2004 AMC 10A Problem 19

Cut the silo's curved surface along a vertical line and flatten it. The surface becomes a rectangle $30\pi$ feet wide and $80$ feet tall, and distances along the surface are unchanged, so the stripe's area is preserved.

In the flattened picture the stripe is a slanted band whose horizontal cross-section at every height is a segment of length $3$. Each revolution of the stripe is a parallelogram with horizontal base $3$ and vertical height $40$ (two revolutions span the $80$-foot height). The area of a parallelogram is base times perpendicular height, so each piece has area $3 \cdot 40 = 120$, and the two pieces together have area
$$2 \cdot 120 = 240.$$

Equivalently, sweeping a horizontal segment of length $3$ through a vertical distance of $80$ covers area $3 \cdot 80 = 240$ regardless of how the band slants.

The answer is $\boxed{\textbf{(C)}\ 240}$.

A cylinder is developable: unrolling it to a flat rectangle preserves lengths and areas, converting a 3-D surface problem into plane geometry. Once flat, a band with constant horizontal width has area (width) $\times$ (total vertical extent), a Cavalieri-style fact that makes the slant, the diameter, and the number of turns irrelevant. When a problem gives data that cancels, that is often the intended lesson.

• Bringing the circumference 30 pi or the slant of the stripe into the computation; the number of revolutions and the diameter are irrelevant.
• Using the stripe's slanted length as the parallelogram's height, or measuring the width perpendicular to the stripe instead of horizontally.