Difference between revisions of "1959 AHSME Problems/Problem 43"
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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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
