[ Special cases of parallelepipeds (other).] Area of the section

The base of a right parallelepiped is a rhombus, the area of which is equal to $Q$. The areas of the diagonal sections are $S 1$ and $S 2$. Find the volume of the parallelepiped.
Let $a$ and $b$ be the diagonals of the base of the given parallelepiped, and $h$ be its height. Since the parallelepiped is a right parallelepiped, its diagonal sections are rectangles. According to the problem,

$$
\frac{\frac{b b}{2}}{2}=Q, a h=S 1, b h=S 2
$$

From the first equality, it follows that $a b=2 Q$. Multiplying the last two equalities term by term, we find that

$$
h=\frac{\sqrt{S_{1} S_{2}}}{\sqrt{2 Q}}.
$$

Let $V$ be the volume of the parallelepiped. Then

$$
V=Q \cdot h=Q \cdot \frac{\sqrt{S_{1} S_{2}}}{\sqrt{2 Q}}=\sqrt{\frac{Q S_{1} S_{2}}{2}}.
$$

## Answer

The largest diagonal of a regular hexagonal prism is $d$ and forms an angle of $30^{\circ}$ with the lateral edge of the prism. Find the volume of the prism.

## Solution

Let the side of the regular hexagon at the base of the given prism be $a$. Then the largest diagonal of the hexagon is $2a$, and it is the orthogonal projection onto the base plane of the largest diagonal of the prism. Let $h$ be the height of the prism, and $V$ be its volume. From the problem, it follows that

$$
2 a=\frac{1}{2} d, h=d \cos 30^{\circ}=\frac{d \sqrt{3}}{2}
$$

If $S$ is the area of the base of the prism, then

$$
S=6 \cdot \frac{a^{2} \sqrt{3}}{4}=\frac{3 a^{2} \sqrt{3}}{2}=\frac{3 a^{2} \sqrt{3}}{32}.
$$

Therefore,

$$
V=S h=\frac{3 a^{2} \sqrt{3}}{32} \cdot \frac{d \sqrt{3}}{2}=\frac{9 a^{4}}{64}
$$

## Answer

$\frac{9 a^{4}}{64}$