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>? | ||
− | + | <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> | ||
− | <math> \ | + | <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> |
− | == Solution == | + | == Solution 1 == |
− | Using | + | 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>. |
− | <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>. |
− | <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> x = 8 \ | + | <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 == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=14|num-a=16}} | |
− | |||
− | |||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | [[Category:Area Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:18, 12 October 2024
Problem
Rhombus is similar to rhombus . The area of rhombus is and . What is the area of rhombus ?
Solution 1
Using the property that opposite angles are equal in a rhombus, and . It is easy to see that rhombus is made up of equilateral triangles and . Let the lengths of the sides of rhombus be .
The longer diagonal of rhombus is . Since is a side of an equilateral triangle with a side length of , . The longer diagonal of rhombus is . Since is twice the length of an altitude of of an equilateral triangle with a side length of , .
The ratio of the longer diagonal of rhombus to rhombus is . Therefore, the ratio of the area of rhombus to rhombus is .
Let be the area of rhombus . Then , so .
Solution 2
Triangle DAB is equilateral so triangles , , , , and are all congruent with angles , and from which it follows that rhombus has one third the area of rhombus i.e. . Note: A quick way to visualize this method is to draw the line and notice the two equilateral triangles and .
Solution 3 (A bit more math involved)
We can extend line , meeting line at . Similarly, we can extend line to meet line at . We can see with some simple math that triangle is a -- triangle, so we can call line as , line as , and line as (because of the -- triangle side proportions).
We can also see that line is a base of rhombus , and line is a height. Since triangle is also a -- triangle, line is also . Since the question told us that the area of rhombus is , we can make the following equation:
Solving for x:
Since the question is to find the area of rhombus , to find the answer, we can just multiply base with the rhombus's height. We'll start by finding the height: instantly we can see that is the height. Since all the sides of a rhombus are equal, and we found earlier that the side length is , if is , that means is the same length as - that is to say,
Now to find the base. We can see that to find the base, we can simply just subtract the length of line from the length of line . Since is , and is , that makes
Now to find : We can see with simple math that triangle is also a -- triangle, which means that . Previously, we found out that is , so:
Now we can find the base:
Multiplying the newly found base by the height we found earlier:
~ilee0820
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AMC 10 Problems and Solutions |
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