1958 AHSME Problems/Problem 43

Problem

$\overline{AB}$ is the hypotenuse of a right triangle $ABC$ . Median $\overline{AD}$ has length $7$ and median $\overline{BE}$ has length $4$ . The length of $\overline{AB}$ is:

$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 5\sqrt{3}\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{13}\qquad \textbf{(E)}\ 2\sqrt{15}$

Solution

$[asy] import geometry; unitsize(50); pair C = (0,0), B = (3,0), A = (0, 4); pair AC = midpoint(A--C), BC = midpoint(B--C); draw(A--B--C--A); draw(A--C, StickIntervalMarker(2, 1)); draw(B--C, StickIntervalMarker(2, 2)); draw(B--AC); draw(A--BC); dot(AC); dot(BC); MP("$A$", A, W); MP("$B$", B, E); MP("$C$", C, W); MP("$E$", AC, W); MP("$D$", BC, S); label("$y$", A--AC, W); label("$y$", AC--C, W); label("$x$", B--BC, S); label("$x$", BC--C, S); draw(rightanglemark(A, C, B)); [/asy]$

By the Pythagorean Theorem, $(2x)^2+y^2=BE^2$ , and $x^2+(2y)^2=AD^2$ .

Plugging in, $4x^2+y^2=16$ , and $x^2+4y^2=49$ .

Adding the equations, $5x^2+5y^2=65$ , and dividing by $5$ , $x^2+y^2=13$ .

Note that the length $AB^2$ is equal to $(2x)^2+(2y)^2=4x^2+4y^2=4(x^2+y^2)$ .

Therefore, the answer is $\sqrt{4(13)}=2\sqrt{13}$ $\fbox{D}$

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
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1958 AHSME Problems/Problem 43

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