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Difference between revisions of "1965 AHSME Problems/Problem 8"

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== Solution ==
 
== Solution ==
−
<math>\fbox{A}</math>
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 +
<asy>
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draw((0,0)--(18,0));
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dot((0,0));
 +
label("A", (-1.5,-0.5));
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dot((18,0));
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label("B",(19.5,-0.5));
 +
 
 +
label("$18$", (9,-1.5));
 +
 
 +
draw((0,0)--(5,10)--(18,0));
 +
dot((5,10));
 +
label("E", (6,11.5));
 +
 
 +
draw((0.918,1.835)--(15.615,1.835));
 +
dot((0.918,1.835));
 +
label("D", (-0.65,2.25));
 +
dot((15.615,1.835));
 +
label("C", (17.25,2.25));
 +
 
 +
</asy>
 +
 
 +
Let the given triangle be <math>\triangle ABE</math> with <math>AB=18</math>. Also, let <math>\overline{CD} \parallel \overline{AB}</math> with <math>C</math> on <math>\overline{BE}</math> and <math>D</math> on <math>\overline{AE}</math>. We know from the problem that <math>[ABCD]=\frac{1}{3}[\triangle ABE]</math>, so <math>[\triangle DCE]=\frac{2}{3}[\triangle ABE]</math>. Because <math>\overline{DC} \parallel \overline{AB}</math>, <math>\triangle DCE \sim \triangle ABE</math> by [[AA similarity]]. Because the ratio of these two triangles' areas is <math>\frac{2}{3}</math>, the ratio of their sides must be <math>\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}</math>. Thus, <math>CD=\frac{\sqrt{6}}{3}AB=\frac{\sqrt{6}}{3}*18=\boxed{6\sqrt{6}}</math>, which is answer choice <math>\fbox{A}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1965|num-b=7|num-a=9}}
 
{{AHSME 40p box|year=1965|num-b=7|num-a=9}}
 +
{{MAA Notice}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Latest revision as of 16:03, 18 July 2024

Problem

One side of a given triangle is 18 inches. Inside the triangle a line segment is drawn parallel to this side forming a trapezoid whose area is one-third of that of the triangle. The length of this segment, in inches, is:

$\textbf{(A)}\ 6\sqrt {6} \qquad \textbf{(B) }\ 9\sqrt {2} \qquad \textbf{(C) }\ 12 \qquad \textbf{(D) }\ 6\sqrt{3}\qquad \textbf{(E) }\ 9$

Solution

[asy] draw((0,0)--(18,0)); dot((0,0)); label("A", (-1.5,-0.5)); dot((18,0)); label("B",(19.5,-0.5)); label("$18$", (9,-1.5)); draw((0,0)--(5,10)--(18,0)); dot((5,10)); label("E", (6,11.5)); draw((0.918,1.835)--(15.615,1.835)); dot((0.918,1.835)); label("D", (-0.65,2.25)); dot((15.615,1.835)); label("C", (17.25,2.25)); [/asy]

Let the given triangle be $\triangle ABE$ with $AB=18$. Also, let $\overline{CD} \parallel \overline{AB}$ with $C$ on $\overline{BE}$ and $D$ on $\overline{AE}$. We know from the problem that $[ABCD]=\frac{1}{3}[\triangle ABE]$, so $[\triangle DCE]=\frac{2}{3}[\triangle ABE]$. Because $\overline{DC} \parallel \overline{AB}$, $\triangle DCE \sim \triangle ABE$ by AA similarity. Because the ratio of these two triangles' areas is $\frac{2}{3}$, the ratio of their sides must be $\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$. Thus, $CD=\frac{\sqrt{6}}{3}AB=\frac{\sqrt{6}}{3}*18=\boxed{6\sqrt{6}}$, which is answer choice $\fbox{A}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

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