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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 29"
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Using [[Heron's Formula]] and <math>R=\frac{abc}{4A}</math>, the answer is <math>\frac{8}{\sqrt{15}}</math>.
 
Using [[Heron's Formula]] and <math>R=\frac{abc}{4A}</math>, the answer is <math>\frac{8}{\sqrt{15}}</math>.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 28|Previous Problem]]
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* [[[[University of South Carolina High School Math Contest/1993 Exam/Problem 30|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 10:42, 23 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 $\frac{8}{\sqrt{15}}$.

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