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

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Math Expert
Joined: 02 Sep 2009
Posts: 58454
The figure above shows a circle inscribed in a square which is in turn  [#permalink]

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31 Oct 2018, 01:37
00:00

Difficulty:

25% (medium)

Question Stats:

82% (02:09) correct 18% (01:36) wrong based on 73 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

Attachment:

phd01.png [ 13.19 KiB | Viewed 695 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, 07:27
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, 14:49
nkin wrote:
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)
Re: The figure above shows a circle inscribed in a square which is in turn   [#permalink] 20 Nov 2018, 14:49
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