1961 AHSME Problems/Problem 10

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 $\boxed{\textbf{(D)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1961 AHSME Problems/Problem 10

Problem 10

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