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

Revision as of 19:45, 28 May 2023

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

Video Solution

https://www.youtube.com/watch?v=lkk3SAjsPaI ~David

2002 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 18: / Line 18: @@
 ===Note===
 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>.
+==Video Solution==
+https://www.youtube.com/watch?v=lkk3SAjsPaI  ~David
 ==See Also==

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