Example 3 A closed figure formed by the intersection of a plane with the surface of a cube is called the "section figure" of the cube. In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $E$ is the midpoint of $A B$, and $F$ is the midpoint of $C C_{1}$. Then the perimeter of the section figure passing through points $D_{1} 、 E 、 F$ is equal to ( ).
(A) $\frac{1}{12}(25+2 \sqrt{13}+9 \sqrt{5})$
(B) $\frac{1}{12}(15+4 \sqrt{13}+9 \sqrt{5})$
(C) $\frac{1}{12}(25+2 \sqrt{13}+6 \sqrt{5})$
(D) $\frac{1}{12}(15+4 \sqrt{13}+6 \sqrt{5})$
Solution 1: As shown in Figure 3, extend $D_{1} F$ and $DC$ to meet at point $P$, draw line $EP$ to intersect $BC$ at point $N$ and the extension of $DA$ at point $S$, connect $D_{1} S$ to intersect $A_{1} A$ at point $M$, then the pentagon $D_{1} M E N F$ is the cross-sectional figure.
By the property of corresponding sides of similar triangles being proportional, we get
$$
A_{1} M=3 M A, C N=2 N B \text {. }
$$

It is easy to find that the perimeter of the cross-sectional pentagon $D_{1} M E N F$ is
$$
\frac{1}{12}(25+2 \sqrt{13}+9 \sqrt{5}) \text {. }
$$

Therefore, the correct choice is (A).
Solution 2: As shown in Figure 4,
draw $E M / / A_{1} S^{\prime} / / D_{1} F$
to intersect $A_{1} A$ at point $M$ (where $S$ is the midpoint of $B_{1} B$), draw $F N$
$/ / C_{1} T / / D_{1} M$ to intersect $B C$
at point $N$ (where $B T=A M$),
then the pentagon $D_{1} M E N F$
is the cross-sectional figure. As in Solution 1, the correct choice is (A).