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

The semiperimeter $s$ of the triangle is $\frac{25+39+40}{2}=52$. Therefore, by Heron's Formula, we can find the area of the triangle as follows: \begin{align*} A &= \sqrt{52(52-25)(52-39)(52-40)} \\ &= \sqrt{52(27)(13)(12)} \\ &= \sqrt{13*4(9*3)(13)(4*3)} \\ &= 13*4*3*3 \\ &= 468 \end{align*} Also, we know that the area of the triangle is $\frac{abc}{4R}$, where $a$, $b$, and $c$ are the sides of the triangle and $R$ is the triangle's circumradius. Thus, we can equate this expression for the area with $468$ to solve for $R$: \begin{align*} \frac{(25)(39)(40)}{4R} &= 468 = 39*12 \\ \frac{25*40}{12} &= 4R \\ R &= \frac{250}{12} = \frac{125}{6} \end{align*} Because the problem asks for the diameter of the circumcircle, our answer is $2R=\boxed{\textbf{(B) }\frac{125}{3}}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 42
Followed by
Problem 44
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