$20 \cdot 45$ The three sides of a triangle are 25, 39, and 40, and its circumcircle diameter is 
(A) $\frac{133}{3}$.
(B) $\frac{125}{3}$.
(C) 42 .
(D) 41 .
(E) 40 .
(10th American High School Mathematics Examination, 1959)
[Solution] Given $R=\frac{a b c}{4 S_{\triangle}}$, where $S_{\triangle}$ is the area of the triangle.
By Heron's formula $S_{\triangle}=\sqrt{p(p-a)(p-b)(p-c)}$, ( $p$ is the semi-perimeter of the triangle)
$$
\begin{array}{ll}
\therefore & R=\frac{25 \cdot 39 \cdot 40}{4 \sqrt{52 \cdot 27 \cdot 13 \cdot 12}}=\frac{125}{6}, \\
\therefore & 2 R=\frac{125}{3} .
\end{array}
$$

Therefore, the correct choice is (B).