$\begin{aligned} & {[\text { Area of the cross-section }]} \\ & {[\text { Prism (other) }]}\end{aligned}$

The height of a right prism is 1, and the base of the prism is a rhombus with a side length of 2 and an acute angle of $30^{\circ}$. A plane that cuts through the prism is drawn through a side of the base, inclined to the base plane at an angle of $60^{\circ}$. Find the area of the cross-section.
Let $A B C D A 1 B 1 C 1 D 1$ be the given prism, the bases $A B C D$ and $A 1 B 1 C 1 D 1$ of which are rhombuses with side length 2, and $\angle D A B=30^{\circ}$, and $A A 1=B B 1=C C 1=D D 1=1$. If $D F$ is the height of the rhombus $A B C D$ dropped to the side $A B$, then by the theorem of three perpendiculars, $D 1 F \perp A B$, so $D F D 1$ is the linear angle of the dihedral angle between the planes of the base $A B C D$ and the diagonal section $A D 1 C 1 B$. Since $D F=A D \sin 30^{\circ}=1$, then $\operatorname{tg} \angle D F D 1=$ $\frac{\mathcal{D}_{1} D}{D F}=1$. Therefore, $\angle D F D 1=45^{\circ}<60^{\circ}$. This means that the given cutting plane intersects the edges $A 1 D 1$ and $B 1 C 1$. Let $M$ and $N$ be the corresponding points of intersection. Since the planes of the bases of the parallelepiped are parallel, as well as the planes of the opposite lateral faces, the quadrilateral $A M N B$ is a parallelogram. Let $M P$ be the perpendicular dropped from point $M$ to the plane of the base $A B C D$. Since the planes $A A 1 D 1 D$ and $A B C D$ are perpendicular, point $P$ lies on their line of intersection $A D$. If $M Q$ is the height of the parallelogram $A M N B$ dropped to the side $A B$, then by the theorem of three perpendiculars, $P Q \perp A B$, so $M Q P$ is the linear angle of the dihedral angle between the planes $A M N B$ and $A B C D$. According to the problem, $\angle M Q P=60^{\circ}$. Therefore,

$$
M Q=\frac{M P}{\operatorname{dn} M M Q Q P}=\frac{A A_{1}}{\operatorname{dn}\left(N^{2}\right.}=\frac{2}{\sqrt{3}} .
$$

Therefore,

$$
S_{A M N B}=A B \cdot M Q=2 \cdot \frac{2}{\sqrt{3}}=\frac{4}{\sqrt{3}}
$$

## Answer

$\frac{4}{\sqrt{3}}$