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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− | * [[University of South Carolina High School Math Contest/1993 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?
![$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2$](http://latex.artofproblemsolving.com/a/d/3/ad39e455fe156ef0c687ce7c7d910ebd4f9c9f27.png)
Solution
Using Heron's Formula and , the answer is
.