Geometry

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Geometry

by teejaycrown » Fri Nov 02, 2012 9:49 pm
A thin of wire 40m long is cut into 2 pieces. One piece is used to form a circle with radius r and the other is used to form a square. No wire is left over. which of the following represent the total area in square metres of the circular and of the square regions in terms of r.

a. pi r square
b. pi r square + 10
c. pi r square + 1/4 pi square r square
d. pi r square +( 40 - 2 pi r) square
e. pi r square+ (10 - 1/2 pi r) square

source : GMAT prep
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by Anurag@Gurome » Fri Nov 02, 2012 9:52 pm
teejaycrown wrote:A thin of wire 40m long is cut into 2 pieces. One piece is used to form a circle with radius r and the other is used to form a square. No wire is left over. which of the following represent the total area in square metres of the circular and of the square regions in terms of r.

a. pi r square
b. pi r square + 10
c. pi r square + 1/4 pi square r square
d. pi r square +( 40 - 2 pi r) square
e. pi r square+ (10 - 1/2 pi r) square

source : GMAT prep
The length of one piece = 2(pi)r and the other is a square. Let us assume that each side of the square is "a".
Perimeter of square = 4a
Now, 2(pi)r + 4a = 40
4a = 40 - 2(pi)r
a = [40 - 2(pi)r]/4 = 10 - (pi)r/2

Total area = Area of Circle + Area of Square
= (pi)r² + a²
= (pi)r² + (10 - (pi)r/2)²

The correct answer is E.
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by GMATGuruNY » Sat Nov 03, 2012 4:57 am
A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other 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. πr²
B. πr² +10
C. πr² + 1/4 π²r²
D. πr² + (40-2πr)²
E. πr² + (10 - (1/2)πr)²
Let r=0, so that the ENTIRE wire is used to form the square.
Perimeter of the square = 40.
Each side of the square = 40/4 = 10.
Area of the square = 10² = 100.
Area of the circle + area of the square = 0 + 100 = 100. This is our target.

Now we plug r=0 into the answers to see which yields our target of 100.

A quick scan reveals that only E works:
πr² + (10 - 1/2πr)² = 0 + 10² = 100.

The correct answer is E.
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