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

Line 1: Line 1:
{{empty}}
 
  
 
== Problem ==
 
== Problem ==

Revision as of 19:18, 15 January 2012

Problem

Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is 24, and $\angle BAD \equal{} 60^\circ$ (Error compiling LaTeX. Unknown error_msg). 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 square of the ratio of any linear length on ABCD to the corresponding linear length on BFDE is equal to the ratio of their areas. because angle BAD=60 degrees, triangle ADB and triangle DBC are equilateral and DB is equal to the other sides of rhombus ABCD. therefore, AC=DB/2*sqrt(3)*2=DB*sqrt(3). DB and AC are the longer diagonal of rhombuses BEDF and ABCD, respectively. so the ratio of their areas is (1/sqrt(3))^2=1/3. area ABCD=24, so area BEDF=8 and the answer is 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