Difference between revisions of "1986 AHSME Problems/Problem 11"
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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> | ||
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==Solution== | ==Solution== | ||
Revision as of 18:58, 1 August 2017
FATSO
In 
and 
. Also, 
is the midpoint of side 
and 
is the foot of the altitude from 
to 
.
The length of 
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]](http://latex.artofproblemsolving.com/4/3/9/439be0c9c9fcdca68ec7ff37d19968d3ec0f7c14.png)
Solution
In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is 
, the median must be 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
