2002 AMC 10A Problem 5

The shaded region is the big circle with the small circles cut out. So we need the big circle's radius, then we subtract.

1. Find the big radius. Draw a line from the center of the figure through the center of one of the outer small circles, out to the big circle. Walking along it from the center:
- through the middle small circle to its edge: $1$,
- straight across one outer small circle: its diameter, $2$,
- and now we are at the big circle, because that outer circle touches it.

So the big radius is $1 + 2 = 3$.

1. Big circle's area: $\pi \times 3^2 = 9\pi$.

1. Small circles: there are $7$ of them (one in the middle, six around it), each of area $\pi \times 1^2 = \pi$. Together: $7\pi$.

4. Shaded area:
$$9\pi - 7\pi = 2\pi.$$

The answer is $\boxed{\textbf{(C)}\ 2\pi}$.

When two circles touch, the line through their centers passes through the touching point, so lengths along that line simply add up. That turns the picture into a one-line measurement. The shaded region is then "big shape minus the pieces removed", the standard move for shaded-area problems.