12.225. The height of a regular triangular pyramid is $H$. The lateral face forms an angle $\alpha$ with the base plane. A plane is drawn through a side of the base and the midpoint of the opposite lateral edge. Find the area of the resulting section.
## Solution.

Let $MO$ be the height of the regular pyramid $MABC$ (Fig. 12.92), $MO=H, F$ be the midpoint of $BC$. Then $AF \perp BC, MF \perp BC, \angle AFM=\alpha$.

In $\triangle MOF\left(\angle MOF=90^{\circ}\right): OF=$ $H \operatorname{ctg} \alpha . K$ is the midpoint of $AM$. In the plane $AMF$, draw $KL \parallel MO$, then $KL \perp AF$ and $KL$ is the midline of $\triangle AOM$.

$$
KL=\frac{1}{2} MO=\frac{H}{2}, AL=LO
$$

Since $AO=2OF$, then $AL=$ $LO=OF, LF=2OF=2H \operatorname{ctg} \alpha$.

Let $LF$ be the projection of $KF$ on the plane $ABC, LF \perp BC$, then $KF \perp BC$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0974.jpg?height=700&width=618&top_left_y=78&top_left_x=656)

Fig. 12.92

In $\triangle BFO\left(\angle BFO=90^{\circ}\right)$ :

$$
BF=OF \operatorname{ctg} \angle OBF=H \operatorname{ctg} \alpha \operatorname{ctg} 30^{\circ}=H \sqrt{3} \operatorname{ctg} \alpha
$$

In $\triangle KLF\left(\angle KLF=90^{\circ}\right)$ :

$$
\begin{aligned}
& KF=\sqrt{KL^{2}+LF^{2}}=\sqrt{\frac{H^{2}}{4}+4H^{2} \operatorname{ctg}^{2} \alpha}=\frac{H}{2} \sqrt{1+16 \operatorname{ctg}^{2} \alpha} \\
& S_{\triangle BKC}=\frac{1}{2} BC \cdot KF=BF \cdot KF=\frac{1}{2} H^{2} \sqrt{3} \operatorname{ctg} \alpha \sqrt{1+16 \operatorname{ctg}^{2} \alpha} \\
& \text { Answer: } \frac{1}{2} H^{2} \sqrt{3} \operatorname{ctg} \alpha \sqrt{1+16 \operatorname{ctg}^{2} \alpha}
\end{aligned}
$$