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2006 AMC 10A Problems/Problem 12

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?

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

Solution

[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 $\boxed{\mathrm{(C) \ }}$.

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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