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Difference between revisions of "2014 AMC 10B Problems/Problem 15"

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==Solution==
 
==Solution==
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Let the length of <math>AD</math> be <math>x</math>, so that the length of <math>AB</math> is <math>2x</math> and <math>\text{[}ABCD\text{]}=2x^2</math>.
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Because <math>ABCD</math> is a rectangle, <math>\angle ADC=90^{\circ}</math>, and so <math>\angle ADC=\angle ECF=\angle FDC=30^{\circ}</math>. Thus <math>\triangle DAE</math> is a <math>30-60-90</math> right triangle; this implies that <math>\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}</math>, so <math>\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}</math>. Now drop the altitude from <math>E</math> of <math>\triangle DEF</math>, forming two <math>30-60-90</math> triangles. Because the length of <math>AD</math> is <math>x</math>, from the properties of a <math>30-60-90</math> triangle the length of <math>AE</math> is <math>\frac{x\sqrt{3}}{3}</math> and the length of <math>DE</math> is thus <math>\frac{2x\sqrt{3}}{3}</math>. Thus the altitude of <math>\triangle DEF</math> is <math>\frac{x\sqrt{3}}{3}</math>, and its base is <math>2x</math>, so its area is <math>\frac{1}{2}(2x)(\frac{x\sqrt{3}}{3})=\frac{x^2\sqrt{3}}{3}</math>.
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To finish, <math>\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(D) }}\sqrt{3}{6}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:13, 20 February 2014

Problem

In rectangle $ABCD$, $DC = 2CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$?

[asy] draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle); draw((0, 0)--(sqrt(3)/3, 1)); draw((0, 0)--(sqrt(3), 1)); label("A", (0, 1), N); label("B", (2, 1), N); label("C", (2, 0), S); label("D", (0, 0), S); label("E", (sqrt(3)/3, 1), N); label("F", (sqrt(3), 1), N); [/asy]

$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ (Error compiling LaTeX. Unknown error_msg)

Solution

Let the length of $AD$ be $x$, so that the length of $AB$ is $2x$ and $\text{[}ABCD\text{]}=2x^2$.

Because $ABCD$ is a rectangle, $\angle ADC=90^{\circ}$, and so $\angle ADC=\angle ECF=\angle FDC=30^{\circ}$. Thus $\triangle DAE$ is a $30-60-90$ right triangle; this implies that $\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}$, so $\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}$. Now drop the altitude from $E$ of $\triangle DEF$, forming two $30-60-90$ triangles. Because the length of $AD$ is $x$, from the properties of a $30-60-90$ triangle the length of $AE$ is $\frac{x\sqrt{3}}{3}$ and the length of $DE$ is thus $\frac{2x\sqrt{3}}{3}$. Thus the altitude of $\triangle DEF$ is $\frac{x\sqrt{3}}{3}$, and its base is $2x$, so its area is $\frac{1}{2}(2x)(\frac{x\sqrt{3}}{3})=\frac{x^2\sqrt{3}}{3}$.

To finish, $\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(D) }}\sqrt{3}{6}}$ (Error compiling LaTeX. Unknown error_msg)

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
All AMC 10 Problems and Solutions

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