Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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=== Solution 3 === | === Solution 3 === | ||
− | We see that if we draw a line to <math>BZ</math> it is half the width of the rectangle so that length would be <math>1</math>, and the resulting triangle is a <math>45-45-90</math> so using the [[Pythagorean Theorem]] we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\boxed{\textbf{ | + | We see that if we draw a line to <math>BZ</math> it is half the width of the rectangle so that length would be <math>1</math>, and the resulting triangle is a <math>45-45-90</math> so using the [[Pythagorean Theorem]] we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\boxed{\textbf{(A) }\frac{1}{2}}</math> which is our answer. |
=== Solution 4 === | === Solution 4 === |
Revision as of 08:05, 19 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 1
By symmetry, is a square.
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 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 |
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