2.46. 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 dm ${}^{2}$ and a diagonal of 8 dm. Find the lateral surface area and the volume of the parallelepiped.
2.46. According to the condition, $A_{1} K=12$ dm, $A K=5$ dm, $A_{1} K \perp A K$ (Fig. 2.40); therefore, $A A_{1}=\sqrt{A_{1} K^{2}+A K^{2}}=13$ (dm). Since $S_{A_{1} L M N}=$ $=24$ dm $^{2}$ and $A A_{1}$ is perpendicular to the section, the required volume is

$$
V=S_{A_{1} L M N} \cdot A A_{1}=24 \cdot 13=312\left(\text { dm }^{3}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-480.jpg?height=361&width=575&top_left_y=656&top_left_x=274)

Fig. 2.40

Next, considering that $A_{1} L M N$ is a rhombus, we have $S_{\text {side }}=4 A A_{1} \cdot A_{1} N$: To find the side of the rhombus, we use the equality

$$
S_{A_{1} L M N}=\frac{1}{2} A_{1} M \cdot L N, \text { or } 24=\frac{1}{2} \cdot 8 \cdot L N,
$$

from which $L N=6$ (dm); then $A_{1} N=\sqrt{3^{2}+4^{2}}=5$ (dm). Thus,

$$
S_{\text {side}}=4 \cdot 13 \cdot 5=260 \text { dm }^{2}
$$

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