2006 AMC 12B Problems/Problem 13

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

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 $\boxed{\text{C}}$ .

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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2006 AMC 12B Problems/Problem 13

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