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

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<math> \textrm{(A) } 6 \qquad \textrm{(B) } 4\sqrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } 6\sqrt {3}</math>
 
<math> \textrm{(A) } 6 \qquad \textrm{(B) } 4\sqrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } 6\sqrt {3}</math>
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[[2006 AMC 12B Problems/Problem 13|Solution]]
 
  
 
== Solution 1==
 
== Solution 1==
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Draw the line <math> \overline{DB}</math> to form an equilateral triangle, since <math>\angle BAD=60</math>, and line segments  
 
Draw the line <math> \overline{DB}</math> to form an equilateral triangle, since <math>\angle BAD=60</math>, and line segments  
 
<math> \overline{AB}</math> and <math> \overline{AD}</math> are equal in length. To find the area of the smaller rhombus, we only need to find the value of any arbitrary base, then square the result. To find the value of the base, use the line we just drew and connect it to point <math>E</math> at a right angle along line <math> \overline{DB}</math>. Call the connected point <math>P</math>, with triangles <math>DPE</math> and <math>EPB</math> being 30-60-90 triangles, meaning we can find the length of <math> \overline{ED}</math> or <math> \overline{EB}</math>. The base of <math>ABCD</math> must be <math>\sqrt{24}</math>, and half of that length must be <math>\sqrt{6}</math>(triangles <math>DPE</math> and <math>EPB</math> are congruent by <math>SSS</math>). Solving for the third length yields <math>\sqrt{8}</math>, which we square to get the answer <math>\boxed{\text{C}}</math>.
 
<math> \overline{AB}</math> and <math> \overline{AD}</math> are equal in length. To find the area of the smaller rhombus, we only need to find the value of any arbitrary base, then square the result. To find the value of the base, use the line we just drew and connect it to point <math>E</math> at a right angle along line <math> \overline{DB}</math>. Call the connected point <math>P</math>, with triangles <math>DPE</math> and <math>EPB</math> being 30-60-90 triangles, meaning we can find the length of <math> \overline{ED}</math> or <math> \overline{EB}</math>. The base of <math>ABCD</math> must be <math>\sqrt{24}</math>, and half of that length must be <math>\sqrt{6}</math>(triangles <math>DPE</math> and <math>EPB</math> are congruent by <math>SSS</math>). Solving for the third length yields <math>\sqrt{8}</math>, which we square to get the answer <math>\boxed{\text{C}}</math>.
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===Solution 3===
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Draw line DB, cutting rhombus BFDE into two triangles which fit nicely into 2/3 of equilateral triangle ABD. Thus the area of BFDE is (2/3)*12=8.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=12|num-a=14}}
 
{{AMC12 box|year=2006|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:49, 16 January 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(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 1

The ratio of any length on $ABCD$ to a corresponding length on $BFDE^2$ is equal to the ratio of their areas. Since $\angle BAD=60$, $\triangle ADB$ and $\triangle DBC$ are equilateral. $DB$, which is equal to $AB$, is the diagonal of rhombus $ABCD$. Therefore, $AC=\frac{DB(2)}{2\sqrt{3}}=\frac{DB}{\sqrt{3}}$. $DB$ and $AC$ are the longer diagonal of rhombuses $BEDF$ and $ABCD$, respectively. So the ratio of their areas is $(\frac{1}{\sqrt{3}})^2$ or $\frac{1}{3}$. One-third the area of $ABCD$ is equal to $8$. So the answer is $\boxed{\text{C}}$.

Solution 2

Draw the line $\overline{DB}$ to form an equilateral triangle, since $\angle BAD=60$, and line segments $\overline{AB}$ and $\overline{AD}$ are equal in length. To find the area of the smaller rhombus, we only need to find the value of any arbitrary base, then square the result. To find the value of the base, use the line we just drew and connect it to point $E$ at a right angle along line $\overline{DB}$. Call the connected point $P$, with triangles $DPE$ and $EPB$ being 30-60-90 triangles, meaning we can find the length of $\overline{ED}$ or $\overline{EB}$. The base of $ABCD$ must be $\sqrt{24}$, and half of that length must be $\sqrt{6}$(triangles $DPE$ and $EPB$ are congruent by $SSS$). Solving for the third length yields $\sqrt{8}$, which we square to get the answer $\boxed{\text{C}}$.

Solution 3

Draw line DB, cutting rhombus BFDE into two triangles which fit nicely into 2/3 of equilateral triangle ABD. Thus the area of BFDE is (2/3)*12=8.

See also

2006 AMC 12B (Problems • Answer Key • Resources)
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

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