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Difficult Math Question #53 - Geometry

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Re: Difficult Math Question #53 - Geometry

by kulksnikhil » Tue Nov 14, 2006 10:30 pm
800guy 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) 2/3
(B) 3/4
(C) 4/3
(D) 3/2
(E) 2

Answer is B 3/4

Perimeter Of S = Perimeter of C
4x = 2Pr (P=Pie =3.142)
r = 2x/P

Area S/Area C = x^2/Pr^2
= x^2/P(2x/P)^2
= x^2/P(4X^2/P^2)
= 1/(4/P)
= P/4

P being 3.142 it is apprx equal to 3/4
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Re: Difficult Math Question #53 - Geometry

by Bharadwaj » Wed Nov 15, 2006 3:25 pm
800guy 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) 2/3
(B) 3/4
(C) 4/3
(D) 3/2
(E) 2
B
Regards,
Bharadwaj
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OA

by 800guy » Wed Nov 15, 2006 5:32 pm
OA:

and the answer would be B...here is the explanation...

Let the side of the square be s..then the perimeter of the square is 4s
Let the radius of the circle be r..then the perimeter of the circle is 2*pi*r

it is given that both these quantities are equal..therefore

4s=2*pi*r

which is then s/r=pi/2

Now the ratio of area of square to area of circle would be

s^2/pi*r^2

(1/pi)*(s/r)^2

= (1/pi)*(pi/2)^2 from the above equality relation

pi=22/7 or 3.14

the value of the above expression is approximate =0.78 which is near to answer B
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