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Perpendicular Medians

Source: 1970 AHSME Problem 28

April 16, 2014
AMC

Problem Statement

In triangle ABCABC, the median from vertex AA is perpendicular to the median from vertex BB. If the lengths of sides ACAC and BCBC are 66 and 77 respectively, then the length of side ABAB is
<spanclass=′latex−bold′>(A)</span>17<spanclass=′latex−bold′>(B)</span>4<spanclass=′latex−bold′>(C)</span>412<spanclass=′latex−bold′>(D)</span>25<spanclass=′latex−bold′>(E)</span>414<span class='latex-bold'>(A) </span>\sqrt{17}\qquad<span class='latex-bold'>(B) </span>4\qquad<span class='latex-bold'>(C) </span>4\dfrac{1}{2}\qquad<span class='latex-bold'>(D) </span>2\sqrt{5}\qquad <span class='latex-bold'>(E) </span>4\dfrac{1}{4}
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