4. A regular triangular pyramid with base edge a and height $h$ is given. Find the area of the section of the pyramid by a plane passing through one of the base edges and the midpoint of the opposite lateral edge.
Solution. Let $ABCD$ be the given pyramid with base $ABC$, let $P$ be the midpoint of $AD$, let the intersection be $BPC$, let $M$ be the midpoint of $BC$, and let $H$ be the centroid of triangle $ABC$. Then $\overline{DH}=h$ and $\overline{AM}=\frac{a \sqrt{3}}{2}$. Let $Q$ be a point on $MA$ such that $PQ \| DH$ (see diagram). Since $P$ is the midpoint of $AD$, it follows that $PQ$ is the midline of triangle $ADH$ and $\overline{PQ}=\frac{h}{2}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_fbbd9d9f1d0d5083b697g-7.jpg?height=519&width=524&top_left_y=978&top_left_x=901)

$$
\overline{AQ}=\overline{QH}=\overline{MH}=\frac{\overline{AM}}{3}=\frac{a \sqrt{3}}{6} \quad \text{and} \quad \overline{MQ}=\frac{a \sqrt{3}}{3}
$$

Since $DH \perp MA$, it follows that $PQ \perp MQ$. According to this,

$$
\overline{PM}^{2}=\overline{MQ}^{2}+\overline{QP}^{2}=\frac{a^{2}}{3}+\frac{h^{2}}{4}
$$

so the desired area is

$$
P=\frac{\overline{BC} \cdot \overline{PM}}{2}=\frac{a}{2} \sqrt{\frac{a^{2}}{3}+\frac{h^{2}}{4}}
$$