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square and rectangle

Expert replies
by LSB » Tue Aug 26, 2008 12:31 pm
Problem from GMATPrep
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Source: — Data Sufficiency |

by bomond » Wed Aug 27, 2008 11:39 am
Let squares side be a and rectangular's side 2x and 3x (because their ratio is 2:3). Their perimeters are equal 4a=10x => a^2=(25x^2)/4

The area of the rectangular 2x*3x=6x^2
The area of the circle a*a=a^2

Ratio =(6x^2)/a^2. Substitute instead a^2=(25x^2)/4. Then 24:25
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by LSB » Wed Aug 27, 2008 1:06 pm
bomond wrote:Let squares side be a and rectangular's side 2x and 3x (because their ratio is 2:3). Their perimeters are equal 4a=10x => a^2=(25x^2)/4

The area of the rectangular 2x*3x=6x^2
The area of the circle a*a=a^2

Ratio =(6x^2)/a^2. Substitute instead a^2=(25x^2)/4. Then 24:25

WOW - You don't have any idea how long I mulled over this during and after the test all while thinking S is a circle. you should have seen some of the formulae I came up with. It was a wild pi party :-)

Anyway - thank you. Your approach makes perfect sense.

By the way - it's actually quite funny. You have a typo talking about circles in your solution as well. There must be something about this question :-)
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by bomond » Thu Aug 28, 2008 4:43 am
LSB wrote: By the way - it's actually quite funny. You have a typo talking about circles in your solution as well. There must be something about this question :-)
I'm sorry. Yes you are right. By the area of the circle I mean square.
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by stern » Thu Aug 28, 2008 11:13 am
a/b = 2/3, 4s = 2(a+b)

if you solve these two eq you get

a= 4s/5, b= 6s/5

area of r/area of s = ab/s^2 = 24/26

so b
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