一、(20 points) As
shown in Figure 3, the
shaded part is a
closed figure within
the semicircle $\odot 0$, it is known that chord $M N$
is parallel to chord $P Q$, and
the distance between the parallel lines $M N$ and $P Q$ is $2 \mathrm{~m}, M N=$
$6 \mathrm{~m}, P Q=8 \mathrm{~m}$. Find the length of $\overparen{M P}$ (accurate to $0.001 \mathrm{~m}$ ).
- 1. As shown in Figure 6, draw a radius $O T$ perpendicular to the parallel chords $M N$ and $P Q$, with the feet of the perpendiculars being $C$ and $D$, respectively. Connect $O P$ and $O M$. Let the distance from the center $O$ to $P Q$ be $s$, and the radius of $\odot O$ be $R$. Then, in the right triangles $\triangle O P D$ and $\triangle O M C$, we can obtain the system of equations:
$$
\left\{\begin{array}{l}
s^{2}+16=R^{2}, \\
(s+2)^{2}+9=R^{2} .
\end{array}\right.
$$

Solving these equations, we get $s=\frac{3}{4}, R=\frac{\sqrt{265}}{4}$.
Let $\angle T O M=\alpha, \angle T O P=\beta$. Then
$\tan \alpha=\frac{3}{h+s}=\frac{12}{11} \Rightarrow \alpha=\arctan \frac{12}{11}$,
$\tan \beta=\frac{4}{s}=\frac{16}{3} \Rightarrow \beta=\arctan \frac{16}{3}$.
Thus, the length of $\overparen{M P}$ is $=R(\beta-\alpha)$
$=\frac{\sqrt{265}}{4}\left(\arctan \frac{16}{3}-\arctan \frac{12}{11}\right)$
$\approx 2.26519511045$.
Therefore, the length of $\overparen{M P}$ is $2.265 \mathrm{~m}$.