3. (17 points) The cross-section of a regular triangular pyramid passes through the midline of the base and is perpendicular to the base. Find the area of the cross-section if the side of the base is 8 and the height of the pyramid is 12.
Answer: 18.

Solution. The section MNP passes through the midline of the base of the pyramid $MN$ and is perpendicular to the base. Therefore, the height $PH$ of the triangle $MNP$ is parallel to the height of the pyramid $DO$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0785b6f179aec901e215g-4.jpg?height=514&width=526&top_left_y=83&top_left_x=822)

We have $OC: OK=2: 1, \quad CH=HK$, therefore, $CH: CO=3: 4$. Using the similarity of triangles $DOC$ and $PHC$, we get that $PH=\frac{3}{4} DO=\frac{3}{4} \cdot 12=9$. Then the area of the section is $S=\frac{1}{2} \cdot PH \cdot MN=\frac{1}{2} \cdot 9 \cdot 4=18$ sq.units.