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Difference between revisions of "2006 AMC 10B Problems/Problem 15"

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The longer [[diagonal]] of rhombus <math>BFDE</math> is <math>BD</math>. Since <math>BD</math> is a side of an equilateral triangle with a side length of <math>s</math>, <math> BD = s </math>. The longer diagonal of rhombus <math>ABCD</math> is <math>AC</math>. Since <math>AC</math> is twice the length of an altitude of of an equilateral triangle with a side length of <math>s</math>, <math> AC = 2 \cdot \frac{s\sqrt{3}}{2} = s\sqrt{3} </math>.
 
The longer [[diagonal]] of rhombus <math>BFDE</math> is <math>BD</math>. Since <math>BD</math> is a side of an equilateral triangle with a side length of <math>s</math>, <math> BD = s </math>. The longer diagonal of rhombus <math>ABCD</math> is <math>AC</math>. Since <math>AC</math> is twice the length of an altitude of of an equilateral triangle with a side length of <math>s</math>, <math> AC = 2 \cdot \frac{s\sqrt{3}}{2} = s\sqrt{3} </math>.
  
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The ratio of the longer diagonal of rhombus <math>BFDE</math> to rhombus <math>ABCD</math> is <math> \frac{s}{s\sqrt{3}} = \frac{\sqrt{3}}{3}</math>. Therefore, the ratio of the [[area]] of rhombus <math>BFDE</math> to rhombus <math>ABCD</math> is <math> \left( \frac{\sqrt{3}}{3} \right) ^2 = \frac{1}{3} </math>
+
The ratio of the longer diagonal of rhombus <math>BFDE</math> to rhombus <math>ABCD</math> is <math> \frac{s}{s\sqrt{3}} = \frac{\sqrt{3}}{3}</math>. Therefore, the ratio of the [[area]] of rhombus <math>BFDE</math> to rhombus <math>ABCD</math> is <math> \left( \frac{\sqrt{3}}{3} \right) ^2 = \frac{1}{3} </math>.
  
 
Let <math>x</math> be the area of rhombus <math>BFDE</math>. Then <math> \frac{x}{24} = \frac{1}{3} </math>, so <math> x = 8 \Longrightarrow \boxed{\mathrm{(C)}}</math>.
 
Let <math>x</math> be the area of rhombus <math>BFDE</math>. Then <math> \frac{x}{24} = \frac{1}{3} </math>, so <math> x = 8 \Longrightarrow \boxed{\mathrm{(C)}}</math>.

Revision as of 15:56, 2 June 2021

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(10)); size(120); 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]

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

Solutions

Solution 1

Using kiouonp of a rhombus, $\angle DAB = \angle DCB = 60 ^\circ$ and $\angle ADC = \angle ABC = 120 ^\circ$. It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$. Let the lengths of the sides of rhombus $ABCD$ be $s$.

The longer diagonal of rhombus $BFDE$ is $BD$. Since $BD$ is a side of an equilateral triangle with a side length of $s$, $BD = s$. The longer diagonal of rhombus $ABCD$ is $AC$. Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$, $AC = 2 \cdot \frac{s\sqrt{3}}{2} = s\sqrt{3}$.

The ratio of the longer diagonal of rhombus $BFDE$ to rhombus $ABCD$ is $\frac{s}{s\sqrt{3}} = \frac{\sqrt{3}}{3}$. Therefore, the ratio of the area of rhombus $BFDE$ to rhombus $ABCD$ is $\left( \frac{\sqrt{3}}{3} \right) ^2 = \frac{1}{3}$.

Let $x$ be the area of rhombus $BFDE$. Then $\frac{x}{24} = \frac{1}{3}$, so $x = 8 \Longrightarrow \boxed{\mathrm{(C)}}$.

Solution 2

Triangle DAB is equilateral so triangles $DEA$, $AEB$, $BED$, $BFD$, $BFC$ and $CFD$ are all congruent with angles $30^\circ$, $30^\circ$ and $120^\circ$ from which it follows that rhombus $BFDE$ has one third the area of rhombus $ABCD$ i.e. $8 \Longrightarrow \boxed{\mathrm{(C)}}$.

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 10 Problems and Solutions

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