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ratio problem?

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by anwarma » Wed Jun 11, 2008 4:46 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

ANS is C

What is parameter of Circle, I thought of it as circumference of a circle but when I made 4s = 2pi*r and then worked it out for ratios on areas, i get the 1/2 which is wrong. Anyone any suggestions where is my approach wrong?
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by AleksandrM » Wed Jun 11, 2008 5:54 pm
You're on the right track:

pi roughly equals to 3

4s = 2pir or 4s = 6r

s = 6r/4 = 3r/2

Area of region S is s^2

Area of region C is pir^2

Ratio of the two areas is s^2/pir^2

OR

(3r/2)^2/3r^2

9r^2/4/3r^2

9r^2/12r^2 [r^2 cancels out]

9/12 = 3/4
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by VerbalAttack » Wed Jun 11, 2008 6:39 pm
Alternatively, you use this method;

4s = 2pir ==> 2s = pir ==> s = (pi/2)r

Squre both sides of equation;
s^2 = (pi/2)^2 r^2 ==> s^2 = (pi/4) pi r^2
As we know pi ~ 3;

(s^2) / (pi r^2) ~ 3/4

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