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

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The figure above shows a circle inscribed in a square which is in turn  [#permalink]

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31 Oct 2018, 00:37
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Difficulty:

25% (medium)

Question Stats:

76% (01:32) correct 24% (00:22) wrong based on 45 sessions

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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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phd01.png [ 13.19 KiB | Viewed 303 times ]

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Re: The figure above shows a circle inscribed in a square which is in turn  [#permalink]

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31 Oct 2018, 06:27
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]

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20 Nov 2018, 13:49
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
Re: The figure above shows a circle inscribed in a square which is in turn &nbs [#permalink] 20 Nov 2018, 13:49
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The figure above shows a circle inscribed in a square which is in turn

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