2004 AMC 10B Problem 9

The square has area $10^2 = 100$ and the circle has area $\pi \cdot 10^2 = 100\pi$.

Place the circle's center at the corner where two sides meet. Those two sides are perpendicular and each has length $10$, the radius, so they cut the disk into four equal quarter-disks, and the square contains exactly one of them (the far sides of the square are at distance $10$ and do not intrude). The overlap of the two regions is therefore a quarter disk of area $25\pi$.

The union counts the overlap once:
$$100 + 100\pi - 25\pi = 100 + 75\pi.$$

The answer is $\boxed{\textbf{(B)}\ 100+75\pi}$.

For a union of two regions, area(union) $=$ area(A) $+$ area(B) $-$ area(overlap). The whole problem is identifying the overlap, and the special placement (center at a vertex, radius equal to the side) makes it a clean $90^\circ$ sector. Look for such "quarter" and "half" configurations whenever a circle is centered on a vertex or a midpoint.

Think of the union as the square plus the part of the disk outside it. The square uses up a $90^\circ$ wedge of the disk, so the outside part is the remaining $270^\circ$, i.e. $\frac{3}{4} \cdot 100\pi = 75\pi$. Total: $100 + 75\pi$.

• Adding the two areas outright (100 + 100 pi, choice (D)) without removing the quarter disk that lies inside the square.
• Subtracting a half disk instead of a quarter disk, as if the center were on a side rather than at a corner.