[ Parallelism of lines and planes]

A plane is drawn through the diagonal of a cube with an edge length of $a$, parallel to the diagonal of one of the cube's faces. Find the area of the resulting section.

#
In the section, a rhombus will be obtained, the diagonals of which are equal to $a \sqrt{2}$ and $a \sqrt{3}$.

## Solution

Let the cutting plane pass through the diagonal $A_{1} C$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ parallel to the diagonal $B_{1} D_{1}$ of the square $A_{1} B_{1} C_{1} D_{1}$. Then the plane of the face $B_{1} B D D_{1}$ intersects the cutting plane along a line parallel to $B_{1} D_{1}$.

Let $M$ and $N$ be the points of intersection of this line with the edges $B B_{1}$ and $D D_{1}$, respectively. By the theorem on the intersection of two parallel planes by a third, $A_{1} M \| C N$ and $M C \| A_{1} N$. Therefore, the quadrilateral $A_{1} M C N$ is a parallelogram. Since the quadrilateral $B_{1} M N D_{1}$ is a rectangle, $M N=B_{1} D_{1}=a \sqrt{2}$.

The orthogonal projection $A_{1} C_{1}$ of the diagonal $A_{1} C$ of the cube onto the base plane $A_{1} B_{1} C_{1} D_{1}$ is perpendicular to $B_{1} D_{1}$. By the theorem of three perpendiculars, the diagonal $A_{1} C$ is perpendicular to the segment $B_{1} D_{1}$, and therefore to the segment $M N$. Thus, the diagonals of the parallelogram $A_{1} M N C$ are mutually perpendicular, i.e., it is a rhombus.

Therefore,

$$
S\left(A_{1} M C N\right)=\frac{1}{2} M N \cdot A_{1} C=\frac{1}{2} a \sqrt{2} \cdot a \sqrt{3}=\frac{1}{2} a^{2} \sqrt{6}
$$

## Answer

$$
a^{2} \sqrt{6} / 2 \text {. } \quad
$$

The base of a right parallelepiped is a parallelogram with an angle of $120^{\circ}$ and sides equal to 3 and 4.

The smaller diagonal of the parallelepiped is equal to the larger diagonal of the base. Find the volume of the parallelepiped.

## Solution

Let $d_{1}$ and $d_{2}$ be the larger and smaller diagonals of the parallelogram with sides equal to 3 and 4, and an angle of $120^{\circ}$ between them. Then

$$
\begin{aligned}
& d_{1}=\sqrt{9+16-2 \cdot 3 \cdot 4 \cdot \cos 120^{\circ}}=\sqrt{25+12}=\sqrt{37} \\
& d_{2}=\sqrt{9+16-2 \cdot 3 \cdot 4 \cdot \cos 60^{\circ}}=\sqrt{25-12}=\sqrt{13}
\end{aligned}
$$

Let $h$ be the height of the parallelepiped, and $V$ be its volume. According to the problem, the smaller diagonal of the parallelepiped is equal to the larger diagonal of the parallelogram, i.e., $\sqrt{37}$, and since the smaller diagonal of the parallelogram is the orthogonal projection of the smaller diagonal of the parallelepiped onto the base plane, then

$$
h=\sqrt{37-13}=\sqrt{24}=2 \sqrt{6} .
$$

Therefore,

$$
V=3 \cdot 4 \cdot \sin 120^{\circ} \cdot 2 \sqrt{6}=6 \sqrt{3} \cdot 2 \sqrt{6}=36 \sqrt{2}
$$

## Answer

$$
36 \sqrt{2} \text {. }
$$