Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 29"

 
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== Problem ==
 
== Problem ==
 +
If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
  
<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
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<center><math>
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\mathrm{(A) \ } 2
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\qquad \mathrm{(B) \ } 8/\sqrt{15}
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\qquad \mathrm{(C) \ } 5/2
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\qquad \mathrm{(D) \ } \sqrt{6}
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\qquad \mathrm{(E) \ }  (\sqrt{6} + 1)/2
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</math></center>
  
 
== Solution ==
 
== Solution ==
 +
Using [[Heron's Formula]] and <math>R=\frac{abc}{4A}</math>, the answer is <math>\frac{8}{\sqrt{15}}</math>.
  
 
== See also ==
 
== See also ==
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam]]

Revision as of 19:18, 22 July 2006

Problem

If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15}  \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ }  (\sqrt{6} + 1)/2$

Solution

Using Heron's Formula and $R=\frac{abc}{4A}$, the answer is $\frac{8}{\sqrt{15}}$.

See also