Difference between revisions of "2013 AMC 10A Problems/Problem 20"
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− | For this square with side length 1, the distance from center to vertex is <math>r = \frac{1}{\sqrt{2}}</math>, hence the area is composed of a semicircle of radius <math>r</math>, plus <math>4</math> times a parallelogram with height <math>\frac{1}{2}</math> and base <math>\frac{\sqrt{2}}{2(1+\sqrt{2})}</math>. That is to say, the total area is <math>\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C )} \frac{\pi}{4} + 2 - \sqrt{2}}</math>. | + | For this square with side length 1, the distance from center to vertex is <math>r = \frac{1}{\sqrt{2}}</math>, hence the area is composed of a semicircle of radius <math>r</math>, plus <math>4</math> times a parallelogram with height <math>\frac{1}{2}</math> and base <math>\frac{\sqrt{2}}{2(1+\sqrt{2})}</math>. That is to say, the total area is <math>\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}} |
Revision as of 18:11, 8 February 2013
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
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 .
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |