1959 AHSME Problems/Problem 43: Difference between revisions
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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 == | |||
Use the formula : Circumradius = abc/4R | Use the formula : Circumradius = abc/4R | ||
Revision as of 13:06, 16 July 2024
Problem
The sides of a triangle are 
, and 
. The diameter of the circumscribed circle is:
Solution
Use the formula : Circumradius = abc/4R
