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A thin piece of wire 40 meters long is cut into two pieces.

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A thin piece of wire 40 meters long is cut into two pieces. [#permalink] New post 21 Jul 2008, 08:16
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?

Πr^2

(Πr^2) +10

Πr^2 + (1/4)(Π^2)(r^2)

Πr^2 + (40-2Πr)^2

Πr^2 + (10-.5Πr^2)^2

(Just so you know, Π = pi)
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Re: Cuttin' Wire [#permalink] New post 21 Jul 2008, 08:25
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ryguy904 wrote:
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?


We make a circle with radius r: the area is Pi*r^2.

How much wire is left? The wire we used for the circle is equal to the circumference of the circle: 2r*Pi. We have 40 - 2r*Pi meters of wire with which to make the square. Thus, 40 - 2r*Pi will be the perimeter of the square, and each side will be one quarter that length: 10 - 0.5r*Pi. The area of the square will be (10 - 0.5r*Pi)^2, and if you add that to Pi*r^2, you'll get the answer.
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Re: Cuttin' Wire [#permalink] New post 21 Jul 2008, 08:29
ryguy904 wrote:
Πr^2 + (10-.5Πr^2)^2


I think you must have mistyped this answer choice- it would be the correct answer if r were not squared. That is, it should read:

Πr^2 + (10-.5Πr)^2
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Re: Cuttin' Wire [#permalink] New post 21 Jul 2008, 08:34
IanStewart wrote:
ryguy904 wrote:
Πr^2 + (10-.5Πr^2)^2


I think you must have mistyped this answer choice- it would be the correct answer if r were not squared. That is, it should read:

Πr^2 + (10-.5Πr)^2


You are correct. I got a little trigger happy typing in the ^2's. LOL

+1 Thank you so much. I couldn't understand the square's area, but you explained that I was simply deducting the perimeter.

Cheers
Re: Cuttin' Wire   [#permalink] 21 Jul 2008, 08:34
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