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

Revision as of 14:01, 2 August 2006

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 $BFDE$ ?

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

Solution

Using properties of a rhombus:

$\angle DAB = \angle DCB = 60 ^\circ$ .

$\angle ADC = \angle ABC = 120 ^\circ$ .

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 $BFDE$ .

$\frac{x}{24} = \frac{1}{3}$

$x = 8 \Rightarrow C$

@@ Line 33: / Line 33: @@
 == See Also ==
 *[[2006 AMC 10B Problems]]
+*[[2006 AMC 10B Problems/Problem 14|Previous Problem]]
+*[[2006 AMC 10B Problems/Problem 16|Next Problem]]

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

Revision as of 14:01, 2 August 2006

Problem

Solution

See Also