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

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== Problem ==
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The sides of a triangle are <math>25,39</math>, and <math>40</math>. The diameter of the circumscribed circle is:
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<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>   
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== Solution ==
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Use the formula : Circumradius = abc/4R
 
Use the formula : Circumradius = abc/4R

Latest revision as of 14:06, 16 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

Use the formula : Circumradius = abc/4R

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