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

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Re: If the perimeter of square region S and the perimeter of circular regi [#permalink]
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Re: If the perimeter of square region S and the perimeter of circular regi [#permalink]
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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
Area = 4pi

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

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Re: If the perimeter of square region S and the perimeter of the [#permalink]
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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Re: If the perimeter of square region S and the perimeter of the [#permalink]
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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If the perimeter of square region S and the perimeter of the [#permalink]
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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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Re: If the perimeter of square region S and the perimeter of the [#permalink]
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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 circular regi [#permalink]
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: $$P_{square}=4x=2\pi{r}=P_{circle}$$ --> $$x=\frac{\pi{r}}{2}$$.

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

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Re: If the perimeter of square region S and the perimeter of circular regi [#permalink]
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]
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 $$= \frac{3}{4}$$

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

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.

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