9. Rhombus $ABCD$ is the base of a pyramid with vertex $S$. All its lateral faces are inclined to the base plane at the same angle of $60^{\circ}$. Points $M, N, K$, and $L$ are the midpoints of the sides of rhombus $ABCD$. A rectangular parallelepiped is constructed on rectangle $MNKL$ as its base. The edges of the upper face of the parallelepiped (opposite to face $MNKL$) intersect the lateral edges of the pyramid $SABCD$ at points $F, P, R$, and $Q$, respectively. The volume of the polyhedron with vertices at points $M, N, K, L, F, P, R, Q$ is $12 \sqrt{3}$, and the radius of the circle inscribed in rhombus $ABCD$ is 2.4. Find the side of rhombus $ABCD$. (16 points)
Solution.

The height of the pyramid $SO = h$. The diagonals of the rhombus $AC = d_1, BD = d_2$.

The height of the parallelepiped is $h / 2$, and the sides of the base of the parallelepiped are $d_1 / 2$ and $d_2 / 2$.

The volume of the parallelepiped is

$$
V_{\Pi} = \frac{d_1}{2} \cdot \frac{d_2}{2} \cdot \frac{h}{2}
$$

The volume of the polyhedron is

$$
V_{\mathrm{M}} = V_{\Pi} - 4 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{d_1}{4} \cdot \frac{d_2}{4} \cdot \frac{h}{2} = \frac{d_1 d_2 h}{8} - \frac{d_1 d_2 h}{48} = \frac{5 d_1 d_2 h}{48}
$$

According to the condition $\frac{5 d_1 d_2 h}{48} = 12 \sqrt{3}$.

Since the radius of the circle inscribed in this rhombus is 2.4, and all lateral faces of the pyramid are inclined to the base plane at an angle of $60^{\circ}$, then $h = 2.4 \sqrt{3}$. Therefore,

$$
\frac{5 d_1 d_2 \cdot 2.4 \sqrt{3}}{48} = 12 \sqrt{3}, \text{ and } d_1 d_2 = 48
$$

Since $\quad S_{ABCD} = \frac{d_1 d_2}{2}, \quad$ and $S_{ABCD} = \frac{4 AB \cdot r_{\mathrm{B}}}{2} = 4.8 \cdot AB, \quad$ then $4.8 \cdot AB = 24 \Rightarrow AB = 5$.

## Answer: 5

![](https://cdn.mathpix.com/cropped/2024_05_06_fb302ee711d03f314b6cg-07.jpg?height=1056&width=854&top_left_y=100&top_left_x=1018)