Difference between revisions of "2006 AMC 10A Problems/Problem 12"
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== Problem == | == Problem == | ||
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Rolly wishes to secure his dog with an 8-foot rope to a [[square (geometry) | square]] shed that is 16 feet on each side. His preliminary drawings are shown. | Rolly wishes to secure his dog with an 8-foot rope to a [[square (geometry) | square]] shed that is 16 feet on each side. His preliminary drawings are shown. | ||
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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> | <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> | ||
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== Solution == | == Solution == | ||
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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 | ||
<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 | <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 | ||
<math>\frac14\cdot(\pi\cdot4^2) = 4\pi</math>. Thus the answer is <math>\mathrm{(C) \ }</math>. | <math>\frac14\cdot(\pi\cdot4^2) = 4\pi</math>. Thus the answer is <math>\mathrm{(C) \ }</math>. | ||
| − | == See | + | == See also == |
| − | + | {{AMC10 box|year=2006|ab=A|num-b=11|num-a=13}} | |
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 07:57, 5 April 2007
Problem
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?
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
square feet of area. Arrangement II allows 
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
. Thus the answer is 
.
See also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
