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Re: The circumference of a circle is [#permalink]
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Carcass wrote:
IlCreatore wrote:
My solution is: since we now that the circumference of the circle is 7/8 the perimeter of the square, we can set 2*pi*r=7/8*4*l, from which we can derive that r = 7/(4*pi)*l. Then the area of a square is l^2, whereas the area of a circle is pi*r^2, where we can substitute r as above. Then, we have to compare l^2 to 49/(16*pi)*l^2 and since 49/16*pi is less than one, the area of the square is greater. So answer should be A. Why B?


OE

Quote:
Because an exterior angle of a triangle is equal to the sum of the two opposite interior angles of the triangle (in this case, the top small triangle), c = a + b.
Therefore, d > c and a + b = c taken together imply that d > a + b. Subtract b from both sides: d – b > a. Quantity B is greater.


Triangles? Should I divide the square in two triangles? I really don't get how a triangle matters in a comparison between a circle and a square
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Re: The circumference of a circle is [#permalink]
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IlCreatore wrote:
Carcass wrote:
IlCreatore wrote:
My solution is: since we now that the circumference of the circle is 7/8 the perimeter of the square, we can set 2*pi*r=7/8*4*l, from which we can derive that r = 7/(4*pi)*l. Then the area of a square is l^2, whereas the area of a circle is pi*r^2, where we can substitute r as above. Then, we have to compare l^2 to 49/(16*pi)*l^2 and since 49/16*pi is less than one, the area of the square is greater. So the answer should be A. Why B?


OE

Quote:
Because an exterior angle of a triangle is equal to the sum of the two opposite interior angles of the triangle (in this case, the top small triangle), c = a + b.
Therefore, d > c and a + b = c taken together imply that d > a + b. Subtract b from both sides: d – b > a. Quantity B is greater.


Triangles? Should I divide the square into two triangles? I really don't get how a triangle matters in a comparison between a circle and a square


Sorry for the mismatch. My pdf is so tight that I wrote the explanation of the previous question. Apologize.

Fixed both the OA and OE
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Re: The circumference of a circle is [#permalink]
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project




The circumference of a circle is \(\frac{7}{8}\) the perimeter of a square.

Quantity A
Quantity B
The area of the square
The area of the circle


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



can someone please explain it ... I didn't get above-mentioned solution

thanks in advance for your kindness
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Re: The circumference of a circle is [#permalink]
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mrk9414 wrote:
Carcass wrote:



The circumference of a circle is \(\frac{7}{8}\) the perimeter of a square.

Quantity A
Quantity B
The area of the square
The area of the circle


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.



can someone please explain it ... I didn't get above-mentioned solution

thanks in advance for your kindness


Let the radius of circle be \(r\) and side of square be \(a\)

Given \(2πr = \frac{7}{8}(4a)\)

i.e. \(\frac{2πr(8)}{7(4)} = a\)

\(a = \frac{4πr}{7}\)

Now,
Col. A: \(a^2\)
Col. B: \(πr^2\)

Put the value of \(a\) in Col. A;

Col. A: \((\frac{4πr}{7})^2\)

Col. B: \(πr^2\)

Col. A: \(\frac{16π^2r^2}{49}\)

Col. B: \(πr^2\)

Dividing both sides by \(πr^2\);

Col. A: \(\frac{16π}{49}\)

Col. B: \(1\)

Multiplying both sides by \(49\);

Col. A: \(16π\)
Col. B: \(49\)

Put \(π = 3.14\),

Col. A: \(50.24\)
Col. B: \(49\)

Hence, option A
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Re: The circumference of a circle is [#permalink]
Hi Carcass, can you rewrite the question for better clarity from \(\frac{7}{8}\) to simply 7/8.

Edit: Basically, there is something wrong with the formatting. Pls check the question and do the needful.
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Re: The circumference of a circle is [#permalink]
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Optimist wrote:
Hi Carcass, can you rewrite the question for better clarity from \(\frac{7}{8}\) to simply 7/8.

Edit: Basically, there is something wrong with the formatting. Pls check the question and do the needful.


Maybe you have some issue to visualize the question.

I do not see any issue in 7/8 formatted in latex
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Re: The circumference of a circle is [#permalink]
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Re: The circumference of a circle is [#permalink]
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