25. Let $U$ be a square with side length 1. If two points are chosen at random on the boundary of $U$, the probability that the distance between these two points is not less than $\frac{1}{2}$ is $\frac{a-b \pi}{c}\left(a, b, c \in \mathbf{Z}_{+},(a, b, c)=1\right)$. Then the value of $a + b + c$ is ( ).
(A) 59
(B) 60
(C) 61
(D) 62
(E) 63
25. A.

As shown in Figure 8, the sides of square $U$ are bisected, i.e.,
$$
\begin{array}{l}
A E=E B=B F \\
=F C=C G=G D \\
=D H=H A=\frac{1}{2} .
\end{array}
$$

By rotational symmetry, assume one point $S$ is on segment $A E$. Then when the other point $T$ is on $B F, F C, C G, G D$, or $D H$, $P\left(|S T| \geqslant \frac{1}{2}\right)=1$.
When point $T$ is on $A E$, $P\left(|S T| \geqslant \frac{1}{2}\right)=0$.
When point $T$ is on $B E$, let $A S=x, B T=y$. Then
$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant \frac{1}{2}, \\
0 \leqslant y \leqslant \frac{1}{2}, \\
x+y \leqslant \frac{1}{2} .
\end{array}\right.
$$

As shown in Figure 9.
It is easy to see that $P\left(|S T| \geqslant \frac{1}{2}\right)=\frac{1}{2}$.
When point $T$ is on $A H$, let $A S=x, A H=y$. Then
$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant \frac{1}{2}, \\
0 \leqslant y \leqslant \frac{1}{2}, \\
x^{2}+y^{2} \geqslant \frac{1}{4} .
\end{array}\right.
$$

It is easy to see that $P\left(|S T| \geqslant \frac{1}{2}\right)=\frac{\frac{1}{4}-\frac{\pi}{16}}{\frac{1}{4}}=\frac{4-\pi}{4}$.
Therefore, $P\left(|S T| \geqslant \frac{1}{2}\right)$
$$
\begin{array}{l}
=\frac{4-\pi}{4} \times \frac{1}{8}+1 \times \frac{5}{8}+0 \times \frac{1}{8}+\frac{1}{2} \times \frac{1}{8} \\
=\frac{26-\pi}{32} .
\end{array}
$$

Thus, $a+b+c=26+1+32=59$.