Difference between revisions of "1961 AHSME Problems/Problem 10"

(Solution to Problem 10)
Line 1: Line 1:
Each side of triangle <math>ABC</math> is 12 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:
+
== Problem 10==
 +
 
 +
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("$3\sqrt{3}$",(31,11));
 +
</asy>
 +
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}}
 
{{MAA Notice}}

Revision as of 14:52, 19 May 2018

Problem 10

Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$, and $E$ is the midpoint of $AD$. The length of $BE$, in the same unit, is:

$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$

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("$3\sqrt{3}$",(31,11)); [/asy] From the Pythagorean Theorem (or by using 30-60-90 triangles), $AD = 6\sqrt{3}$. That means $DE = 3\sqrt{3}$. Using the Pythagorean Theorem again, $BE = \sqrt{63}$, which is answer choice $\boxed{\textbf{(D)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png