Example 1 Let the base edge length of the regular tetrahedron $V-A B C$ be 4, and the side edge length be 8. Construct a section $A E D$ through $A$ that intersects the side edges $V B, V C$. Find the minimum perimeter of the section $\triangle A E D$.
Solve As shown in Figure 17-3, the surface of the tetrahedron is cut along the side edge $V A$, and its unfolded diagram is shown in Figure 17-4. The perimeter of $\triangle A E D$ is minimized if and only if points $D, E$ lie on line $A A^{\prime}$. 

Let $\angle B V C=\alpha$, then $\angle A V A^{\prime}=3 \alpha . \cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$

In $\triangle V B C$, by the cosine rule, we get $\cos \alpha=\frac{V B^{2}+V C^{2}-B C^{2}}{2 \cdot V B \cdot V C}=\frac{2 \times 8^{2}-4^{2}}{2 \times 8^{2}}=\frac{7}{8}$

Therefore, $\cos 3 \alpha=4 \times\left(\frac{7}{8}\right)^{3}-3 \times \frac{7}{8}=\frac{7}{2^{7}}$

In $\triangle A V A^{\prime}$, by the cosine rule, we get
$$
A A^{\prime}=\sqrt{V A^{2}+V A^{\prime 2}-2 V A \cdot V A^{\prime} \cdot \cos 3 \alpha}=\sqrt{8^{2}+8^{2}-2 \cdot 8 \cdot 8 \cdot \frac{7}{2^{7}}}=11
$$

Thus, the minimum perimeter of $\triangle A D E$ is 11.