Four. (15 points) As shown in Figure 5, in a "V-shaped" groove $A B C D E F$, $\angle A, \angle B, \angle D, \angle E$ are all right angles. Given that $P M, P N$ are perpendicular to $A B, D E$ respectively, with the feet of the perpendiculars being $M, N$. Find the length of the broken line $M P N$ (accurate to 0.01).


As shown in Figure 11, connect $F C$ intersecting $P N$ at point $Q$, then $Q$ is the midpoint of $F C$. Take the midpoint $R$ of $A B$, and connect $R Q$. Then $R Q$ is the midline of trapezoid $A B C F$. Draw $P S \perp R Q$, with the foot of the perpendicular at $S$, and draw $C G \perp E F$ and $C H \perp A F$, with the feet of the perpendiculars at $G$ and $H$ respectively.
In the right triangle $\triangle C G F$, let $\angle C F G=\alpha$. Then
$$
\tan \alpha=\frac{C G}{G F}=\frac{4}{10-8}=2 \Rightarrow \alpha \approx 63.4349^{\circ} \text {. }
$$

It is easy to see that $\angle H F C=\alpha$.
Since $F E / / P N$, we have $\angle F Q P=\angle C F G=\alpha$.
Also, since $A F / / R Q$, then
$$
\angle R Q F=180^{\circ}-\angle H F C=180^{\circ}-\alpha \text {. }
$$

In the right triangle $\triangle P S Q$,
$$
\begin{array}{l}
P S=M R=1, \\
\angle P Q S=\angle R Q F-\angle F Q P \\
=180^{\circ}-2 \alpha=53.1302^{\circ},
\end{array}
$$
$$
\begin{array}{l}
S Q=\frac{P S}{\tan 53.1302^{\circ}}=0.750, \\
P Q=\frac{P S}{\sin 53.1302^{\circ}}=1.250 .
\end{array}
$$

Thus, the length of the broken line MPN is
$$
\begin{array}{l}
(R Q-S Q)+(P Q+Q N) \\
=6-0.750+1.250+9 \approx 15.50 .
\end{array}
$$