2004 AMC 10B Problem 9

1. Find the two areas. The square is $10 \times 10 = 100$. The circle is $\pi \cdot 10^2 = 100\pi$.
2. If we just add them, the part that lies in both shapes is counted twice. So find that shared part.
3. The circle's center is a corner of the square. The two sides of the square leaving that corner are at a right angle, and each is $10$ long, the same as the radius. So the part of the circle inside the square is exactly a $90^\circ$ slice out of $360^\circ$, a quarter circle. That quarter circle fits entirely inside the square, because the square's far sides are $10$ away from the center, the same as the radius.
4. The shared part is a quarter circle, with area $\frac{1}{4} \cdot 100\pi = 25\pi$.
5. Count it only once:
$$100 + 100\pi - 25\pi = 100 + 75\pi.$$

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

Area of two shapes together = first area + second area - the part they share. The whole job is finding the shared part. A circle centered at a corner of a square, with radius equal to the side, always shares exactly a quarter of itself with the square.

Build the union as "the square, plus the part of the circle sticking out." The square takes up a $90^\circ$ slice of the circle, so the part sticking out is the other $270^\circ$, which is $\frac{3}{4}$ of the circle: $\frac{3}{4} \cdot 100\pi = 75\pi$. The total is $100 + 75\pi$.

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