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If the area of square S and the area of circle C are equal,

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If the area of square S and the area of circle C are equal, [#permalink] New post 29 Jan 2008, 15:09
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If the area of square S and the area of circle C are equal, then the ratio of the perimeter of S to the circumference of C is closest to?

9/8

or

4/3

I did some aproximation so I wound up w/ the wrong answer... id like to see your approach. Thx.
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Re: Another Geometry [#permalink] New post 29 Jan 2008, 15:19
GMATBLACKBELT wrote:
If the area of square S and the area of circle C are equal, then the ratio of the perimeter of S to the circumference of C is closest to?

9/8

or

4/3

I did some aproximation so I wound up w/ the wrong answer... id like to see your approach. Thx.


The ratio of perimters is: 2*PI*r : 8r

I think that would be closer to 4/3?
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Re: Another Geometry [#permalink] New post 29 Jan 2008, 15:21
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\(\frac98\)

\(a^2=\pi r^2\) --> \(\frac{a}{r}=\sqrt{\pi}\)

\(ratio=\frac{4a}{2\pi r}=\frac{2}{\pi}*\frac{a}{r}=\frac{2}{\pi}*\sqrt{\pi}=\frac{2}{sqrt{\pi}}\)

I compare square of numbers:

ratio^2 ~ 4/3.14

(9/8)^2=81/64 ~ 4/3.2 - ok

(4/3)^2=16/9 ~ 5.2/3.1
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Re: Another Geometry [#permalink] New post 29 Jan 2008, 15:28
walker wrote:
\(\frac98\)

\(a^2=\pi r^2\) --> \(\frac{a}{r}=\sqrt{\pi}\)

\(ratio=\frac{4a}{2\pi r}=\frac{2}{\pi}*\frac{a}{r}=\frac{2}{\pi}*\sqrt{\pi}=\frac{2}{sqrt{\pi}}\)

I compare square of numbers:

ratio^2 ~ 4/3.14

(9/8)^2=81/64 ~ 4/3.2 - ok

(4/3)^2=16/9 ~ 5.2/3.1


I tried this, i messed it up though.

Thx.

9/8 is the OA.
Re: Another Geometry   [#permalink] 29 Jan 2008, 15:28
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If the area of square S and the area of circle C are equal,

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