2013 AMC 10A Problems/Problem 20
Contents
Problem
A unit square is rotated about its center. What is the area of the region swept out by the interior of the square?
Solution 1
First, we need to see what this looks like. Below is a diagram.
For this square with side length 1, the distance from center to vertex is , hence the area is composed of a semicircle of radius , plus times a parallelogram with height and base . That is to say, the total area is .
Solution 2
Let be the center of the square and be the intersection of and . The desired area consists of the unit square, plus 4 regions congruent to the region bounded by arc , , and , plus 4 triangular regions congruent to right triangle . The area of the region bounded by arc , , and is (Area of Circle-Area of Square)/8. Since the circle has radius , the area of the region is , so 4 times the area of that region is . Now we find the area of . . Since is a 45-45-90 right triangle, the area of is $\dfrac{BC^2}{2}=\dfrac{({\dfrac{\sqrt {2}}{2}-\dfrac{1}{2})^2}{2}$ (Error compiling LaTeX. ! File ended while scanning use of \@genfrac.), so 4 times the area of is . Finally, the area of the whole region is \dfrac{\pi}{4}+2-\sqrt {2}$.
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AMC 10 Problems and Solutions |
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