1959 AHSME Problems/Problem 43: Difference between revisions
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== Problem == | |||
The sides of a triangle are <math>25,39</math>, and <math>40</math>. The diameter of the circumscribed circle is: | |||
<math>\textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40</math> | |||
== Solution == | |||
The semiperimeter <math>s</math> of the triangle is <math>\frac{25+39+40}{2}=52</math>. 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 <math>\frac{abc}{4R}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are the sides of the triangle and <math>R</math> is the triangle's [[circumradius]]. Thus, we can equate this expression for the area with <math>468</math> to solve for <math>R</math>: | |||
\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 <math>2R=\boxed{\textbf{(B) }\frac{125}{3}}</math>. | |||
== See also == | |||
{{AHSME 50p box|year=1959|num-b=42|num-a=44}} | |||
{{MAA Notice}} | |||
Latest revision as of 11:09, 22 July 2024
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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| All AHSME Problems and Solutions | ||
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