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If the perimeter of square region S and the perimeter of circular regi

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If the perimeter of square region S and the perimeter of circular regi  [#permalink]

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New post Updated on: 06 Nov 2018, 02:49
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If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to:


A. \(\frac {3}{2}\)

B. \(\frac {4}{3}\)

C. \(\frac {3}{4}\)

D. \(\frac {2}{3}\)

E. \(\frac {1}{2}\)

Originally posted by Dumpling on 22 Feb 2006, 12:30.
Last edited by Bunuel on 06 Nov 2018, 02:49, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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If the perimeter of square region S and the perimeter of circular regi  [#permalink]

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New post 20 Jun 2010, 07:03
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chintzzz wrote:
I got an answer of B but the official answer is different. please explain

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


Given: \(P_{square}=4x=2\pi{r}=P_{circle}\) --> \(x=\frac{\pi{r}}{2}\).

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

Answer: C.

Similar question: https://gmatclub.com/forum/if-the-perim ... 27004.html
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Re: Ratio  [#permalink]

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New post 10 Mar 2011, 09:51
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 circular regi  [#permalink]

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New post 09 Mar 2014, 02:07
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Please refer diagram below:
Attachments

pi.jpg
pi.jpg [ 34.5 KiB | Viewed 7990 times ]

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Re: If the perimeter of square region S and the perimeter of circular regi  [#permalink]

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New post 09 Apr 2015, 22:12
Hi All,

This question is perfect for TESTing VALUES.

We're told that the PERIMETER of a square is equal to the PERIMETER (meaning the 'circumference') of a circle. We're asked to figure out the approximate ratio of the area of the square to the area of the circle.

Let's TEST VALUES. Since we're dealing with a circle, let's work "pi" into our math right from the beginning....

Perimeter = 4pi

For the Square:
Perimeter = 4pi
Side length = pi
Area = (pi)^2

For the Circle:
Perimeter = 4pi
Radius = 2
Area = 4pi

Area of Square/Area of Circle = (pi)^2/4pi = pi/4 = about 3.14/4 = about 3/4

Final Answer:

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Re: If the perimeter of square region S and the perimeter of the  [#permalink]

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New post 30 Jun 2015, 09:39
Bunuel wrote:
chintzzz wrote:
I got an answer of B but the official answer is different. please explain

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


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

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

Answer: C.



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

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New post 30 Jun 2015, 09:45
xLUCAJx wrote:
Bunuel wrote:
chintzzz wrote:
I got an answer of B but the official answer is different. please explain

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


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

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

Answer: C.



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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If the perimeter of square region S and the perimeter of the  [#permalink]

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New post 28 Sep 2015, 00:24
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Bunuel wrote:
chintzzz wrote:
I got an answer of B but the official answer is different. please explain

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


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

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

Answer: C.


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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Re: If the perimeter of square region S and the perimeter of the  [#permalink]

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New post 28 Sep 2015, 02:07
2
DropBear wrote:
Bunuel wrote:
chintzzz wrote:
I got an answer of B but the official answer is different. please explain

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


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

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

Answer: C.


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 circular regi  [#permalink]

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New post 07 May 2016, 06:04
I reached till (22 * 7)/ 4...Is there a quick way to estimate/ calculate the final answer?


Bunuel wrote:
Stiv wrote:
If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to:

A. \(\frac {3}{2}\)

B. \(\frac {4}{3}\)

C. \(\frac {3}{4}\)

D. \(\frac {2}{3}\)

E. \(\frac {1}{2}\)



Given: \(P_{square}=4x=2\pi{r}=P_{circle}\) --> \(x=\frac{\pi{r}}{2}\).

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

Answer: C.
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Re: If the perimeter of square region S and the perimeter of circular regi  [#permalink]

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New post 07 May 2016, 10:03
Hi ameyaprabhu,

The questions that you'll face on the Official GMAT are almost all based on patterns of some type (even if you don't immediately realize that a pattern is there). In the Quant section, TESTing VALUES can often be used to define a pattern, so you might try using that approach in these types of situations. My explanation (2 posts above your post) shows how to approach this prompt in that way.

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Re: If the perimeter of square region S and the perimeter of circular regi  [#permalink]

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New post 16 Dec 2018, 08:53
Dumpling wrote:
If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to:


A. \(\frac {3}{2}\)

B. \(\frac {4}{3}\)

C. \(\frac {3}{4}\)

D. \(\frac {2}{3}\)

E. \(\frac {1}{2}\)


Let Perimeter of Square \(= 4a\) and Area of Square \(= a^2\)

Let Perimeter of Circle = \(2\)\(\pi\)\(r\) and Area of Circle \(= \pi\)\(r^2\)

Perimeter of Square and Circle are equal.

\(4a =\) \(2\)\(\pi\)\(r\) \(=> a = \frac{r\pi}{2}\)

Ratio of Area of Square to Area of Circle \(= \frac{a^2}{r^2\pi}\)

Substituting value of \("a"\) in above expression we get;

\(\frac{(r\pi/2)^2}{r^2\pi}\) \(= \frac{(r^2\pi^2/4)}{r^2\pi}\) \(= \frac{r^2\pi^2}{4r^2\pi}\)

Hence Ratio of Area of Square to Area of Circle = \(\frac{\pi}{4}\)

\(\pi\) is approx \(= 3\)

Hence required ratio \(= \frac{3}{4}\)

Answer C
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Re: If the perimeter of square region S and the perimeter of circular regi  [#permalink]

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New post 16 Dec 2018, 10:16

Solution



Given:
    • The perimeter of square region S and the perimeter of circular region C are equal.

To find:
    • The ratio of the area of S to the area of C is closest to which option among the given ones.

Approach and Working:

    • Let side of square region S is a and radius of circular region C is r.
      o Perimeter of square = Perimeter of circle
      o 4a= 2*pi*r
      o 2a= pi*r
      o \(a= \frac{{pi * r}}{2}\)

    • Area of S : Area of C = \(a^2 :pi * r^2\)
      o = \(\frac{{pi^2 * r^2}}{4} : pi * r^2\)
      o = \(\frac{pi}{4}\)= ¾

Hence, the correct answer is option C.

Answer: C
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Re: If the perimeter of square region S and the perimeter of circular regi  [#permalink]

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New post 17 Jan 2020, 08:09
Dumpling wrote:
If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to:


A. \(\frac {3}{2}\)

B. \(\frac {4}{3}\)

C. \(\frac {3}{4}\)

D. \(\frac {2}{3}\)

E. \(\frac {1}{2}\)


\(S = C\)

\(4a = 2πr\)

\(2a = πr\)

\(2a = \frac{22r}{7}\)

\(a = \frac{11r}{7}\)

If \(r = 7\) , \(a = 11\)

So, Area of \(S = 7*7\) ; Area of \(C = 2*\frac{22}{7}*7\)

Or, Area S = 49 & Area C = 44

Area S / Area C = \(\frac{49}{44} = 1.11xxxx\)

Now, Check the options Denominator > Numerator ( Reject options C, D & E)

Option (A) can be rejected as 3/2 = 1.5 , left with option (B) , Our ANswer
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Re: If the perimeter of square region S and the perimeter of circular regi   [#permalink] 17 Jan 2020, 08:09
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