Difference between revisions of "1958 AHSME Problems/Problem 43"

Revision as of 12:34, 23 February 2018

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

$\fbox{D}$

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == Solution ==
-<asy>
-import geometry;
-unitsize(50);
-pair A = (0,0), B = (3,0), C = (0, 4);
-pair AB = midpoint(A--B), AC = midpoint(A--C);
-draw(A--B--C--A);
-draw(A--B, StickIntervalMarker(2, 1));
-draw(A--C, StickIntervalMarker(2, 2));
-draw(C--AB);
-draw(B--AC);
-dot(AB);
-dot(AC);
-MP("$A$", A, W);
-MP("$B$", B, E);
-MP("$C$", C, W);
-MP("$M$", AB, S);
-MP("$N$", AC, W);
-label("$x$", A--AB, S);
-label("$x$", AB--B, S);
-label("$y$", A--AC, W);
-label("$y$", AC--C, W);
-draw(rightanglemark(C, A, B));
-</asy>

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Difference between revisions of "1958 AHSME Problems/Problem 43"

Revision as of 12:34, 23 February 2018

Problem

Solution

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