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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]

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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. $$\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2$$
[Reveal] Spoiler: OA
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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]

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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]

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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]

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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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16 Jan 2012, 01:21
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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]

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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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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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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]

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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 PAPER 40 MT LONG [#permalink]

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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]

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25 Dec 2013, 01:52
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gmarchanda wrote:
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]

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01 Jun 2014, 22:44
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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
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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02 Jun 2014, 00:17
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: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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10 Nov 2014, 20:41
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In this problem, its not at all mentioned that "wire 40 meters long is cut into two equal pieces"

We have to take a variable for distribution of 40 meters in square & circle.

Refer diagram below:

Attachment:

sqa.png [ 3.51 KiB | Viewed 14654 times ]

Let the perimeter of the square = d

then circumference of the circle = 40-d

$$2\pi r = 40-d$$

$$d = 40 - 2\pi r$$ ..................... (1)

Area of Circle $$= \pi r^2$$ .............. (2)

Each side of square $$= \frac{d}{4} = \frac{40 - 2\pi r}{4} = 10 - \frac{\pi r}{2}$$ .................... From (1)

Area of square $$= (10 - \frac{\pi r}{2})^2$$

Total Area $$= \pi r^2 + (10 - \frac{\pi r}{2})^2$$

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

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15 Nov 2015, 21:00
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Quote:
A thin piece of 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 term of r?

A) πr²
B) πr² + 10
C) πr² + 1/4(π² * r²)
D) πr² + (40 - 2π * r)²
E) πr² + (10 - 1/2π * r)²

Another approach is to plug in a value for r and see what the output should be.

Let's say r = 0. That is, the radius of the circle = 0
This means, we use the ENTIRE 40-meter length of wire to create the square.
So, the 4 sides of this square will have length 10, which means the area = 100

So, when r = 0, the total area = 100

We'll now plug r = 0 into the 5 answer choices and see which one yields an output of 100

A) π(0²) = 0 NOPE
B) π(0²) + 10 = 10 NOPE
C) π(0²) + 1/4(π² * 0²) = 0 NOPE
D) π(0²) + (40 - 2π0)² = 1600 NOPE
E) π(0²) + (10 - 1/2π(0))² = 100 PERFECT!

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

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16 Nov 2015, 00:27
anilnandyala 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. $$\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2$$

Given: 40 meters long wire is cut into two pieces. One circle with radius r and the rest is used to form a square
Required: Total area of the circle and the square

Since we are given the length of the wire, we should find the perimeters.

Perimeter of the circle = $$2*\pi*r$$
Hence left over wire after removing the circle = $$40 - (2*\pi*r)$$

We need to form a square from this length.
In other words, this is the perimeter of the square and each side of a square is equal.

Hence side of square = $$\frac{1}{4}(40 - (2*\pi*r))$$ = $$10 - \frac{1}{2}\pi*r$$

Area of the square = $$(10 - \frac{1}{2}\pi*r)^2$$

Total required area = $$\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2$$
Option E
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A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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25 Nov 2015, 20:08
It doesn't say HOW LONG the wire is cut sooooooo

R=0 Perimeter of the Square = 40

Area of the square = 10^2

Still took me 2 minutes just to understand the question though.

Oh wow, gmatprep had the same idea as me.
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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26 Nov 2015, 01:02
DJ1986 wrote:
It doesn't say HOW LONG the wire is cut sooooooo

R=0 Perimeter of the Square = 40

Area of the square = 10^2

Still took me 2 minutes just to understand the question though.

Oh wow, gmatprep had the same idea as me.

Hi DJ1986,
Quiet an interesting way to tackle the question
But if we go by what you say, there is no circle at all.

Whereas the question says "One piece is used to form a circle with radius r, and the other is used to form a square"
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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27 Nov 2015, 12:07
TeamGMATIFY wrote:
DJ1986 wrote:
It doesn't say HOW LONG the wire is cut sooooooo

R=0 Perimeter of the Square = 40

Area of the square = 10^2

Still took me 2 minutes just to understand the question though.

Oh wow, gmatprep had the same idea as me.

Hi DJ1986,
Quiet an interesting way to tackle the question
But if we go by what you say, there is no circle at all.

Whereas the question says "One piece is used to form a circle with radius r, and the other is used to form a square"

I thought about that and decided that maybe the circle is microscopically so small it may as well be rounded to zero or maybe the person just imagined cutting a wire and making a circle but never actually did. Either way 0+whatever=whatever
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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28 Mar 2017, 13:00
let's take a brief look, the whole wire was used, so we need to take into account perimetr of the both figures when calculating the area

Pr^2+((40-2Pr)/4)^2
We need to divide the latter by 4 because perimeter of a squae has 4 sides

Re: A thin piece of wire 40 meters long is cut into two pieces.   [#permalink] 28 Mar 2017, 13:00

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