Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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== Problem == | == Problem == | ||
− | In | + | In rectangle <math>ADEH</math>, points <math>B</math> and <math>C</math> trisect <math>\overline{AD}</math>, and points <math>G</math> and <math>F</math> trisect <math>\overline{HE}</math>. In addition, <math>AH=AC=2</math>, and <math>AD=3</math>. What is the area of quadrilateral <math>WXYZ</math> shown in the figure? |
<asy> | <asy> | ||
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</asy> | </asy> | ||
− | <math>\ | + | <math>\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{\sqrt{2}}{2}\qquad\textbf{(C) } \frac{\sqrt{3}}{2}\qquad\textbf{(D) } \sqrt{2} \qquad\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad</math> |
== Solution == | == Solution == |
Revision as of 12:07, 17 December 2021
Contents
Problem
In rectangle , points and trisect , and points and trisect . In addition, , and . What is the area of quadrilateral shown in the figure?
Solution
Solution 1
Draw . , so is a . Hence , and .
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Drawing lines as shown above and piecing together the triangles, we see that is made up of 12 squares congruent to . Hence .
Solution 3
We see that if we draw a line to it is half the width of the rectangle so that length would be , and the resulting triangle is a so using the Pythagorean Theorem we can get that each side is so the area of the middle square would be which is our answer.
Solution 4
Since and are trisection points and , we see that . Also, , so triangle is a right isosceles triangle, i.e. . By symmetry, triangles , , and are also right isosceles triangles. Therefore, , which means triangle is also a right isosceles triangle. Also, triangle is a right isosceles triangle.
Then , and . Hence, .
By symmetry, quadrilateral is a square, so its area is
~made by AoPS (somewhere) -put here by qkddud~
Video Solution by the Beauty of Math
https://www.youtube.com/watch?v=GX33rxlJz7s
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.