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

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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 $BFDE$ ?

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

Solutions

Solution 1

Using kiouonp of 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 $x = 8 \Longrightarrow \boxed{\mathrm{(C)}}$ .

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. $8 \Longrightarrow \boxed{\mathrm{(C)}}$ .

@@ Line 18: / Line 18: @@
 == Solutions ==
 === Solution 1 ===
-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>.
+Using kiouonp 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>

2006 AMC 10B (Problems • Answer Key • Resources)
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