a thin piece of wire 40 meter 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 represents the total area, in square meters, of the circular and the square regions in terms of r?
answer is pie r^2 + (10-1/2pie r^2)
please show me how to even begin solving this problem, or what you do in solving a problem like this.
gmat prep question
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- jayhawk2001
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The piece that forms the circle is of length 2*pi*r. So remaining isyvonne12 wrote:a thin piece of wire 40 meter 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 represents the total area, in square meters, of the circular and the square regions in terms of r?
answer is pie r^2 + (10-1/2pie r^2)
please show me how to even begin solving this problem, or what you do in solving a problem like this.
40 - 2*pi*r
If perimeter of square is 40 - 2*pi*r, each side is 1/4th of that which is
10 - pi*r/2
So, total area = pi*r^2 + (10 - pi*r/2)^2
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Say the length of first piece of wire is "t". This t = 2*pie*r (circumference of the circle)
Now the remainder of the wire after the circle is formed = 40 - t or in other words 40 - 2*pie*r (substituting for t)
Perimeter of square = 40 - 2*pie*r
Side of square = (40-2*pie*r)/4
Area of square = [(40-2*pie*r)/4]^2
= (10-r*pie/2)^2
Hence area of circle plus square = (10-r*pie/2)^2 + (pie*r)^2
Now the remainder of the wire after the circle is formed = 40 - t or in other words 40 - 2*pie*r (substituting for t)
Perimeter of square = 40 - 2*pie*r
Side of square = (40-2*pie*r)/4
Area of square = [(40-2*pie*r)/4]^2
= (10-r*pie/2)^2
Hence area of circle plus square = (10-r*pie/2)^2 + (pie*r)^2