2005 AMC 12A Problems/Problem 7
Contents
[hide]Problem
Square is inside the square so that each side of can be extended to pass through a vertex of . Square has side length and . What is the area of the inner square ?
Solution
Arguable the hardest part of this question is to visualize the diagram. Since each side of can be extended to pass through a vertex of , we realize that must be tilted in such a fashion. Let a side of be .
Notice the right triangle (in blue) with legs and hypotenuse . By the Pythagorean Theorem, we have . Thus,
Solution
You can also notice that the four triangles are congruent because the right angles of square cause them to be similar, and the hypotenuse of the triangles are the same because they are the sides of the square. Then you have
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 12 Problems and Solutions |
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