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Difference between revisions of "2006 AMC 12B Problems/Problem 13"

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== Solution ==
 
== Solution ==
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The ratio of any length on ABCD to a corresponding length on BFDE squared is equal to the ratio of their areas. Since <math>\angle BAD=60 </math>, <math>\triangle ADB</math> and <math>\triangle DBC</math> are equilateral. DB, which is equal to AB, is the diagonal of rhombus ABCD. Therefore, <math>AC=\frac{DB(2)}{2\sqrt{3}}=\frac{DB}{\sqrt{3}}</math>. DB and AC are the longer diagonal of rhombuses BEDF and ABCD, respectively. So the ratio of their areas is <math>(\frac{1}{\sqrt{3}})^2</math> or <math>\frac{1}{3}</math>. One-third the area of ABCD is equal to 8. So the answer is <math>\boxed{C}</math>.
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The ratio of any length on <math>ABCD</math> to a corresponding length on <math>BFDE^2</math> is equal to the ratio of their areas. Since <math>\angle BAD=60 </math>, <math>\triangle ADB</math> and <math>\triangle DBC</math> are equilateral. <math>DB</math>, which is equal to <math>AB</math>, is the diagonal of rhombus <math>ABCD</math>. Therefore, <math>AC=\frac{DB(2)}{2\sqrt{3}}=\frac{DB}{\sqrt{3}}</math>. <math>DB</math> and <math>AC</math> are the longer diagonal of rhombuses <math>BEDF</math> and <math>ABCD</math>, respectively. So the ratio of their areas is <math>(\frac{1}{\sqrt{3}})^2</math> or <math>\frac{1}{3}</math>. One-third the area of <math>ABCD</math> is equal to <math>8</math>. So the answer is <math>\boxed{\text{C}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2006|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:02, 11 April 2016

Problem

Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is 24, and $\angle BAD = 60^\circ$. What is the area of rhombus $BFDE$?

[asy] defaultpen(linewidth(0.7)+fontsize(11)); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)/2); draw(B--C--D--A--B--F--D--E--B); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); [/asy]

$\textrm{(A) } 6 \qquad \textrm{(B) } 4\sqrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } 6\sqrt {3}$

Solution

Solution

The ratio of any length on $ABCD$ to a corresponding length on $BFDE^2$ is equal to the ratio of their areas. Since $\angle BAD=60$, $\triangle ADB$ and $\triangle DBC$ are equilateral. $DB$, which is equal to $AB$, is the diagonal of rhombus $ABCD$. Therefore, $AC=\frac{DB(2)}{2\sqrt{3}}=\frac{DB}{\sqrt{3}}$. $DB$ and $AC$ are the longer diagonal of rhombuses $BEDF$ and $ABCD$, respectively. So the ratio of their areas is $(\frac{1}{\sqrt{3}})^2$ or $\frac{1}{3}$. One-third the area of $ABCD$ is equal to $8$. So the answer is $\boxed{\text{C}}$.

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 12 Problems and Solutions

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