2013 AMC 10A Problems/Problem 20

Problem

A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?

$\textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8}$

For this square with side length 1, the distance from center to vertex is $r = \frac{1}{\sqrt{2}}$ , hence the area is composed of a semicircle of radius $r$ , plus $4$ times a parallelogram with height $\frac{1}{2}$ and base $\frac{\sqrt{2}}{2(1+\sqrt{2})}$ . That is to say, the total area is $\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}$ .

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2013 AMC 10A Problems/Problem 20

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