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# If the perimeter of square region S and the perimeter of the

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Intern
Joined: 16 Jun 2010
Posts: 17
If the perimeter of square region S and the perimeter of the  [#permalink]

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Updated on: 05 Aug 2012, 02:30
1
5
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Difficulty:

35% (medium)

Question Stats:

73% (01:19) correct 27% (01:37) wrong based on 274 sessions

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If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to

A. 3/2
B. 4/3
C. 3/4
D. 2/3
E. 1/2

Originally posted by chintzzz on 19 Jun 2010, 22:23.
Last edited by Bunuel on 05 Aug 2012, 02:30, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 48110
If the perimeter of square region S and the perimeter of the  [#permalink]

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20 Jun 2010, 08:03
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2
chintzzz wrote:

If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to
A.3/2
B.4/3
C.3/4
D.2/3
E.1/2

$$P_{square}=4x=2\pi{r}=P_{circle}$$ --> $$x=\frac{\pi{r}}{2}$$.

$$\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})$$.

Similar question: https://gmatclub.com/forum/if-the-perim ... 27004.html
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Joined: 07 Jun 2010
Posts: 48
Location: United States

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19 Jun 2010, 23:44
perimeter of square = perimeter of circle

4 x = 2 pie r

x= side of square

solving above and putting pie = 22/7

7x = 11r

so x/r = 77/7

so (area of square)/(area of circle) = (x^2)/(pie r^2)

we already know x/r = 11/7 so (x^2)/(r^2) = (11^2)/(7^2)

so (x^2)/(pie r^2) = (11^2)/[(7^2) * (22/11)]

solving we will get 11/14 which is near to 3/4
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10 Mar 2011, 10:51
1
If Perimeter of square = Perimeter of Circle, then:
4a = 2$$\pi$$r, where a = side of square and r is radius of the circle
a/r = $$\pi$$/2

Area of S/Area of C = $$a^2$$/ $$\pi$$ $$r^2$$
= $$(\pi/2)^2$$ * 1/$$\pi$$
= $$\pi$$/4 = 3.14/4
= $$\approx$$ 3/4 = C
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Re: If the perimeter of square region S and the perimeter of the  [#permalink]

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30 Jun 2015, 10:39
Bunuel wrote:
chintzzz wrote:

If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to
A.3/2
B.4/3
C.3/4
D.2/3
E.1/2

$$P_{square}=4x=2\pi{r}=P_{circle}$$ --> $$x=\frac{\pi{r}}{2}$$.

$$\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})$$.

Can you explain this portion: Pcircle --> x=πr2.
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Joined: 02 Sep 2009
Posts: 48110
Re: If the perimeter of square region S and the perimeter of the  [#permalink]

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30 Jun 2015, 10:45
xLUCAJx wrote:
Bunuel wrote:
chintzzz wrote:

If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to
A.3/2
B.4/3
C.3/4
D.2/3
E.1/2

$$P_{square}=4x=2\pi{r}=P_{circle}$$ --> $$x=\frac{\pi{r}}{2}$$.

$$\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})$$.

Can you explain this portion: Pcircle --> x=πr2.

Equating the perimeter of square (4x, where x is the length of the side) and the perimeter of the circle (circumference = $$2\pi{r}$$), we get $$4x=2\pi{r}$$, which gives $$x=\frac{\pi{r}}{2}$$.

Hope it's clear.
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WE: Operations (Military & Defense)
If the perimeter of square region S and the perimeter of the  [#permalink]

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28 Sep 2015, 01:24
1
Bunuel wrote:
chintzzz wrote:

If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to
A.3/2
B.4/3
C.3/4
D.2/3
E.1/2

$$P_{square}=4x=2\pi{r}=P_{circle}$$ --> $$x=\frac{\pi{r}}{2}$$.

$$\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})$$.

Apologies for what is probably a very simple question, but could someone please explain this step here?

$$\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}$$

Thank you
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Posts: 48110
Re: If the perimeter of square region S and the perimeter of the  [#permalink]

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28 Sep 2015, 03:07
2
DropBear wrote:
Bunuel wrote:
chintzzz wrote:

If the perimeter of square region S and the perimeter of the circular region C are equal, then the ratio of the area of S to area of C is closes to
A.3/2
B.4/3
C.3/4
D.2/3
E.1/2

$$P_{square}=4x=2\pi{r}=P_{circle}$$ --> $$x=\frac{\pi{r}}{2}$$.

$$\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})$$.

Apologies for what is probably a very simple question, but could someone please explain this step here?

$$\frac{x^2}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}$$

Thank you

$$x=\frac{\pi{r}}{2}$$.

$$\frac{A_{square}}{A_{circle}}=\frac{x^2}{\pi{r^2}}=\frac{(\frac{\pi{r}}{2})^2}{\pi{r^2}}=\frac{\frac{\pi^2 r^2}{4}}{\pi{r^2}}=\frac{\pi^2{r^2}}{4\pi{r^2}}=\frac{\pi}{4}\approx(\frac{3}{4})$$.

Hope it helps.
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Re: If the perimeter of square region S and the perimeter of the  [#permalink]

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29 Jul 2018, 06:08
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