Difference between revisions of "2002 AMC 10A Problems/Problem 19"

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?

$\text{(A)}\ 2\pi/3 \qquad \text{(B)}\ 2\pi \qquad \text{(C)}\ 5\pi/2 \qquad \text{(D)}\ 8\pi/3 \qquad \text{(E)}\ 3\pi$

Solution

Part of what Spot can reach is $\frac{240}{360}=\frac{2}{3}$ of a circle with radius 2, which gives him $\frac{8\pi}{3}$ . He can also reach two $\frac{60}{360}$ parts of a unit circle, which combines to give $\frac{\pi}{3}$ . The total area is then $3\pi$ , which gives $\boxed{\text{(E)}}$ .

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

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 ==Solution==
-Spot can reach a <math>2\pi-\frac{2\pi}{3}=\frac{4\pi}{3}</math> angle of a unit circle; hence he can reach a region with area <math>\boxed{\frac{2\pi}{3}\Rightarrow\text{(A)}}</math>.
+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>.
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Difference between revisions of "2002 AMC 10A Problems/Problem 19"

Revision as of 23:37, 26 December 2008

Problem

Solution

See Also