1962 AHSME Problems/Problem 39: Difference between revisions

Revision as of 22:01, 10 November 2013

Problem

Two medians of a triangle with unequal sides are $3$ inches and $6$ inches. Its area is $3 \sqrt{15}$ square inches. The length of the third median in inches, is:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6}$

Solution

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Revision as of 22:26, 9 November 2013 view source Fadebekun (talk \| contribs) 300 edits Created page with "==Problem== Two medians of a triangle with unequal sides are <math>3</math> inches and <math>6</math> inches. Its area is <math>3 \sqrt{15}</math> square inches. The length of th..."		Revision as of 22:01, 10 November 2013 view source Fadebekun (talk \| contribs) 300 edits →Solution Newer edit →
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1962 AHSME Problems/Problem 39: Difference between revisions

Revision as of 22:01, 10 November 2013

Problem

Solution