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

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== Problem ==
 
== Problem ==
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[[Image:2006 AMC 10A-12.GIF]]
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<!-- [[Image:2006 AMC 10A-12.GIF]] -->
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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.
  
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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.
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<asy>
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size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10));
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D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle);
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D((16,-8)--(24,-8));
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label('Dog', (24, -8), SE);
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MP('I', (8,-8), (0,0));
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MP('8', (16,-4), W);
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MP('8', (16,-12),W);
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MP('8', (20,-8), N);
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label('Rope', (20,-8),S);
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D((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle);
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D((16,-24)--(24,-24));
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MP("II", (8,-28), (0,0));
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MP('4', (16,-22), W);
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MP('8', (20,-24), N);
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label("Dog",(24,-24),SE);
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label("Rope", (20,-24), S);
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</asy>
  
 
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,\,\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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<math> \textbf{(A) } I,\,\textrm{ by }\,8\pi\qquad \textbf{(B) } I,\,\textrm{ by }\,6\pi\qquad \textbf{(C) } II,\,\textrm{ by }\,4\pi\qquad \textbf{(D) } II,\,\textrm{ by }\,8\pi\qquad \textbf{(E) } II,\,\textrm{ by }\,10\pi </math>
  
 
== Solution ==
 
== Solution ==
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[[Image:2006 AMC 10A-12b.GIF]]
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<!-- [[Image:2006 AMC 10A-12b.GIF]] -->
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+
<asy>
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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  
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size(150); pathpen = linewidth(0.7); defaultpen(linewidth(0.7)+fontsize(10));
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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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filldraw(arc((16,-8),8,-90,90)--cycle, rgb(0.9,0.9,0.6));
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<math>\frac14\cdot(\pi\cdot4^2) = 4\pi</math>. Thus the answer is <math>\mathrm{(C) \ }</math>.
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filldraw(arc((16,-26),8,-90,90)--cycle, rgb(0.9,0.9,0.6));
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filldraw(arc((16,-22),4,90,180)--(16,-22)--cycle, rgb(0.9,0.9,0.6));
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D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle);
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D((16,-8)--(24,-8));
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label('Dog', (24, -8), SE);
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MP('I', (8,-8), (0,0));
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MP('8', (16,-4), W);
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MP('8', (16,-12),W);
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MP('8', (20,-8), N);
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label('Rope', (20,-8),S);
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pair sD = (0,-2);
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D(shift(sD)*((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle));
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D(shift(sD)*((16,-24)--(24,-24)));
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MP("II", (8,-28)+sD, (0,0));
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MP('4', (16,-22)+sD, W);
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MP('8', (20,-24)+sD, N);
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label("Dog",(24,-24)+sD,SE);
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label("Rope", (20,-24)+sD, S);
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</asy>
 +
Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement <math>I</math> allows the dog  
 +
<math>\frac12\cdot(\pi\cdot8^2) = 32\pi</math> square feet of area. Arrangement <math>II</math> allows <math>32\pi</math> square feet plus a little more on the top part of the fence. So we already know that Arrangement <math>II</math> 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>\boxed{\textbf{(C) } II,\,\textrm{ by }\,4\pi}</math>.
  
 
== See also ==
 
== See also ==
Line 19: Line 59:
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]
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{{MAA Notice}}

Latest revision as of 09:14, 17 December 2021

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.

[asy] size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,-8)--(24,-8)); label('Dog', (24, -8), SE); MP('I', (8,-8), (0,0)); MP('8', (16,-4), W); MP('8', (16,-12),W); MP('8', (20,-8), N); label('Rope', (20,-8),S); D((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle); D((16,-24)--(24,-24)); MP("II", (8,-28), (0,0)); MP('4', (16,-22), W); MP('8', (20,-24), N); label("Dog",(24,-24),SE); label("Rope", (20,-24), S); [/asy]

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

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

Solution

[asy] size(150); pathpen = linewidth(0.7); defaultpen(linewidth(0.7)+fontsize(10)); filldraw(arc((16,-8),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-26),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-22),4,90,180)--(16,-22)--cycle, rgb(0.9,0.9,0.6)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,-8)--(24,-8)); label('Dog', (24, -8), SE); MP('I', (8,-8), (0,0)); MP('8', (16,-4), W); MP('8', (16,-12),W); MP('8', (20,-8), N); label('Rope', (20,-8),S); pair sD = (0,-2); D(shift(sD)*((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle)); D(shift(sD)*((16,-24)--(24,-24))); MP("II", (8,-28)+sD, (0,0)); MP('4', (16,-22)+sD, W); MP('8', (20,-24)+sD, N); label("Dog",(24,-24)+sD,SE); label("Rope", (20,-24)+sD, S); [/asy] 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 $\boxed{\textbf{(C) } II,\,\textrm{ by }\,4\pi}$.

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

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