1959 AHSME Problems/Problem 43

Problem

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

Solution

Use the formula : Circumradius = abc/4R

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1959 AHSME Problems/Problem 43

Problem

Solution