5. Let the base edge length of the regular tetrahedron $V-ABC$ be 4, and the side edge length be 8. The section through $A$ intersecting the side edges $VB$ and $VC$ is $\triangle AED$. Then the minimum perimeter of the section $\triangle AED$ is ( ).
(A) $12 \frac{1}{5}$
(B) 11
(C) 12
(D) $11 \frac{1}{5}$
5. B.

From the edge $V A$, the surface of the tetrahedron is unfolded as shown in Figure 2. The perimeter of $\triangle A D E$ is minimized if and only if $D$ and $E$ lie on the line segment $A A_{1}$.
Let $\angle B V C=\alpha, \angle A V A_{1}=3 \alpha$. Then $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$.
In $\triangle V B C$, by the cosine rule, $\cos \alpha=\frac{V B^{2}+V C^{2}-B C^{2}}{2 V B \cdot V C}=\frac{7}{8}$. Therefore, $\cos 3 \alpha=\frac{7}{2^{7}}$. In $\triangle A V A_{1}$, by the cosine rule, $A A_{1}=\sqrt{V A^{2}+A_{1} V^{2}-2 V A \cdot A_{1} V \cos 3 \alpha}=11$.