1961 AHSME Problems/Problem 10
Problem 10
Each side of 
is 
units. 
is the foot of the perpendicular dropped from 
on 
,
and 
is the midpoint of 
. The length of 
, in the same unit, is:
Solution
![[asy] draw((0,0)--(50,0)--(25,43.301)--cycle); draw((25,43.301)--(25,0)); dot((0,0)); label("$B$",(0,0),SW); dot((50,0)); label("$C$",(50,0),SE); dot((25,43.301)); label("$A$",(25,43.301),N); dot((25,0)); label("$D$",(25,0),S); dot((25,21.651)); label("$E$",(25,21.651),E); draw((25,21.651)--(0,0)); label("$12$",(10,25)); label("$6$",(12.5,-5)); label("$6$",(37.5,-5)); label("$12$",(40,25)); draw((25,3)--(28,3)--(28,0)); label("$3\sqrt{3}$",(30.5,11)); [/asy]](http://latex.artofproblemsolving.com/e/b/5/eb57508fc932a2c87cec459ea45cad89ccf91abd.png)
Note that is an equilateral triangle. From the Pythagorean Theorem (or by using 30-60-90 triangles), 
. That means 
. Using the Pythagorean Theorem again, 
, which is answer choice 
.
See Also
| 1961 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
