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Difference between revisions of "1959 AHSME Problems/Problem 43"

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== Solution ==
 
== Solution ==
  
−
Use the formula : Circumradius = abc/4R
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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 $25,39$, 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 (Problems • Answer Key • Resources)
Preceded by
Problem 42 		Followed by
Problem 44
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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