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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]  21 Dec 2010, 07:20
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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. $$\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2$$
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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]  21 Dec 2010, 07:38
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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 - $$\pi{r^2}+(10-\frac{\pi{r}}{2})^2$$.

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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]  23 Jan 2011, 15:04
Bunuel...how did u get side of the square as (40-2pi*r)/4 ?
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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]  23 Jan 2011, 15:08
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ajit257 wrote:
Bunuel...how did u get side of the square as (40-2pi*r)/4 ?

$$40-2\pi{r}$$ meters of wire are left for the square means that $$40-2\pi{r}$$ is the perimeter of the square so the side of it will be $$\frac{40-2\pi{r}}{4}=10-\frac{\pi{r}}{2}$$.
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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]  23 Jan 2011, 15:12
ah ...thanks Bunuel
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wire [#permalink]  16 Jan 2012, 01:21
A thin piece of 40 m wire is cut in two. One piece is a circle with radius of r. Other is a square. No wire is left over. Which represents total area of circle and square in terms of r?

From a 40m rope, two equal parts of 20 represent, respectively, the circumference of Circle C and the perimeter of Square D.

1. Express the perimeter of D in terms of the circumference of C.

Perimeter + Circumference  = 40

Perimeter = 40 - C

2. Substitute accepted identities for perimeter and circumference.

4s = 40 - 2(pi)(r)

3.  Express the side of the square in terms of r.

s = 10 - (pi)(r)/2

4. Write area of circle and square in terms of r.

Area(C) + Area(D)
= (pi)(r)^2 + s^2
= (pi)(r)^2 +  (10 - (pi)(r)/2)^2

5.  Verify

2(pi)(r) = 20

r = 10/pi
= 3.18 (approx. to hundredth)

s   = (10 - (pi)(r)/2)
= 10-  [3.14(3.18)]/2
= 5

The actual value of the side of the square derived from stating area of square in terms of r is consistent with its known perimeter.

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Re: 40 meters wire problem [#permalink]  16 Jan 2012, 01:52
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PhilosophusRex 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.

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) $$\pi R^2 + (10 - 1/2\pi R)^2$$

4) or 5) is a given, I just dont see how you make the calculation necessary.

Very minimal calculations are required if you assume a convenient value for R e.g. R = 7/2 since $$\pi = 22/7$$
Circumference in this case $$= 2\pi*R = 2*(22/7)*(7/2) = 22$$
So length of leftover wire = 18
Side of square = 18/4
Area of square = $$(9/2)^2$$

Now put R = 7/2 in the second half of the options you want to check.
i.e. say you want to check option (E)
$$(10 - 1/2\pi R)^2 = [10 - (1/2)*(22/7)*(7/2)]^2 = (9/2)^2$$

Option (E) is correct since the area of the square is (9/2)^2 when R = 7/2 (as shown above)
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wire [#permalink]  05 Mar 2012, 07:38
Tried to solve this using the numbers given but I am not able to proceed. Please help. Just curious to know. Thanks.

Say the length of each piece = 20.

So Circumference =20
perimeter of square = 4*side
So length of side = 20/4 = 5.

How do I proceed further?
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Re: wire [#permalink]  05 Mar 2012, 07:53
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ENAFEX wrote:
Tried to solve this using the numbers given but I am not able to proceed. Please help. Just curious to know. Thanks.

Say the length of each piece = 20.

So Circumference =20
perimeter of square = 4*side
So length of side = 20/4 = 5.

How do I proceed further?

The length of each piece = 20;

Circumference of the circle is $$2\pi{r}=20$$ --> $$r=\frac{10}{\pi}$$ --> $$area=\pi{r^2}=\frac{100}{\pi}$$;
Perimeter of the square $$4*side=20$$ --> $$side=5$$ --> $$area=5^2=25$$;

The total area is $$\frac{100}{\pi}+25$$. Now, you should substitute the value of $$r=\frac{10}{\pi}$$ in the answer choices and see which one gives $$\frac{100}{\pi}+25$$. Answer choice E works.

Hope it helps.
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  28 Aug 2013, 08:42
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Bumping for review and further discussion.
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  01 Dec 2013, 20:52
Let the perimeter of circle be x---->2pi*r=x----(1)
Let the perimeter of square be 40-x---->4S=40-x
Substitute value of x from(1)
----->S=(40/4)-(2pi*r/4)
----->S=10-pi*r/2
----->$$S^2=(10-pi*r/2)^2$$
Therefore total area is $$pi*r^2+(10-pi*r/2)^2$$
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  11 Dec 2013, 19:31
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?

Originally, I said that the perimeter of both objects was S (for the square) and 40-S for the circle. Why couldn't I solve the problem using S and 40-S? What is the thought process for this question or others that require similar reasoning?

Anyways, we are told that the wire is cut to form the shape (i.e. the perimeter of the square and the circle) The perimeter of the square is L*4 and the perimeter of the circle is pi*d. The perimeter for the square is 40-pi*d where pi*d represents the perimeter. To find the area, we can get the length of one of the sides of the square so we must find a leg length. (40-pi*d)/4 = 10-[(pi*d)/4]. The area of the circle is simply pi*r^2

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
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  12 Dec 2013, 01:46
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WholeLottaLove 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?

Originally, I said that the perimeter of both objects was S (for the square) and 40-S for the circle. Why couldn't I solve the problem using S and 40-S? What is the thought process for this question or others that require similar reasoning?

Anyways, we are told that the wire is cut to form the shape (i.e. the perimeter of the square and the circle) The perimeter of the square is L*4 and the perimeter of the circle is pi*d. The perimeter for the square is 40-pi*d where pi*d represents the perimeter. To find the area, we can get the length of one of the sides of the square so we must find a leg length. (40-pi*d)/4 = 10-[(pi*d)/4]. The area of the circle is simply pi*r^2

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

Not sure I understand your question but your solution is basically the same as this: a-thin-piece-of-paper-40-mt-long-106671.html#p839272
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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]  24 Dec 2013, 23:35
Bunuel 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?

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 - $$\pi{r^2}+(10-\frac{\pi{r}}{2})^2$$.

Bunuel- the perimeter of the circle and the square are the same, so: 2(pi)r=4a (a being one side of the square).
a= (pi)r/2
area of square in terms of r: a^2= (pi)^2*r^2/4
wont this mean that option c is also correct? what am I missing?
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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]  25 Dec 2013, 01:52
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Bunuel 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?

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 - $$\pi{r^2}+(10-\frac{\pi{r}}{2})^2$$.

Bunuel- the perimeter of the circle and the square are the same, so: 2(pi)r=4a (a being one side of the square).
a= (pi)r/2
area of square in terms of r: a^2= (pi)^2*r^2/4
wont this mean that option c is also correct? what am I missing?

Hello.

The question only says: A thin piece of wire 40 meters long is cut into two pieces, NOT into same length pieces. Thus, your assumption 2(pi)r=4a is wrong.

Hope it's clear.
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  01 Jun 2014, 22:44
A little confusion here:
As the total length is 2πr + 4x = 40
=> πr + 2x = 20

I assumed length 10 as the circumference of the circle and 10 as the perimeter of the square.

=> πr=10 and x=5 and computed the total area, which is
πr²+x²
=> 100/π + 25 ----------(1)

I then substituted this values in the answer choice to find out the matches with (1), choices C and E seem to match. Can someone please tell me what am I doing wrong.

Thanks,
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  02 Jun 2014, 00:17
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mamboe wrote:
A little confusion here:
As the total length is 2πr + 4x = 40
=> πr + 2x = 20

I assumed length 10 as the circumference of the circle and 10 as the perimeter of the square.

=> πr=10 and x=5 and computed the total area, which is
πr²+x²
=> 100/π + 25 ----------(1)

I then substituted this values in the answer choice to find out the matches with (1), choices C and E seem to match. Can someone please tell me what am I doing wrong.

Thanks,
a

Why did you assume that? We are NOT told that the piece is cut into the same length pieces. We are just told that the piece is cut into two pieces.
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Re: wire [#permalink]  11 Jun 2014, 06:58
Bunuel wrote:
ENAFEX wrote:
Tried to solve this using the numbers given but I am not able to proceed. Please help. Just curious to know. Thanks.

Say the length of each piece = 20.

So Circumference =20
perimeter of square = 4*side
So length of side = 20/4 = 5.

How do I proceed further?

The length of each piece = 20;

Circumference of the circle is $$2\pi{r}=20$$ --> $$r=\frac{10}{\pi}$$ --> $$area=\pi{r^2}=\frac{100}{\pi}$$;
Perimeter of the square $$4*side=20$$ --> $$side=5$$ --> $$area=5^2=25$$;

The total area is $$\frac{100}{\pi}+25$$. Now, you should substitute the value of $$r=\frac{10}{\pi}$$ in the answer choices and see which one gives $$\frac{100}{\pi}+25$$. Answer choice E works.

Hope it helps.

Hi Bunuel,

Then in this case even answer choice C works. Right?
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  27 Oct 2014, 13:02
maggie27 wrote:
Bunuel wrote:
ENAFEX wrote:
Tried to solve this using the numbers given but I am not able to proceed. Please help. Just curious to know. Thanks.

Say the length of each piece = 20.

So Circumference =20
perimeter of square = 4*side
So length of side = 20/4 = 5.

How do I proceed further?

The length of each piece = 20;

Circumference of the circle is $$2\pi{r}=20$$ --> $$r=\frac{10}{\pi}$$ --> $$area=\pi{r^2}=\frac{100}{\pi}$$;
Perimeter of the square $$4*side=20$$ --> $$side=5$$ --> $$area=5^2=25$$;

The total area is $$\frac{100}{\pi}+25$$. Now, you should substitute the value of $$r=\frac{10}{\pi}$$ in the answer choices and see which one gives $$\frac{100}{\pi}+25$$. Answer choice E works.

Hope it helps.

Hi Bunuel,

Then in this case even answer choice C works. Right?

Can someone answer this ? Choice C & E both work with this method....how to reject one?
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  10 Nov 2014, 11:21
Bunuel : Was unnecessarily complicating the question. Spent 5 min on it and still ended up guessing it wrong.. Might be the pressure of the practice test made me miss the whole point of the question. Definitely the best example to say Practice and FOCUS is the key on GMAT..
Re: A thin piece of wire 40 meters long is cut into two pieces.   [#permalink] 10 Nov 2014, 11:21

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