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

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A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   29 Mar 2009, 06:28
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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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A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   21 Dec 2010, 08:20
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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. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)

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 wire 40 meters long is cut into two pieces. One piece [#permalink] New post   10 Nov 2014, 21:41
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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
sqa.png [ 3.51 KiB | Viewed 185865 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 \(= \pi r^2 + (10 - \frac{\pi r}{2})^2\)

Answer = E
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   15 Nov 2015, 22:00
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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!

Answer: E

Cheers,
Brent
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   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 wire 40 meters long is cut into two pieces. One piece [#permalink] New post   23 Jan 2011, 16:08
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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 wire 40 meters long is cut into two pieces. One piece [#permalink] New post   16 Jan 2012, 02:21
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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: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   16 Jan 2012, 02:52
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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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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   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: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   05 Mar 2012, 08:53
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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. One piece [#permalink] New post   25 Dec 2013, 00:35
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 - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Answer: E.



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 wire 40 meters long is cut into two pieces. One piece [#permalink] New post   25 Dec 2013, 02:52
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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 - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Answer: E.



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. One piece [#permalink] New post   16 Nov 2015, 01:27
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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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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   06 May 2017, 17:14
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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. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)



We are given that 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.

Since the circumference of a circle with radius r is 2πr, the amount of wire used to form the circle is 2πr. Thus, we have 40 – 2πr left over to form the square. In other words, the perimeter of the square is 40 – 2πr. However, since we need to calculate the total area of the circular and the square regions, we need to determine the side of the square in terms of r. Since the perimeter of the square is 40 – 2πr, the side of the square is:

side = (40 – 2πr)/4

side = 10 – (1/2)πr

Now we can determine the areas of the circle and the square.

Area of circle = πr2

Area of square = side^2 = (10 – (1/2)πr)^2

Thus, the combined area of the circle and square is πr2 + (10 – (1/2)πr)2.

Answer: E
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   07 Sep 2017, 05:18
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Hi GMATters,

Check out my video explanation for this question here!



Enjoy!

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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   21 Dec 2017, 23:35
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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 - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Answer: E.



Could you please share some extra questions which are similar to this?
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   22 Dec 2017, 00:07
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shivamtibrewala 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 - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Answer: E.



Could you please share some extra questions which are similar to this?


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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   22 Dec 2017, 00:07
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shivamtibrewala 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 - \(\pi{r^2}+(10-\frac{\pi{r}}{2})^2\).

Answer: E.



Could you please share some extra questions which are similar to this?


23. Geometry




24. Coordinate Geometry




25. Triangles




26. Polygons




27. Circles




28. Rectangular Solids and Cylinders




29. Graphs and Illustrations




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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   21 Apr 2021, 05:19
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Solution:

Area of a circle = πr^2 and

Wire used = Circumference of the wire

=2πr meters

Wire left for a square =40−2πr meters

Side of this square will be 40−2πr/4

=10−(πr/2)

=>Area of the square = (10−πr/2)^2.

The total area = πr^2+(10−πr/2)^2.

(option e)

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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink] New post   28 Jun 2022, 20:28
Hi,

I was wondering what would happen if they asked the same question but instead of square they used rectangle?
so perimeter of rectange + Circumference of circle = 40
2 (l + w ) = 40 - 2 pi r
l + w = (40 - 2 pi r)/2
Area of circle + area of rec

Pi r^2 + l x w - what would i write under l x w ?

thanks!
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece [#permalink]
28 Jun 2022, 20:28
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