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2006 AMC 10A Problem 12

Problem 12 of 25IntermediateGeometry

Rolly wishes to secure his dog with an 88-foot rope to a square shed that is 1616 feet on each side. His preliminary drawings are shown. Which of these arrangements gives the dog the greater area to roam, and by how many square feet?

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Solution

In arrangement I the dog is tied at the middle of a side and sweeps a half-disk of radius 88: area 12π82=32π.\frac12 \pi \cdot 8^2 = 32\pi. The rope reaches exactly to the corners, so nothing wraps. In arrangement II the dog is tied 44 feet from a corner. It sweeps the same 32π32\pi half-disk, and after the rope reaches the corner, 44 feet remain to sweep a quarter-disk of radius 44: 14π42=4π.\frac14 \pi \cdot 4^2 = 4\pi. So II gives 36π,36\pi, exceeding I by 4π.4\pi. Thus, the correct answer is C.

More practice

Concepts: sector · circle area

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.