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Difference between revisions of "2006 AMC 10A Problems/Problem 12"

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Which of these arrangements give the dog the greater [[area]] to roam, and by how many square feet?
 
Which of these arrangements give the dog the greater [[area]] to roam, and by how many square feet?
  
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<math> \mathrm{(A) \ } I, by 8\pi\qquad \mathrm{(B) \ } I, by 6\pi\qquad \mathrm{(C) \ } II, by 4\pi\qquad \mathrm{(D) \ } II, by 8\pi\qquad \mathrm{(E) \ } II, by 10\pi </math>
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<math> \mathrm{(A) \ } I,\,\textrm{ by }\,8\pi\qquad \mathrm{(B) \ } I,\,\textrm{ by }\,6\pi\qquad \mathrm{(C) \ } II,\,\textrm{ by }\,4\pi\qquad \mathrm{(D) \ } II,\,\textrm{ by }\,8\pi\qquad \mathrm{(E) \ } II,\,\textrm{ by }\,10\pi </math>
 
== Solution ==
 
== Solution ==
 
Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement I allows the dog  
 
Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement I allows the dog  
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<math>.5*(\pi*8^2) = 32\pi</math> square feet of area. Arrangement II allows <math>32\pi</math> square feet plus a little more on the top part of the fence. So we already know that Arrangement II allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is  
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<math>\frac12\cdot(\pi\cdot8^2) = 32\pi</math> square feet of area. Arrangement II allows <math>32\pi</math> square feet plus a little more on the top part of the fence. So we already know that Arrangement II allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is  
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<math>.25*(\pi*4^2) = 4\pi</math>. Thus the answer is <math>\mathrm{(C) \ }</math>.
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<math>\frac14\cdot(\pi\cdot4^2) = 4\pi</math>. Thus the answer is <math>\mathrm{(C) \ }</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 22:04, 25 February 2007

Problem

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Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.

Which of these arrangements give the dog the greater area to roam, and by how many square feet?

$\mathrm{(A) \ } I,\,\textrm{ by }\,8\pi\qquad \mathrm{(B) \ } I,\,\textrm{ by }\,6\pi\qquad \mathrm{(C) \ } II,\,\textrm{ by }\,4\pi\qquad \mathrm{(D) \ } II,\,\textrm{ by }\,8\pi\qquad \mathrm{(E) \ } II,\,\textrm{ by }\,10\pi$

Solution

Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement I allows the dog $\frac12\cdot(\pi\cdot8^2) = 32\pi$ square feet of area. Arrangement II allows $32\pi$ square feet plus a little more on the top part of the fence. So we already know that Arrangement II allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is $\frac14\cdot(\pi\cdot4^2) = 4\pi$. Thus the answer is $\mathrm{(C) \ }$.

See Also

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