1959 AHSME Problems/Problem 43
Problem
The sides of a triangle are 
, and 
. The diameter of the circumscribed circle is:
Solution
The semiperimeter 
of the triangle is 
. 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 
, where 
, 
, and 
are the sides of the triangle and 
is the triangle's circumradius. Thus, we can equate this expression for the area with 
to solve for :
\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 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 42 |
Followed by Problem 44 | |
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