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The figure above shows a circle inscribed in a square which is in turn

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Bunuel
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The figure above shows a circle inscribed in a square which is in turn [#permalink] New post   31 Oct 2018, 01:37
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The figure above shows a circle inscribed in a square which is in turn inscribed within a larger circle. What is the ratio of the area of the larger circle to that of the smaller circle?


A. √2

B. π/2

C. π^2/(4√2)

D. 2

E. π/√2


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nkin
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Re: The figure above shows a circle inscribed in a square which is in turn [#permalink] New post   31 Oct 2018, 07:27
Small circle radius = r,
Half of diagonal of square = sqrt(r^2 + r^2) = \(\sqrt{2}\)r = Radius of larger circle

Ratio = pi*\((sqrt(2)r)^2\)/pi*\(r^2\) = 2

Option D
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Re: The figure above shows a circle inscribed in a square which is in turn [#permalink] New post   20 Nov 2018, 14:49
nkin wrote:
Small circle radius = r,
Half of diagonal of square = sqrt(r^2 + r^2) = \(\sqrt{2}\)r = Radius of larger circle

Ratio = pi*\((sqrt(2)r)^2\)/pi*\(r^2\) = 2

Option D


I'm a little lost. I thought the diagonal of a square was d=s√2 so wouldn't d=s√2/2 be the radius of the larger circle ? Where did you get sqrt(r^2 + r^2)
Please help me understand where I'm not seeing what's obvious
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Re: The figure above shows a circle inscribed in a square which is in turn [#permalink] New post   14 Sep 2020, 07:07
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Bunuel wrote:

The figure above shows a circle inscribed in a square which is in turn inscribed within a larger circle. What is the ratio of the area of the larger circle to that of the smaller circle?


A. √2

B. π/2

C. π^2/(4√2)

D. 2

E. π/√2


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Attachment:
phd01.png


Let the sides of the square be 2 , thus radius of the smaller circle is 1

So, Area of the smaller circle is \(π*1^2 = π\)

Diameter of the bigger circle is \(2\sqrt{2}\) , so radius of the bigger circle is \(\sqrt{2}\)

Hence, Area of the Larger circle is \(π*\sqrt{2}*\sqrt{2} = 2π\)

Quote:
What is the ratio of the area of the larger circle to that of the smaller circle?


Thus, ratio of the area of the larger circle to that of the smaller circle is \(\frac{2π}{π} = 2\), Answer must be (D) 2
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Rogermoris
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Re: The figure above shows a circle inscribed in a square which is in turn [#permalink] New post   14 Sep 2020, 09:36
Let us assume that the small circle radius = r,
Diagonal of square = \(\sqrt{(r^2 + r^2)} = \sqrt{2}r\)

Half the diagonal is the radius of larger circle.

Ratio =\( \frac{pi*(\sqrt{2})^2*r^2}{pi*r^2} = 2\)

Option D
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Re: The figure above shows a circle inscribed in a square which is in turn [#permalink] New post   14 Sep 2020, 16:14
ratio = A_large / A_small = ?
A_large = pi*R^2
A_small = pi*r^2

pi cancels.
ratio = R^2 / r^2 = (R/r)^2

45-45-90 triangle with no values
R = rt(2)/2 -> allows square s=1 -> then r=1/2

R/r = [rt(2)/2] / [1/2] = rt(2)

(R/r)^2 = 2

Ans: D
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Re: The figure above shows a circle inscribed in a square which is in turn [#permalink]
14 Sep 2020, 16:14
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