2006 AMC 10A Problems/Problem 17
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~
Solution 5 (Proof Bash)
By symmetry, quadrilateral is a square.
First step, proving that .
We can tell quadrilateral is a parallelogram because and .
By knowing that, we can say that .
Finally, we can now prove by AA, with a ratio of 2:1.
Since and . Then is a 45-45-90 triangle.
This will make making and a 45-45-90 triangle.
This will make, . Since is the length of the square, our answer will be
~ghfhgvghj10
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 | |
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