Difference between revisions of "2006 AMC 10B Problems/Problem 4"

 
(Added problem and solution)
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== Problem ==
 
== Problem ==
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Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
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[[Image:2006amc10b04.gif]]
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<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 </math>
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== Solution ==
 
== Solution ==
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The area painted red is equal to the area of the smaller cirle, and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.
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So:
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<math> A_{red}=\pi(\frac{1}{2})^2=\frac{\pi}{4} </math>
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<math> A_{blue}=\pi(\frac{3}{2})^2-\pi(\frac{1}{2})^2=2\pi </math>
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So the desired ratio is:
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<math> \frac{2\pi}{\frac{\pi}{4}}=8\Rightarrow D </math>
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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 19:06, 13 July 2006

Problem

Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?

2006amc10b04.gif

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9$

Solution

The area painted red is equal to the area of the smaller cirle, and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.

So:

$A_{red}=\pi(\frac{1}{2})^2=\frac{\pi}{4}$

$A_{blue}=\pi(\frac{3}{2})^2-\pi(\frac{1}{2})^2=2\pi$

So the desired ratio is:

$\frac{2\pi}{\frac{\pi}{4}}=8\Rightarrow D$

See Also

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