1959 AHSME Problem 43

The sides of a triangle are 25,39, and 40. The diameter of the circumscribed circle is: \textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.