Difference between revisions of "1986 AHSME Problems/Problem 11"

Revision as of 18:57, 1 August 2017

Problem

In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$ . Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$ . The length of $HM$ is

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]$

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 7.5\qquad \textbf{(E)}\ 8$

Solution

In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is $13$ , the median must be $6.5$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 25: / Line 25: @@
 ==Solution==
-In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is <math>13</math>, the hypotenuse must be <math>6.5</math>.
+In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is <math>13</math>, the median must be <math>6.5</math>.
 == See also ==

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Difference between revisions of "1986 AHSME Problems/Problem 11"

Revision as of 18:57, 1 August 2017

Problem

Solution

See also