12.206. A section is made through the vertex of a regular triangular pyramid and the midpoints of two sides of the base. Find the area of the section and the volume of the pyramid, given the side $a$ of the base and the angle $\alpha$ between the section and the base.
Solution.

Let $D$ and $E$ be the midpoints of sides $AB$ and $AC$ of the base of a regular pyramid $SABC$ (Fig. 12.73), $\triangle DSE$ - a section of this pyramid, $SO$ the height of the pyramid, $K$ the point of intersection of $AO$ and $DE$. Then $OK \perp DE$, $DE=\frac{a}{2}$, $AK=\frac{a \sqrt{3}}{4}$, $OA=\frac{a \sqrt{3}}{3}$, $OK=OA-AK=\frac{a \sqrt{3}}{12}$.

The line $KO$ is the projection of $KS$ onto the plane of the base, therefore, $KS \perp DE$, $\angle SKO$ is the angle between the section and the base, $\angle SKO=\alpha$.

$\triangle DOE$ is the projection of $\triangle DSE$ onto the plane of the base of the pyramid, hence

$$
S_{\triangle DSE}=\frac{S_{\triangle DOE}}{\cos \angle SKO}=\frac{\frac{1}{2} DE \cdot OK}{\cos \alpha}=\frac{a^{2} \sqrt{3}}{48 \cos \alpha}
$$

In $\triangle SOK\left(\angle SOK=90^{\circ}\right): SO=OK \operatorname{tg} \angle SKO=\frac{a \sqrt{3}}{12} \operatorname{tg} \alpha$.

Volume of the pyramid: $V=\frac{1}{3} S_{\triangle ABC} \cdot SO=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot \frac{a \sqrt{3}}{12} \operatorname{tg} \alpha=\frac{a^{3} \operatorname{tg} \alpha}{48}$.

Answer: $\frac{a^{2} \sqrt{3}}{48 \cos \alpha} ; \frac{a^{3} \operatorname{tg} \alpha}{48}$.