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

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Math Expert
Joined: 02 Sep 2009
Posts: 53020

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07 Sep 2018, 00:12
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Difficulty:

15% (low)

Question Stats:

88% (01:52) correct 13% (01:55) wrong based on 32 sessions

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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 [ 4.25 KiB | Viewed 471 times ]

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07 Sep 2018, 08:03
Bunuel wrote:

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