$\left[\frac{\text { Properties of sections }}{\text { Cube }}\right]$

Given a cube ABCDA1B1C1D1 with edge $a$. Find the area of the section of this cube by a plane passing through vertex $C$ and the midpoints of edges $C 1 B 1$ and $C 1 D 1$.
Let $M$ and $N$ be the midpoints of the edges $C_1 B_1$ and $C_1 D_1$ of the cube $A B C D A_1 B_1 C_1 D_1$. In the section, an isosceles triangle $C M N$ is obtained. If $K$ is the point of intersection of the segments $M N$ and $A_1 C_1$, then $C_1 K \perp M N$, and $C_1 K$ is the orthogonal projection of the inclined $C K$ on the plane $A_1 B_1 C_1 D_1$. By the theorem of three perpendiculars, $C K \perp M N$, so $C K$ is the height of the triangle $C M N$. Since $M N$ is the midline of the triangle $B_1 C_1 D_1$,

$$
M N = \frac{1}{2} B_1 D_1 = \frac{1}{2} a \sqrt{2}, \quad C_1 K = \frac{1}{4} A_1 C_1 = \frac{1}{4} a \sqrt{2}.
$$

By the Pythagorean theorem from the right triangle $C C_1 K$, we find that

$$
C K = \sqrt{C_1 C^2 + C_1 K^2} = \sqrt{a^2 + \frac{a^2}{8}} = \frac{3 a \sqrt{2}}{4}.
$$

Therefore,

$$
S_{\triangle C M N} = \frac{1}{2} M N \cdot C K = \frac{1}{2} \cdot \frac{a \sqrt{2}}{2} \cdot \frac{3 a \sqrt{2}}{4} = \frac{3 a^2}{8}.
$$

$\frac{3}{8} a^2$.