2002 AMC 10B Problem 5

To find the shaded area you need the big circle's size, so find its radius first.

1. In the figure the two small circles sit side by side, touching each other, and each one touches the big circle. A straight line through their two centers runs all the way across the big circle, through its center.
2. Along that line, the big circle's diameter is made of the two small diameters placed end to end:
$$\text{big diameter} = 6 + 4 = 10, \qquad \text{big radius} = 5.$$
3. The shaded region is the big circle with the two small circles cut out:
$$\pi \cdot 5^2 - \pi \cdot 3^2 - \pi \cdot 2^2 = 25\pi - 9\pi - 4\pi = 12\pi.$$

The answer is $\boxed{\textbf{(E)}\ 12\pi}$.

When circles touch, the line joining their centers passes through the point where they touch. That lets you add up lengths along one straight line, which here gave the big diameter directly. Most shaded-area problems then come down to "whole shape minus the pieces cut out."

Because the small circles are tangent to each other and each is tangent to the big circle, the line through their centers passes through both points where they touch the big circle and through the big circle's center. That line cuts across the big circle along a diameter made of the two small diameters laid end to end:
$$\text{diameter} = 2\cdot 3 + 2\cdot 2 = 10, \qquad \text{radius} = 5.$$

The shaded region is the big disk with the two small disks removed:
$$\pi\cdot 5^2 - \pi\cdot 3^2 - \pi \cdot 2^2 = 25\pi - 9\pi - 4\pi = 12\pi.$$

The answer is $\boxed{\textbf{(E)}\ 12\pi}$.