A thin piece of wire 40 meters long is cut into two pieces. One piece : GMAT Problem Solving (PS)
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# A thin piece of wire 40 meters long is cut into two pieces. One piece

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A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink]

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14 Jun 2009, 23:35
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Difficulty:

55% (hard)

Question Stats:

59% (02:38) correct 41% (01:33) wrong based on 71 sessions

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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 square regions in terms of R.

1) $$\pi R^2$$
2) $$\pi R^2 + 10$$
3) $$\pi R^2 + 1/4\pi^2R^2$$
4) $$\pi R^2 + (40 - 2\pi R)^2$$
5) $$\pi R^2 + (10 - 1/2\pi R)^2$$

[Reveal] Spoiler:
4) or 5) is a given, I just dont see how you make the calculation necessary.
[Reveal] Spoiler: OA
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink]

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15 Jun 2009, 00:15
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KUDOS

Given:
40 meters long wire cut into two pieces.

Now Length of piece one = circumference of a circle with radius R = $$2\pi R$$

Length of piece two = perimeter of a square created using the remaining wire = $$40-2\pi R$$

$$\Rightarrow \text{length of each side of square} = \frac{40-2\pi R}{4}$$

$$= \frac{4(10-\frac{1}{2}\pi R)}{4}$$

$$= 10-\frac{1}{2}\pi R$$

Now
$$\text{Area of circle} = \pi R^2$$
$$\text {Area of square} = (10-\frac{1}{2}\pi R)^2$$

$$\text {Sum of the areas}=\pi R^2 + (10-\frac{1}{2}\pi R)^2$$
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink]

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15 Jun 2009, 02:01
Option 'E' is correct.
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink]

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07 Jan 2012, 10:49
GREAT EXPLANATION NOOKWAY. THANK YOU
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink]

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10 May 2016, 11:10
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece   [#permalink] 10 May 2016, 11:10
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