2006 AMC 10A Problem 12

Question

Rolly wishes to secure his dog with an

8

-foot rope to a square shed that is

16

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?

\mathrm{I},

by

8\pi

\mathrm{I},

by

6\pi

\mathrm{II},

by

4\pi

\mathrm{II},

by

8\pi

\mathrm{II},

by

10\pi

In each case the dog sweeps a half-disk of radius

8

on the open side

In arrangement II the rope also wraps around the near corner

In arrangement I the dog is tied at the middle of a side and sweeps a half-disk of radius

8

: area

\frac12 \pi \cdot 8^2 = 32\pi.

The rope reaches exactly to the corners, so nothing wraps.

In arrangement II the dog is tied

4

feet from a corner. It sweeps the same

32\pi

half-disk, and after the rope reaches the corner,

4

feet remain to sweep a quarter-disk of radius

4

:

\frac14 \pi \cdot 4^2 = 4\pi.

So II gives

36\pi,

exceeding I by

4\pi.

Thus, the correct answer is C.

Problem 12 in other years