Difference between revisions of "2006 AMC 10A Problems/Problem 12"
(→Solution) |
m (added category and link to previous and next problem) |
||
| Line 14: | Line 14: | ||
== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
| + | |||
| + | *[[2006 AMC 10A Problems/Problem 11|Previous Problem]] | ||
| + | |||
| + | *[[2006 AMC 10A Problems/Problem 13|Next Problem]] | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 14:46, 4 August 2006
Problem
Missing diagram
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 give the dog the greater area to roam, and by how many square feet?
Solution
Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement I allows the dog .5*(pi*8^2) = 32pi square feet of area. Arrangement II allows 32pi square feet plus a little more on the top part of the fence. So we already know that Arrangement II allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is .25*(pi*4^2) = 4pi. Thus the answer is (c)
