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

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CEO
Joined: 29 Mar 2007
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If the area of square S and the area of circle C are equal, [#permalink]

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29 Jan 2008, 16:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Senior Manager
Joined: 26 Jan 2008
Posts: 263

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29 Jan 2008, 16:19
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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Kudos [?]: 117 [0], given: 16

CEO
Joined: 17 Nov 2007
Posts: 3585

Kudos [?]: 4483 [1], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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29 Jan 2008, 16:21
1
KUDOS
Expert's post
$$\frac98$$

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

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

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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CEO
Joined: 29 Mar 2007
Posts: 2554

Kudos [?]: 500 [0], given: 0

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29 Jan 2008, 16:28
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.

Kudos [?]: 500 [0], given: 0

Re: Another Geometry   [#permalink] 29 Jan 2008, 16:28
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