Difference between revisions of "2002 AMC 10A Problems/Problem 19"
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==Solution== | ==Solution== | ||
− | Spot can reach | + | Part of what Spot can reach is <math>\frac{240}{360}=\frac{2}{3}</math> of a circle with radius 2, which |
+ | gives him <math>\frac{8\pi}{3}</math>. He can also reach two <math>\frac{60}{360}</math> parts of a unit circle, which combines to give <math>\frac{\pi}{3}</math>. The total area is then <math>3\pi</math>, which gives <math>\boxed{\text{(E)}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 23:37, 26 December 2008
Problem
Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?
Solution
Part of what Spot can reach is of a circle with radius 2, which gives him . He can also reach two parts of a unit circle, which combines to give . The total area is then , which gives .
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |