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A thin piece of wire 40 meters long 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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New post 30 Apr 2017, 18:01
Hi - I did exactly the 20-20 split, but doesn't C also work?

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

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New post 01 May 2017, 01:32
mdacosta wrote:
Hi - I did exactly the 20-20 split, but doesn't C also work?

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.


Yes, C works too. When plugging numbers, it might happen that two or more choices give "correct" answer. If this happens, just pick some other number and check again these "correct" options only.
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Re: A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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New post 06 May 2017, 16:14
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. [#permalink]

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New post 11 May 2017, 04:48
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?

Solution:
Approach 1:
A circle with radius r will have the circumference (the length of wire used) = 2πr
The ares of the circle is πr^2

So, to crate the square you have = 40 - 2πr
This will be the perimeter of the square. So, the length of each side of the square = (40 - 2πr)/4. = 10 - (1/2)πr
The area of the square is = (10 - (1/2)πr)^2

The answer is (E).


Approach 2:
Plug in values for the variable. Let r = 0 (It does mean that there is no circle at all.....but it is completely safe to assume this)

So, that means the complete 40 meter wire is used to create the square.
The perimeter of the square is 40.
=> The length of each side = 40/4 = 10
So, the area is 10^2 = 100

Now, check the answers.
A) π(0²) = 0
B) π(0²) + 10 = 10
C) π(0²) + 1/4(π² * 0²) = 0
D) π(0²) + (40 - 2π0)² = 1600
E) π(0²) + (10 - 1/2π(0))² = 100 PERFECT!

The answer is (E).
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A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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New post 11 May 2017, 05:50
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. \(\pi*r^2 + (10 - \frac{1}{2}\pi*r)^2\)


40 meter wire is divided into 2 parts ...
1st part is used to form a circle with radius r. So its perimeter = \(2\pi*r\)
Length of left part of wire = \(40-2*\pi*r\)
Side of square = \((40-2\pi*r)/4\) = \(10-\pi*r/2\)

Total area = \(\pi*r^2 + (10- \pi*r/2)^2\)
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A thin piece of wire 40 meters long is cut into two pieces. [#permalink]

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New post 07 Sep 2017, 04:18
Hi GMATters,

Check out my video explanation for this question here!



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

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New post 21 Dec 2017, 22:35
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.



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

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New post 21 Dec 2017, 23:07
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.



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

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New post 21 Dec 2017, 23:07
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.



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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: A thin piece of wire 40 meters long is cut into two pieces.   [#permalink] 21 Dec 2017, 23:07

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