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

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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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

Please refer diagram below:

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:

GMAT assassins aren't born, they're made,
Rich
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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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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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I reached till (22 * 7)/ 4...Is there a quick way to estimate/ calculate the final answer?

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.

GMAT assassins aren't born, they're made,
Rich
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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

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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\(S = C\)

\(4a = 2πr\)

\(2a = πr\)

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

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

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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