8,9

Given a cube $A B C D A_1 B_1 C_1 D_1$ with edge $a$. On the rays $C_1 C, C_1 B_1$, and $C_1 D_1$, segments $C_1 M$, $C_1 N$, and $C_1 K$ are laid out, each equal to $\frac{5}{2} a$. Construct the section of this cube by the plane passing through points $M, N, K$ and find the area of the resulting section.
Lines $M K$ and $C D$ lie in the plane of the face $D C C 1 D 1$. The point $F$ of their intersection lies in the cutting plane (since it belongs to the line $M K$, which lies in this plane) and in the plane of the face $A B C D$ (since it belongs to the line $C D$). Therefore, point $F$ is located on the line of intersection of the cutting plane with the plane of the face $A B C D$. Similarly, the point $E$ of intersection of the lines $B C$ and $M N$ is also located on the line of intersection of these planes. Thus, the cutting plane intersects the plane of the face $A B C D$ along the line $E F$. From the similarity of triangles $M C F$ and $M C 1 K$ it follows that

![](https://cdn.mathpix.com/cropped/2024_05_06_b09eb9152e3973547418g-20.jpg?height=120&width=618&top_left_y=2673&top_left_x=720)

$$
D F=C F-C D=\frac{3}{2} a-a=\frac{1}{2} a
$$

Similarly, $C E=\frac{3}{2} a, B E=\frac{1}{2} a$. Therefore, $E F \| B D$. Let $P$ and $Q$ be the points of intersection of the line $E F$ with the edges $A B$ and $A D$. Then the quadrilateral $B P F D$ is a parallelogram. Therefore, $B P=D F=\frac{1}{2} a$. Similarly, $D Q=$ $B E=\frac{1}{2} a$. Therefore, $P$ and $Q$ are the midpoints of $A B$ and $A D$. Therefore, $P Q$ is the midline of the triangle $A B D, P Q=\frac{1}{2}$ $B D=\frac{a \sqrt{2}}{2}$. In the same way, we construct the point $R$ of intersection of the cutting plane with the edge $A A 1$ and prove that $R$ is the midpoint of $A A 1$ and $Q R=P R=\frac{a \sqrt{2}}{2}$. Thus, the constructed section is an equilateral triangle $P Q R$ with side $\frac{a \sqrt{2}}{2}$. Therefore,

$$
S_{\triangle P Q R}=\frac{\left(\frac{(a \sqrt{2}}{2}\right)^{2} \sqrt{3}}{4}=\frac{a^{2} \sqrt{3}}{8}
$$

## Answer