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

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==Problem==
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==FATSO==
  
 
In <math>\triangle ABC, AB = 13, BC = 14</math> and <math>CA = 15</math>. Also, <math>M</math> is the midpoint of side <math>AB</math> and <math>H</math> is the foot of the altitude from <math>A</math> to <math>BC</math>.  
 
In <math>\triangle ABC, AB = 13, BC = 14</math> and <math>CA = 15</math>. Also, <math>M</math> is the midpoint of side <math>AB</math> and <math>H</math> is the foot of the altitude from <math>A</math> to <math>BC</math>.  
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\textbf{(D)}\ 7.5\qquad
 
\textbf{(D)}\ 7.5\qquad
 
\textbf{(E)}\ 8    </math>
 
\textbf{(E)}\ 8    </math>
 
 
 
  
 
==Solution==
 
==Solution==

Revision as of 18:58, 1 August 2017

FATSO

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 $6.5$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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