2013 AMC 10A Problem 20

Question

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?

1 - \dfrac{\sqrt2}{2} + \dfrac{\pi}{4}

\dfrac{1}{2} + \dfrac{\pi}{4}

2 - \sqrt2 + \dfrac{\pi}{4}

\dfrac{\sqrt2}{2} + \dfrac{\pi}{4}

1 + \dfrac{\sqrt2}{4} + \dfrac{\pi}{8}

Work with one quadrant of the swept region

Decompose that quadrant into a sector and two right-triangle pieces

Consider one quarter of the swept region and multiply its area by

4

.

The sector has angle

45^\circ

and radius

\frac{\sqrt2}{2}

, so its area is

\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}

.

The two right-triangle pieces in that quarter have areas

\frac12\cdot\frac12\cdot\frac12=\frac18

and

\frac12\left(\frac{\sqrt2-1}{2}\right)^2

.

Multiplying the quarter-area sum by

4

gives

2-\sqrt2+\frac{\pi}{4}

.

Thus, C is the correct answer.

Problem 20 in other years