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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]  14 Oct 2006, 23:10
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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. $$\pi*r^2$$
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. $$\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-thin-piece-of-paper-40-mt-long-106671.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Dec 2013, 01:55, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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E

((40 - 2pi r)/4*(40 - 2pi r)/4) + pir^2

(1600 - 80pir -80pir + 4pi^2r^2)/16 + pir^2

100 - 10pir + (4pi^2^r^2 / 16) + pir^2

100 - 10pir + pi^2r^2/4 + pir^2

pir^2 + (10 - 1/2pir)^2
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Re: Gmat Powerprep Word Problem [#permalink]  15 Oct 2006, 02:04
\$uckafr33 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 square regions in terms of r?

a) pi r^2

b) pi r^2 +10

c) pi r^2 + 1/4 pi^2 r^2

d) pi r^2 + (40 - 2 pi r)^2

e) pi r^2 + (10 - 1/2 pi r)^2

Let the first peice be x meters in length, other side = 40-x
circumference of circle = x; 2pi r = x;
area of circle + area of square
pi r^2 + ((40-x)/4)^2;
pi r^2 + (10-2pi r/4)^2
pi r^2 + (10-pi r/2)^2;

E
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OA is E. Thanks guys!
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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?

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

TWO PIECES = X,40-X

CIRCOM OF CIRCLE = 2Pi r = x thus circom of square = 40-2Pir and thus area = (10-1/2pir)^2

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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  11 Dec 2013, 19:01
Hello from the GMAT Club BumpBot!

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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  12 Dec 2013, 01:55
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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. $$\pi*r^2$$
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 - $$\pi{r^2}+(10-\frac{\pi{r}}{2})^2$$.

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] 12 Dec 2013, 01:55
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