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

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Rhombus <math>ABCD</math> is similar to rhombus <math>BFDE</math>. The area of rhombus <math>ABCD</math> is <math>24</math> and <math> \angle BAD = 60^\circ </math>. What is the area of rhombus <math>BFDE</math>?  
 
Rhombus <math>ABCD</math> is similar to rhombus <math>BFDE</math>. The area of rhombus <math>ABCD</math> is <math>24</math> and <math> \angle BAD = 60^\circ </math>. What is the area of rhombus <math>BFDE</math>?  
  
<!-- [[Image:2006amc10b15.gif]] -->
 
 
<asy> defaultpen(linewidth(0.7)+fontsize(10)); size(120);
 
<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 A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3));
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label("$F$", F, dir(point--F));
 
label("$F$", F, dir(point--F));
 
</asy>
 
</asy>
<math> \mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 4\sqrt{3}\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 6\sqrt{3} </math>
 
  
== Solution ==
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<math> \textbf{(A) } 6\qquad \textbf{(B) } 4\sqrt{3}\qquad \textbf{(C) } 8\qquad \textbf{(D) } 9\qquad \textbf{(E) } 6\sqrt{3} </math>
Using properties of a [[rhombus]], <math> \angle DAB = \angle DCB = 60 ^\circ </math> and <math> \angle ADC = \angle ABC = 120 ^\circ </math>. It is easy to see that rhombus <math>ABCD</math> is made up of [[equilateral triangle]]s <math>DAB</math> and <math>DCB</math>. Let the lengths of the sides of rhombus <math>ABCD</math> be <math>s</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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== Solution 1 ==
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Using the property that opposite angles are equal in a [[rhombus]], <math> \angle DAB = \angle DCB = 60 ^\circ </math> and <math> \angle ADC = \angle ABC = 120 ^\circ </math>. It is easy to see that rhombus <math>ABCD</math> is made up of [[equilateral triangle]]s <math>DAB</math> and <math>DCB</math>. Let the lengths of the sides of rhombus <math>ABCD</math> be <math>s</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>
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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>.
  
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>.
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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>.
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Let <math>x</math> be the area of rhombus <math>BFDE</math>. Then <math> \frac{x}{24} = \frac{1}{3} </math>, so <math> x = \boxed{\textbf{(C) }8}</math>.
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== Solution 2 ==
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Triangle DAB is equilateral so triangles <math>DEA</math>, <math>AEB</math>, <math>BED</math>, <math>BFD</math>, <math>BFC</math> and <math>CFD</math> are all congruent with angles <math>30^\circ</math>, <math>30^\circ</math> and <math>120^\circ</math> from which it follows that rhombus <math>BFDE</math> has one third the area of rhombus <math>ABCD</math> i.e. <math>8 \Longrightarrow \boxed{\textbf{(C) }8} </math>.
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Area Problems]]
 
[[Category:Area Problems]]
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{{MAA Notice}}

Revision as of 14:20, 26 January 2022

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 $BFDE$?

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

Solution 1

Using the property that opposite angles are equal in 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 $x = \boxed{\textbf{(C) }8}$.

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. $8 \Longrightarrow \boxed{\textbf{(C) }8}$.

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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