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Difference between revisions of "2006 AMC 10A Problems/Problem 17"

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<math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad</math>
 
<math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad</math>
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[[Image:2006_AMC10A-17.png]]
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__TOC__
 
== Solution ==
 
== Solution ==
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== See Also ==
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=== Solution 1 ===
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*[[2006 AMC 10A Problems]]
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It is not difficult to see by [[symmetry]] that <math>WXYZ</math> is a [[square]].
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[[Image:2006_AMC10A-17a.png]]
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Draw <math>\overline{BZ}</math>. Clearly <math>BZ = \frac 12AH = 1</math>. Then <math>\displaystyle \triangle BWZ</math> is [[isosceles]], and is a <math>45-45-90 \triangle</math>. Hence <math>WZ = \frac{1}{\sqrt{2}}</math>, and <math>[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac 12\ \mathrm{(A)}</math>.
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There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
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=== Solution 2 ===
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[[Image:2006_AMC10A-17b.png]]
  
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*[[2006 AMC 10A Problems/Problem 16|Previous Problem]]
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Draw the lines as shown above, and count the squares. There are 12, so we have <math>\frac{2\cdot 3}{12} = \frac 12</math>.
  
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*[[2006 AMC 10A Problems/Problem 18|Next Problem]]
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== See Also ==
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{{AMC10 box|year=2006|num-b=16|num-a=18|ab=A}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 18:58, 15 September 2007

Problem

In rectangle $ADEH$, points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$. In addition, $AH=AC=2$. What is the area of quadrilateral $WXYZ$ shown in the figure?

$\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) \ } \frac{2\sqrt{2}}{2}\qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad$

2006 AMC10A-17.png

Solution

Solution 1

It is not difficult to see by symmetry that $WXYZ$ is a square.

2006 AMC10A-17a.png

Draw $\overline{BZ}$. Clearly $BZ = \frac 12AH = 1$. Then $\displaystyle \triangle BWZ$ is isosceles, and is a $45-45-90 \triangle$. Hence $WZ = \frac{1}{\sqrt{2}}$, and $[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac 12\ \mathrm{(A)}$.

There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.

Solution 2

2006 AMC10A-17b.png

Draw the lines as shown above, and count the squares. There are 12, so we have $\frac{2\cdot 3}{12} = \frac 12$.

See Also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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