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2006 AMC 10A Problems/Problem 17

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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?

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

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See Also

Solution

Solution 1

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

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

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 • 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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_10A_Problems/Problem_17"

Category: Introductory Geometry Problems

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