Difference between revisions of "1961 AHSME Problems/Problem 10"
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| − | Each side of | + | == Problem== |
| + | |||
| + | Each side of <math>\triangle ABC</math> is <math>12</math> units. <math>D</math> is the foot of the perpendicular dropped from <math>A</math> on <math>BC</math>, | ||
| + | and <math>E</math> is the midpoint of <math>AD</math>. The length of <math>BE</math>, in the same unit, is: | ||
| + | |||
| + | <math>\textbf{(A)}\ \sqrt{18} \qquad | ||
| + | \textbf{(B)}\ \sqrt{28} \qquad | ||
| + | \textbf{(C)}\ 6 \qquad | ||
| + | \textbf{(D)}\ \sqrt{63} \qquad | ||
| + | \textbf{(E)}\ \sqrt{98}</math> | ||
| + | |||
| + | ==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> | ||
| + | Note that <math>\triangle ABC</math> is an [[equilateral triangle]]. From the [[Pythagorean Theorem]] (or by using 30-60-90 triangles), <math>AD = 6\sqrt{3}</math>. That means <math>DE = 3\sqrt{3}</math>. Using the Pythagorean Theorem again, <math>BE = \sqrt{63}</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1961|num-b=9|num-a=11}} | ||
| + | |||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 22:27, 26 May 2018
Problem
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. 
