$\left[\begin{array}{l}\text { Properties of sections } \\ {[\text { Regular tetrahedron }]}\end{array}\right]$

A plane is drawn through the midpoint of an edge of a regular tetrahedron, perpendicular to the adjacent edge. Find the area of the resulting section if the edge of the tetrahedron is $a$.
Let $K$ be the midpoint of edge $AD$ of a regular tetrahedron $ABCD$. If a plane passing through point $K$ and perpendicular to line $CD$ intersects edges $CD$ and $BD$ at points $M$ and $L$, then $KM \perp CD$ and $LM \perp CD$, since lines $KM$ and $LM$ lie in a plane perpendicular to line $CD$. Let $N$ be the midpoint of segment $DC$. Then $AN$ and $BN$ are the heights of the equilateral triangles $ACD$ and $BCD$. Therefore, $KM \parallel AN$ and $LM \parallel BN$. Consequently, $KM$ and $LM$ are the midlines of triangles $AND$ and $BND$. Triangle $KML$ is similar to triangle $ANB$ with a similarity ratio of $\frac{1}{2}$. Therefore, $S_{\triangle KML} = \frac{1}{4} S_{\triangle ANB}$. In the isosceles triangle $ANB$, the sides are known:

$$
AB = a, \quad BN = AN = \sqrt{AC^2 - CN^2} = \sqrt{a^2 - \frac{a^2}{4}} = \frac{a \sqrt{3}}{2}
$$

Therefore,

$$
S_{\triangle ANB} = \frac{1}{2} AB \sqrt{AN^2 - \frac{AB^2}{4}} = \frac{1}{2} a \sqrt{\frac{3a^2}{4} - \frac{a^2}{4}} = \frac{a^2 \sqrt{2}}{4}
$$

Thus, $S_{\Delta KML} = \frac{\mathbf{a}^2 \sqrt{2}}{16}$

## Answer

$\frac{\mathbf{a}^2 \sqrt{2}}{16}$