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Master | Next Rank: 500 Posts
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by grandh01 » Wed Aug 29, 2012 2:43 pm
If the perimeter of square region S and
the perimeter of circular region C are
equal, then the ratio of the area of S to
the area of C is closest to
(A)3/2
(B)4/3
(C)3/4
(D)2/3
(E)1/2

OA is C
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by SmartAssJun » Wed Aug 29, 2012 4:53 pm
grandh01 wrote:If the perimeter of square region S and
the perimeter of circular region C are
equal, then the ratio of the area of S to
the area of C is closest to
(A)3/2
(B)4/3
(C)3/4
(D)2/3
(E)1/2

OA is C
Let X be the perimeter of both the sqaure and the circular region
The radius of the circular region is then ''X/2pi'' so the area is X^2/4pi^2*pi =X^2/4pi
The side length of the sqaure is X/4 so the area is X^2/16
4pi/16= pi/4 is most aproximately = 3/4
Therefore it's C
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Master | Next Rank: 500 Posts
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by everything's eventual » Wed Aug 29, 2012 5:42 pm
Let side of square be x and radius of circle be r

Perimeters are equal.

4x = 2*pi*r

Therefore, x = (pi*r)/2 -------(1)

We need ratio of areas.

Therefore, we need x^2 / pi*r^2 ------- (2)

Substitute (1) in (2) and you will get pi/4 which is apprx equal to 3/4

Therefore (c) is the answer.
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