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.

Draw a line through the center of the figure and the center of one surrounding circle. Along that line, from the big circle's center outward, we cross the radius of the central unit circle ($1$), then the full diameter of a surrounding unit circle ($2$), and arrive at the big circle. So the big radius is $1 + 2 = 3$ and its area is $\pi \cdot 3^2 = 9\pi$.

Seven unit circles (one in the middle, six around it) are removed, each of area $\pi$. The shaded area is
$$9\pi - 7\pi = 2\pi.$$

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