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, 08:20

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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]

21 Dec 2010, 08:38

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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.

Answer: E.
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Re: A THIN PIECE OF PAPER 40 MT LONG [#permalink]

23 Jan 2011, 16:04

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, 16:08

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, 16:12

ah ...thanks Bunuel
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wire [#permalink]

16 Jan 2012, 02:21

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, 02:52

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, 08:38

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, 08:53

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ENAFEX wrote:

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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Question [#permalink]

02 May 2012, 03:49

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?
1)\pi r^2
2)\pi r^2 + 10
3) \pi r^2 + \frac{1}{4} \pi^2 r^2
4) \pi r^2 + (40 - \pi r)^2
5)\pi r^2 + (10 - \frac{1}{2} \pi r) ^2

I answered correctly by guessing, I was lucky. So, I would appreciate if someone explains the problem.
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Re: Question [#permalink]

02 May 2012, 04:03

Stiv 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?
1)\pi r^2
2)\pi r^2 + 10
3) \pi r^2 + \frac{1}{4} \pi^2 r^2
4) \pi r^2 + (40 - \pi r)^2
5)\pi r^2 + (10 - \frac{1}{2} \pi r) ^2

I answered correctly by guessing, I was lucky. So, I would appreciate if someone explains the problem.

Merging similar topics. Please refer to the solutions above.
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Re: Question [#permalink] 02 May 2012, 04:03

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A thin piece of wire 40 meters long is cut into two pieces.

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