2002 AMC 10B Problem 5

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

Tangent circles have collinear centers, so a chain of tangent circles measures out lengths along a straight line. Here the chain spans the big circle exactly, converting the picture into a single diameter computation. Shaded-region problems are almost always "big area minus the pieces removed."

• Using 10 as the big radius (getting 100 pi) or taking 3 + 2 = 5 as the diameter, instead of recognizing 5 as the radius.
• Subtracting only one of the two small circles, which gives $16\pi$ or $21\pi$ (not among the choices, a hint that something is off).