10,11 A regular triangular pyramid is cut by a plane perpendicular to the base and dividing two sides of the base in half. Find the area of the section of the pyramid by this plane, given that the side of the base is 2, and the height of the pyramid is 4.

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Let the specified plane pass through the midpoints $M$ and $N$ of sides $AB$ and $AC$ of the base $ABC$ of a regular triangular pyramid $DABC$ and intersect the lateral edge $AD$ at point $P$, and the median $AF$ of the equilateral triangle $ABC$ at point $K$. The plane of triangle $ADF$ passes through the perpendicular $DO$ (the height of the pyramid) to the plane of the base $ABC$. Therefore, the plane $ADF$ is perpendicular to the plane of the base of the pyramid. The intersecting plane is also perpendicular to the plane of the base of the pyramid (by condition) and intersects the plane $ADF$ along the line $PK$. Thus, the line $PK$ is perpendicular to the plane of the base of the pyramid and $KP \parallel DO$. Further, we have:

$$
\begin{gathered}
AK=\frac{1}{2} AF, \quad AO=\frac{2}{3} AF, \quad \frac{AK}{AO}=\frac{\frac{1}{\sqrt{2}} AF}{AF}=\frac{3}{1} \\
KP=\frac{\partial D \cdot AK}{AC}=\frac{13}{1}=3, \quad MN=\frac{1}{2} BC=1
\end{gathered}
$$

Since $KP$ is the height of triangle $MPN$, we have

$$
S_{\triangle MPN}=\frac{1}{2} MN \cdot KP=\frac{1}{2} \cdot 1 \cdot 3=\frac{3}{2}.
$$

## Answer

1.5

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