2002 AMC 10A Problem 19

At the tether vertex the hexagon occupies an interior angle of $120^\circ$, so Spot can sweep the remaining $360^\circ - 120^\circ = 240^\circ$ with the full rope of length $2$:
$$\frac{240}{360} \cdot \pi \cdot 2^2 = \frac{8\pi}{3}.$$

When the rope is pulled along one side of the hexagon, it reaches the neighboring vertex with $2 - 1 = 1$ yard to spare. Bending around that vertex it sweeps the exterior angle of the hexagon, $60^\circ$, with radius $1$. This happens on both sides:
$$2 \cdot \frac{60}{360} \cdot \pi \cdot 1^2 = \frac{\pi}{3}.$$

Total area: $\frac{8\pi}{3} + \frac{\pi}{3} = 3\pi$. The answer is $\boxed{\textbf{(E)}\ 3\pi}$.

Tethered-animal problems decompose into sectors: one large sector where the rope is free, then smaller sectors each time the rope bends around a corner with the remaining length. The sweep angle at each corner equals the polygon's exterior angle. Draw the hexagon and trace the rope physically; every bend is a new sector.

• Forgetting the two small sectors where the rope bends around the adjacent corners, giving 8pi/3.
• Using $300^\circ$ (treating the hexagon's angle as $60^\circ$) for the main sector.