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

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<math>\frac{4\sqrt{3\sqrt{3}}}{3} \cdot 2\sqrt{\sqrt{3}} = \frac{8\sqrt{9}}{3} = \frac{24}{3} = \boxed{\textbf{(C) }8} </math>
 
<math>\frac{4\sqrt{3\sqrt{3}}}{3} \cdot 2\sqrt{\sqrt{3}} = \frac{8\sqrt{9}}{3} = \frac{24}{3} = \boxed{\textbf{(C) }8} </math>
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~ilee0820
  
 
== See Also ==
 
== See Also ==

Revision as of 17:57, 7 October 2024

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]

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

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. $8 \Longrightarrow \boxed{\textbf{(C) }8}$. Note: A quick way to visualize this method is to draw the line $DB$ and notice the two equilateral triangles $\triangle ADB$ and $\triangle DBC$.

Solution 3

We can extend line $DE$, meeting line $AB$ at $G$. Similarly, we can extend line $DE$ to meet line $BC$ at $H$. We can see with some simple math that triangle $ADG$ is a $30$-$60$-$90$ triangle, so we can call line $AG$ as $x$, line $DG$ as $x\sqrt{3}$, and line $AD$ as $2x$ (because of the $30$-$60$-$90$ triangle side proportions).

We can also see that line $AD$ is a base of rhombus $ABCD$, and line $DH$ is a height. Since triangle $DHC$ is also a $30$-$60$-$90$ triangle, line $DH$ is also $x\sqrt{3}$. Since the question told us that the area of rhombus $ABCD$ is $24$, we can make the following equation:

$2x \cdot x\sqrt{3} = 24$

Solving for x:

$2x^2\sqrt{3} = 24$

$x^2\sqrt{3} = 12$

$x^2 = \frac{12}{\sqrt{3}}$

$x^2 = 4\sqrt{3}$

$x = 2\sqrt{\sqrt{3}}$


Since the question is to find the area of rhombus $BFDE$, to find the answer, we can just multiply base $DE$ with the rhombus's height. We'll start by finding the height: instantly we can see that $GB$ is the height. Since all the sides of a rhombus are equal, and we found earlier that the side length is $2x$, if $AG$ is $x$, that means $GB$ is the same length as $AG$ - that is to say,

$GB = AG = 2\sqrt{\sqrt{3}}$


Now to find the base. We can see that to find the base, we can simply just subtract the length of line $EG$ from the length of line $DG$. Since $DG$ is $x\sqrt{3}$, and $x$ is $2\sqrt{\sqrt{3}}$, that makes

$DG = 2\sqrt{\sqrt{3}} \cdot \sqrt{3} = 2\sqrt{3\sqrt{3}}$


Now to find $EG$: We can see with simple math that triangle $EGB$ is also a $30$-$60$-$90$ triangle, which means that $EG = \frac{GB}{\sqrt{3}}$. Previously, we found out that $GB$ is $2\sqrt{\sqrt{3}}$, so:

$EG = \frac{2\sqrt{\sqrt{3}}}{\sqrt{3}} = \frac{2\sqrt{3\sqrt{3}}}{3}$


Now we can find the base:

$DG - EG = 2\sqrt{3\sqrt{3}} - \frac{2\sqrt{3\sqrt{3}}}{3} = \frac{4\sqrt{3\sqrt{3}}}{3}$


Multiplying the newly found base by the height we found earlier:

$\frac{4\sqrt{3\sqrt{3}}}{3} \cdot 2\sqrt{\sqrt{3}} = \frac{8\sqrt{9}}{3} = \frac{24}{3} = \boxed{\textbf{(C) }8}$ ~ilee0820

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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