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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 28 Oct 2006, 03:25
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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 the square regions in terms of r?

A) pir^2
B) pir^2+10
C) pir^2+(1/4)pi^2r^2
D) pir^2+(40-2pir)^2
E) pir^2+(10-1/2pir)^2

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-thin-piece-of-paper-40-mt-long-106671.html
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 [#permalink] New post 28 Oct 2006, 03:54
E


((40 - 2pir)/4)^2 + pir^2

(10 - 1/2pir) ^2 + pir^2
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink] New post 28 Nov 2012, 01:41
Is there any other way to solve this question?
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink] New post 28 Nov 2012, 03:20
nelz007 wrote:
Is there any other way to solve this question?


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. \pi*r^2
B. \pi*r^2 + 10
C. \pi*r^2 + \frac{1}{4}*\pi^2*r^2
D. \pi*r^2 + (40 - 2\pi*r)^2
E. \pi*r^2 + (10 - \frac{1}{2}\pi*r)^2

The area of a circle will be - \pi{r^2} and 2\pi{r} meters of wire will be used;
There will be 40-2\pi{r} meters of wire left for a square. Side of this square will be \frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}, hence the area of the square will be (10-\frac{\pi{r}}{2})^2.

The total area will be - \pi{r^2}+(10-\frac{\pi{r}}{2})^2.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-thin-piece-of-paper-40-mt-long-106671.html
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Re: A thin piece of wire 40 meters long is cut into two pieces.   [#permalink] 28 Nov 2012, 03:20
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