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

Latest revision as of 13:18, 12 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 (A bit more math involved)

We can extend line $DE$ , meeting line $AB$ at $G$ . Similarly, we can extend line $DF$ 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

@@ Line 2: / Line 2: @@
 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);
 pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3));
@@ Line 14: / Line 13: @@
 label("$F$", F, dir(point--F));
 </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 ==
+<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>
+== Solution 1 ==
+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>
+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>.
+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 = \boxed{\textbf{(C) }8}</math>.
+== Solution 2 ==
+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>.
+Note: A quick way to visualize this method is to draw the line <math>DB</math> and notice the two equilateral triangles <math>\triangle ADB</math> and <math>\triangle DBC</math>.
+== Solution 3 (A bit more math involved) ==
+We can extend line <math>DE</math>, meeting line <math>AB</math> at <math>G</math>. Similarly, we can extend line <math>DF</math> to meet line <math>BC</math> at <math>H</math>. We can see with some simple math that triangle <math>ADG</math> is a <math>30</math>-<math>60</math>-<math>90</math> triangle, so we can call line <math>AG</math> as <math>x</math>, line <math>DG</math> as <math>x\sqrt{3}</math>, and line <math>AD</math> as <math>2x</math> (because of the <math>30</math>-<math>60</math>-<math>90</math> triangle side proportions).
+We can also see that line <math>AD</math> is a base of rhombus <math>ABCD</math>, and line <math>DH</math> is a height. Since triangle <math>DHC</math> is also a <math>30</math>-<math>60</math>-<math>90</math> triangle, line <math>DH</math> is also <math>x\sqrt{3}</math>. Since the question told us that the area of rhombus <math>ABCD</math> is <math>24</math>, we can make the following equation:
+<math>2x \cdot x\sqrt{3} = 24</math>
+Solving for x:
+<math>2x^2\sqrt{3} = 24</math>
+<math>x^2\sqrt{3} = 12</math>
+<math>x^2 = \frac{12}{\sqrt{3}}</math>
+<math>x^2 = 4\sqrt{3}</math>
+<math>x = 2\sqrt{\sqrt{3}}</math>
+Since the question is to find the area of rhombus <math>BFDE</math>, to find the answer, we can just multiply base <math>DE</math> with the rhombus's height. We'll start by finding the height: instantly we can see that <math>GB</math> is the height. Since all the sides of a rhombus are equal, and we found earlier that the side length is <math>2x</math>, if <math>AG</math> is <math>x</math>, that means <math>GB</math> is the same length as <math>AG</math> - that is to say,
+<math>GB = AG = 2\sqrt{\sqrt{3}}</math>
+Now to find the base. We can see that to find the base, we can simply just subtract the length of line <math>EG</math> from the length of line <math>DG</math>. Since <math>DG</math> is <math>x\sqrt{3}</math>, and <math>x</math> is <math>2\sqrt{\sqrt{3}}</math>, that makes
+<math>DG = 2\sqrt{\sqrt{3}} \cdot \sqrt{3} = 2\sqrt{3\sqrt{3}}</math>
+Now to find <math>EG</math>: We can see with simple math that triangle <math>EGB</math> is also a <math>30</math>-<math>60</math>-<math>90</math> triangle, which means that <math>EG = \frac{GB}{\sqrt{3}}</math>. Previously, we found out that <math>GB</math> is <math>2\sqrt{\sqrt{3}}</math>, so:
+<math>EG = \frac{2\sqrt{\sqrt{3}}}{\sqrt{3}} = \frac{2\sqrt{3\sqrt{3}}}{3}</math>
+Now we can find the base:
+<math>DG - EG = 2\sqrt{3\sqrt{3}} - \frac{2\sqrt{3\sqrt{3}}}{3} = \frac{4\sqrt{3\sqrt{3}}}{3}</math>
+Multiplying the newly found base by the height we found earlier:
+<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>
+~ilee0820
 == See Also ==
-{{AMC10 box|year=2006|ab=B|num-b=7|num-a=9}}
+{{AMC10 box|year=2006|ab=B|num-b=14|num-a=16}}
 [[Category:Introductory Geometry Problems]]
 [[Category:Area Problems]]
+{{MAA Notice}}

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