10,11

The side of the base $ABC$ of the pyramid $TABCD$ is 4, the lateral edge $TA$ is perpendicular to the plane of the base. Find the area of the section of the pyramid by a plane passing through the midpoints of the edges $AC$ and $BT$ and parallel to the median $BD$ of the face $BCT$, given that the distance from the vertex $T$ to this plane is $\frac{1}{2}$.
The plane of the face $BCT$ passes through the line $BD$, which is parallel to the cutting plane, meaning it intersects the cutting plane along a line passing through the midpoint $N$ of the edge $BT$ parallel to $BD$. This line intersects the edge $CT$ at its midpoint $K$, so $\frac{TK}{KC}=\frac{1}{3}$. Let the line passing through the midpoint $M$ of the edge $AC$ and the point $K$ intersect the line $AT$ at the point $P$, and the line $PN$ intersect the edge $AB$ at the point $L$. Then the quadrilateral $MKNL$ is the section mentioned in the problem.

By Menelaus' theorem, $\frac{AM}{MI} \cdot \frac{CK}{KN} \cdot \frac{TP}{AP}=1$, or $\frac{1}{1} \cdot \frac{3}{1} \cdot \frac{TP}{AP}=1$, from which $\frac{TP}{AP}=\frac{1}{3}$. Similarly, $\frac{AL}{LB} \cdot \frac{BN}{NT} \cdot \frac{TP}{AP}=1$, or $\frac{AL}{LB} \cdot \frac{1}{1} \cdot \frac{1}{3}=1$, from which $\frac{AL}{LB}=3$.

Let $K_1$ and $N_1$ be the orthogonal projections of the points $K$ and $N$ onto the base plane $ABC$. Then $MK_1N_1L$ is the orthogonal projection of the section $MKLN$ onto the base plane. If $S_{MKNL}=S$, and $S_{MK_1LN_1}=s$, then $S=\frac{s}{\cos \alpha}$,
where $\alpha$ is the angle between the section plane and the base plane of the pyramid.

It is known that

$$
\frac{AM}{AC}=\frac{1}{2}, \frac{AL}{AB}=\frac{3}{1}, \frac{AK_1}{AC}=\frac{TK}{TC}=\frac{1}{3}, \frac{AN_1}{AB}=\frac{TN}{TB}=\frac{1}{2}
$$

therefore

$$
\begin{gathered}
s=S_{\triangle AML} - S_{\triangle AK_1N_1}=\frac{AM}{AC} \cdot \frac{AL}{AB} S_{\triangle ABC} - \frac{AK_1}{AC} \cdot \frac{AN_1}{AB} S_{\triangle ABC}= \\
=\left(\frac{1}{2} \cdot \frac{3}{4} - \frac{1}{4} \cdot \frac{1}{2}\right) \frac{t^2 \sqrt{3}}{4}=\frac{1}{4} \cdot 4 \sqrt{3}=\sqrt{3}
\end{gathered}
$$

Let $AE$ be the perpendicular dropped from point $A$ to $ML$. Then by the theorem of three perpendiculars, $PE \perp ML$, meaning $AEP$ is the linear angle of the dihedral angle between the cutting plane and the base plane of the pyramid, i.e., $\angle AEP=\alpha$. Moreover, the perpendiculars $TH$ and $AF$, dropped from points $T$ and $A$ to the line $PE$, are perpendiculars to the cutting plane. From the similarity of triangles $PHT$ and $PFA$, we find that $AF=TH \cdot \frac{PA}{PT}=\frac{1}{2} \cdot 3=\frac{3}{2}$.

By the cosine theorem

$$
ML=\sqrt{AM^2 + AL^2 - 2 \cdot AM \cdot AL \cos 60^\circ}=\sqrt{4+9-6}=\sqrt{7}
$$

and since

$$
S_{\triangle AML}=\frac{1}{2} \cdot \frac{3}{4} S_{\triangle ABC}=\frac{3}{8} \cdot 4 \sqrt{3}=\frac{3}{2} \sqrt{3}
$$

then

$$
AE=\frac{2 S_{\Delta+ML}}{ML}=\frac{3 \sqrt{3}}{\sqrt{7}}
$$

From the right triangle $AFE$, we find that

$$
\sin \alpha=\frac{AF}{AE}=\frac{\frac{3}{2}}{\frac{3 \sqrt{3}}{\sqrt{7}}}=\frac{\sqrt{7}}{2 \sqrt{3}}
$$

therefore $\cos \alpha=\sqrt{1-\frac{7}{12}}=\frac{\sqrt{5}}{2 \sqrt{3}} \cdot$ Consequently,

$$
S=\frac{s}{\cos \alpha}=\frac{\sqrt{3}}{\frac{\sqrt{5}}{2 \sqrt{3}}}=\frac{6}{\sqrt{5}}
$$

## Answer

$\frac{6}{\sqrt{5}}$