2006 AMC 12B Problems/Problem 13

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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 (ProblemsAnswer KeyResources)
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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