Example 7 As shown in Figure 8, each face of the tetrahedron $ABCD$ is an acute triangle, and $AB=CD$ $=a, AC=BD=b, AD=$ $BC=c$, plane $\pi$ intersects edges $AB$, $BC$, $CD$, $DA$ at points $P$, $Q$, $R$, $S$ respectively. The minimum value of the perimeter of quadrilateral $PQRS$ is ( ).
(A) $2 a$
(B) $2 b$
(C) $2 c$
(D) $a+b+c$
Explanation: As shown in Figure 9, the lateral faces of the tetrahedron are unfolded into a plane figure. Since all faces of the tetrahedron are acute triangles, and 
$$
\begin{array}{l}
A B=C D, A C=B D, A D=B C, \\
\text { thus, in the unfolded plane figure, } \\
A D \Perp B C \\
\text { and } A^{\prime} D^{\prime} \text {, } \\
\end{array}
$$
$$
A B \Perp C D^{\prime} \text {, }
$$

where $A$ and $A^{\prime}$, $D$ and $D^{\prime}$ are the same points in the tetrahedron, hence $A, C, A^{\prime}, D, B, D^{\prime}$ are collinear in threes, and $A A^{\prime}=D D^{\prime}=2 B D$.

Furthermore, the quadrilateral $P Q R S$ in the unfolded figure becomes the broken line $S P Q R S^{\prime}$ (where $S$ and $S^{\prime}$ are the same point in the tetrahedron), thus, when $P, Q, R$ are on $S S^{\prime}$, $S P+P Q+Q R+R S^{\prime}$ is minimized, i.e., the perimeter of quadrilateral $P Q R S$ is minimized.
Since $S A=S^{\prime} A^{\prime}$, the minimum value is
$$
l=S S^{\prime}=D D^{\prime}=2 B D=2 b \text {. }
$$

Therefore, the correct choice is (B).