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>? | ||
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<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)); | ||
Line 14: | Line 13: | ||
label("$F$", F, dir(point--F)); | label("$F$", F, dir(point--F)); | ||
</asy> | </asy> | ||
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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> | |
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− | + | == 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 | + | 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 = | + | 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>. | ||
== Solution 2 == | == 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>. | |
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+ | 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>. | ||
== See Also == | == See Also == |
Latest revision as of 22:01, 20 August 2022
Contents
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
.
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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