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

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Part of what Spot can reach is <math>\frac{240}{360}=\frac{2}{3}</math> of a circle with radius 2, which  
 
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>.
 
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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===Note===
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We can clearly see that the area must be more than <math>\frac{8\pi}{3}</math>, and the only such answer is <math>\boxed{\text{(E)}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:34, 9 September 2018

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 $\boxed{\text{(E)}}$.

Note

We can clearly see that the area must be more than $\frac{8\pi}{3}$, and the only such answer is $\boxed{\text{(E)}}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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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