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Geometry problem

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by rwrangler » Wed Dec 12, 2007 3:45 pm
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 ...

a)7/9
b)8/9
c)9/8
d)4/3
e)2/1

Not sure how to do this.
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by Suyog » Wed Dec 12, 2007 4:33 pm
Area of square and circle is equal

Lets say area is 100

Square - Area = 100 side = 10 and perimeter = 40
Circle - Area = pi*r^2 = 100, radius = 5.6 approx perimeter = 2pir = 31 approx

ratio of 40/32 = 4/3 approx.

Ans D.
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by gmatguy16 » Wed Dec 12, 2007 5:36 pm
given s^2 = n * r ^2 say n = pi
s/r = sqrt (n)
now we need to know ratio= r = 4 s /(2 n r) = 2 /sqrt( n)
r ^ 2 = 4 /n = 14/11 = 1.27 ...
hence r = approx 9/8
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Re: Geometry problem

by jayhawk2001 » Wed Dec 12, 2007 6:29 pm
rwrangler 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 ...

a)7/9
b)8/9
c)9/8
d)4/3
e)2/1

Not sure how to do this.
s^2 = pi * r^2
s/r = sqrt(pi)

So, 4s / (2*pi*r) = 2 / sqrt(pi)

sqrt(pi) = sqrt(3.14) ~ 1.7

So, we know 2/1.7 > 1 and <2> 1.33. Hence, C
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by rwrangler » Wed Dec 12, 2007 6:34 pm
answer to OA is C.

Thanks guys.
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by Suyog » Wed Dec 12, 2007 7:55 pm
Sorry guys...
pi*r^2 = 100 gives r = 5.64 and 2pir = 2 * 3.14 * 5.62 = 35.14

40/35.14 = 1.12 = 9/8

Thanks!
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