2005 AMC 12A Problems/Problem 7
Contents
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 2
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
Solution by wxl18
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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