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 squared 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 $\boxed{C}$ .

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

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2006 AMC 12B Problems/Problem 13

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