2005 AMC 10B Problem 8

Every tile looks the same, so work out one tile and then multiply.

1. White part of one tile. There are four quarter circles, each with radius $\frac{1}{2}$. Four quarters make one whole circle, so the white area is the area of one circle of radius $\frac{1}{2}$:
$$\pi \cdot \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}.$$
2. Shaded part of one tile. The tile is $1$ foot by $1$ foot, so its area is $1$. The shaded part is what is not white: $1 - \frac{\pi}{4}$.
3. Number of tiles. The floor is $8$ by $10$, so it holds $8 \cdot 10 = 80$ tiles.
4. Total shaded area.
$$80 \cdot \left(1 - \frac{\pi}{4}\right) = 80 - 20\pi.$$

The answer is $\boxed{\textbf{(A)}\ 80-20\pi}$.

With a repeating pattern, study a single copy and multiply by the number of copies. Inside the tile, the four matching quarter circles combine into one full circle, so you only need the circle area formula. When the white shape is the easy one, find it and subtract from the total.

Look at a single tile. The four white pieces are quarter circles of the same radius $\frac{1}{2}$, so together they have the area of one whole circle:
$$\pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}.$$

The tile has area $1$, so the shaded part of one tile is $1 - \frac{\pi}{4}$.

The floor holds $8 \cdot 10 = 80$ tiles, giving a total shaded area of
$$80\left(1 - \frac{\pi}{4}\right) = 80 - 20\pi.$$

The answer is $\boxed{\textbf{(A)}\ 80-20\pi}$.