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GMAT Prep question

by rossmj » Fri Apr 17, 2009 7:34 am
A thin peice of wire 40m long is cut into 2 pieces. One piece is used to form a circle with radius r, and the other piece is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?
A. pie r^2
B. pie r^2 +10
C. pie r^2 + (1/4) pie^2 r^2
D. pie r^2 + (40-2pie r)^2
E. pie r^2 + (10-(1/2)pie r)^2
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by truplayer256 » Fri Apr 17, 2009 7:44 am
In this problem, we know that the total circumference of the circle and the perimeter of the square must add up to 40 since the length of the wire is 40. So let's assume that s represents one side of a square.

4s=Perimeter of a square.
2*pi*r= Perimeter or circumference of a circle.

2*pi*r+4s=40<-----(Equation A)
Total area of the square and the circle can be represented by the following:

pi*r^2+s^2<------ (Equation B)

Since we want the total area to be in terms of r, we must write s in terms of r. From equation A, we can see that s=(40-2*pi*r)/4. Now, all we have to do is plug this value of s in terms of r into Equation B and we get:

pi*r^2+((40-2*pi*r)/(4))^(2) or pi*r^2+(10-(pi*r)/2))^(2) E.
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by rossmj » Fri Apr 17, 2009 7:50 am
Thank you Truplayer256, I forgot to divide by 4 and kept coming up with D. Thought I was losing my mind.
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