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Is the area of circle C greater than the area of square S?

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Is the area of circle C greater than the area of square S?  [#permalink]

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New post 21 May 2016, 03:50
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58% (02:06) correct 42% (01:58) wrong based on 117 sessions

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Is the area of circle C greater than the area of square S?  [#permalink]

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New post 21 May 2016, 05:13
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Bunuel wrote:
Is the area of circle C greater than the area of square S?

(1) The diameter of C is equal to the diagonal of square S.
(2) The perimeter of S is less than the circumference of C.



Hi,
Area of a circle is dependent on radius, r, and That of square on side, say a..
so if we know the ratio between radius and the side we can answer the Q..

(1) The diameter of C is equal to the diagonal of square S.
This tells us that \(2r=\sqrt{2}a\)..
Suff..

(2) The perimeter of S is less than the circumference of C.
since we are dealing with inequality sign, always better to check for the values..
\(4a<2*pi*r................. a<pi*\frac{r}{2}\).....................
since both sides are +ive, square both sides...
\(a^2<(pi*\frac{r}{2})^2.........................a^2<pi*r^2*\frac{pi}{4}\).......................
since pi/4 <1, we can safely say \(a^2<pi*r^2\) OR area of square < area of circle
Suff

D
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Is the area of circle C greater than the area of square S?  [#permalink]

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New post 27 Sep 2017, 19:37
We need to find if Area(C) >Area(S)

Let the side of the square is 'a' and radius of the circle be 'r'.

Stmt1: diameter (d)=\(\sqrt{2}a\)
on solving this we get, radius r = \(\sqrt{2}a/2\)
Using this relation we can deduce a relationship between the areas, hence we don't need to solve any further. This statement is sufficient.

Stmt2: \(4a<2 * pi * r\)
=> \(2a<pi*r\)
On squaring both sides...
=> \(a^2 < (pi^2 * r^2)/4\)

Again a clear relationship between a and r can be found here.. Hence this statement is sufficient.

The answer is D.
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Re: Is the area of circle C greater than the area of square S?  [#permalink]

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New post 08 Aug 2019, 18:54
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Re: Is the area of circle C greater than the area of square S?   [#permalink] 08 Aug 2019, 18:54
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