1970 AHSME Problems/Problem 28

Problem

In triangle $ABC$, the median from vertex $A$ is perpendicular to the median from vertex $B$. If the lengths of sides $AC$ and $BC$ are $6$ and $7$ respectively, then the length of side $AB$ is

$\text{(A) } \sqrt{17}\quad \text{(B) } 4\quad \text{(C) } 4\tfrac{1}{2}\quad \text{(D) } 2\sqrt{5}\quad \text{(E) } 4\tfrac{1}{4}$

Solution

$\fbox{A}$ let the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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