16. Given a tetrahedron $ABCD$, the areas of faces $ABC$ and $ABD$ are 4 and 7, respectively, and the dihedral angle between them is $60^{\circ}$. Then the area of the section through edge $AB$ and the center of the inscribed sphere of the tetrahedron is $\qquad$.
Assuming $A B \perp$ plane $C D F, A B=2$, then $C F=4, D F=7$. The section $A B E$ is the bisector plane of the dihedral angle $C-A B-D$. By the angle bisector theorem, we get
$$
\begin{array}{l}
\overrightarrow{F E}=\frac{4}{11} \overrightarrow{F D}+\frac{7}{11} \overrightarrow{F C} \\
\Rightarrow \overrightarrow{F E}^{2}=\frac{1}{121}\left(16 \cdot 49+49 \cdot 16+56 \cdot 4 \cdot 7 \cdot \frac{1}{2}\right) \\
=\frac{3 \cdot 16 \cdot 49}{121} \Rightarrow E F=\frac{28 \sqrt{3}}{11} .
\end{array}
$$

Therefore, the area of the section through the edge $A B$ and the center of the inscribed sphere of the tetrahedron is $\frac{28 \sqrt{3}}{11}$.