume of the parallelepiped $V=S_{M N K L} \cdot C C_{1}=312$ dm$^3$. Let $K M=8$ dm. Then $N L=\frac{2 S_{M N K L}}{K M}=6$ dm. Let $T$ be the point of intersection of the diagonals of the rhombus $M N K L$. Then $K L=\sqrt{K T^{2}+T L^{2}}=5$ dm, and the perimeter of the perpendicular cross-section $P_{M N K L}=20$ dm. The lateral surface area $S=P_{M N K L} \cdot C C_{1}=260$ dm$^2$.

Answer: 260 dm$^2$; 312 dm$^3$.