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

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==Solution==
 
==Solution==
 +
<center><asy>
 +
draw(polygon(6));
 +
draw(Arc((1/2,sqrt(3)/2),2,-60,180));
 +
draw(Arc((-1/2,sqrt(3)/2),1,180,240));
 +
draw(Arc((1,0),1,240,300));
 +
draw((-1/2,sqrt(3)/2)--(-3/2,sqrt(3)/2), dotted);
 +
draw((1,0)--(3/2,-sqrt(3)/2),dotted);
 +
</asy></center>
 
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>.

Revision as of 14:16, 14 September 2020

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

[asy] draw(polygon(6)); draw(Arc((1/2,sqrt(3)/2),2,-60,180)); draw(Arc((-1/2,sqrt(3)/2),1,180,240)); draw(Arc((1,0),1,240,300)); draw((-1/2,sqrt(3)/2)--(-3/2,sqrt(3)/2), dotted); draw((1,0)--(3/2,-sqrt(3)/2),dotted); [/asy]

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 (Problems • Answer Key • Resources)
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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