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The radius of the larger circle is 2 inches longer than the radius of

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The radius of the larger circle is 2 inches longer than the radius of  [#permalink]

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New post 07 Sep 2018, 00:12
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The radius of the larger circle is 2 inches longer than the radius of the smaller circle. If the radius of the smaller circle is r then, in terms of r, what is the ratio of the unshaded area to the total area?


A. \(\frac{r^2}{r^2 + 4r + 4}\)

B. \(\frac{\pi r}{4\pi r + 4}\)

C. \(\frac{r^2}{4r + 4}\)

D. \(\frac{r}{r^2 + 4}\)

E. \(\frac{\pi r^2 + 4r + 4}{\pi r^2}\)


Attachment:
image006.jpg
image006.jpg [ 4.25 KiB | Viewed 440 times ]

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Re: The radius of the larger circle is 2 inches longer than the radius of  [#permalink]

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New post 07 Sep 2018, 08:03
Bunuel wrote:
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The radius of the larger circle is 2 inches longer than the radius of the smaller circle. If the radius of the smaller circle is r then, in terms of r, what is the ratio of the unshaded area to the total area?


A. \(\frac{r^2}{r^2 + 4r + 4}\)

B. \(\frac{\pi r}{4\pi r + 4}\)

C. \(\frac{r^2}{4r + 4}\)

D. \(\frac{r}{r^2 + 4}\)

E. \(\frac{\pi r^2 + 4r + 4}{\pi r^2}\)


Attachment:
image006.jpg


Area of un-shaded portion(White)=Area of smaller circle with radius 'r'=\(\pi*r^2\)
Total area=Area of bigger circle with radius 'r+2'= \(\pi*(r+2)^2\)

Required ratio=\(\frac{\pi*r^2}{\pi*(r+2)^2}\)=\(\frac{r^2}{r^2 + 4r + 4}\)

Ans. (A)
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The radius of the larger circle is 2 inches longer than the radius of  [#permalink]

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New post 09 Sep 2018, 05:32
Area of unshaded portion = \(πr^2\)
Total area = \(π(r+2)^2\) = \(π(r^2+4r+4)\)
Ratio = \(\frac{πr^2}{π(r^2+4r+4)}\) = \(\frac{r^2}{(r^2+4r+4)}\)
Answer A.
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The radius of the larger circle is 2 inches longer than the radius of &nbs [#permalink] 09 Sep 2018, 05:32
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