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

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Intern
Joined: 02 Nov 2017
Posts: 18
Re: A thin piece of wire 40 meters long is cut into two pieces. One piece  [#permalink]

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

Could you please share some extra questions which are similar to this?
Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: A thin piece of wire 40 meters long is cut into two pieces. One piece  [#permalink]

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

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

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Posts: 53020
Re: A thin piece of wire 40 meters long is cut into two pieces. One piece  [#permalink]

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

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

For more check Ultimate GMAT Quantitative Megathread
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Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: A thin piece of wire 40 meters long is cut into two pieces. One piece  [#permalink]

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21 Dec 2017, 23:09
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 - $$\pi{r^2}+(10-\frac{\pi{r}}{2})^2$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/a-thin-piece ... 06671.html
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Intern
Joined: 05 Apr 2018
Posts: 3
Re: A thin piece of wire 40 meters long is cut into two pieces. One piece  [#permalink]

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23 Jun 2018, 01:20
Hi friends,

Today I came across this question in a GMAT prep exam and I really can't figure out what I am doing wrong. I consistently arrive to answer C though I perfectly understand how to get to option E. Here is my reasoning:

Area of the circle: $$π*r^{2}$$

Area of the square:
Since there are 20 meters of wire to make the square, then each side of the square must have 5 meters long.
If $$2*π*r=20$$, then $$r=\frac{10}{π}$$
In order to represent the side of the square (5 meters long) in terms of $$r$$, I say that $$side of the square=r*\frac{π}{2}=\frac{10}{π}*\frac{π}{2}=5$$. So $$area of the square=(r*\frac{π}{2})^2=\frac{1}{4}∗π^2∗r^2$$

Total area: $$πr^{2}+\frac{1}{4}∗π^2∗r^2$$

Option C

It would help me a lot if somebody could tell me where is the flaw in my method.
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece  [#permalink]

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18 Oct 2018, 06:08
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Re: A thin piece of wire 40 meters long is cut into two pieces. One piece   [#permalink] 18 Oct 2018, 06:08

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