5. $\frac{3 \sqrt{55}}{64}$.

As shown in Figure 2, the tetrahedron is cut along the edge $D A$ and unfolded into a pentagon.

When points $A, E, F, A_{1}$ are collinear, the perimeter of the section $\triangle A E F$ is minimized. At this time, the isosceles $\triangle D A A_{1}$ and $\triangle D B C$ have the same angle bisector, so $A A_{1} \parallel B C$. Therefore,
$$
\angle A B D=\angle D B C=\angle D E F=\angle A E B \text {, }
$$
which means $\triangle A B E$ is an isosceles triangle, and $\triangle A B E, \triangle A_{1} C F$, and $\triangle A E F$ are all similar to the triangular faces of the tetrahedron.
Let $B E=x, E F=y$.
From $\frac{B E}{A B}=\frac{E F}{D E}=\frac{B C}{D B}$
$\Rightarrow \frac{x}{1}=\frac{y}{2-x}=\frac{1}{2}$
$\Rightarrow x=\frac{1}{2}, y=\frac{3}{4}$.
Thus, $A E=A_{1} F=1, E F=\frac{3}{4}$.
In $\triangle A E F$,
$h=\sqrt{1-\left(\frac{3}{8}\right)^{2}}=\frac{\sqrt{55}}{8}$.
Therefore, $S=\frac{1}{2} E F \cdot h=\frac{3 \sqrt{55}}{64}$.
5. Given a regular tetrahedron $D-ABC$ with a base edge length of 1 and a side edge length of 2. A plane is drawn through point $A$ intersecting the side edges $BD$ and $CD$ at points $E$ and $F$ respectively. When the perimeter of $\triangle AEF$ is minimized, the area of $\triangle AEF$ is $\qquad$.