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As the figure above shows, four identical circles are inscri

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Joined: 10 Feb 2011
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As the figure above shows, four identical circles are inscri [#permalink]

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17 Feb 2011, 15:04
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185. As the figure above shows, four identical circles are inscribed in a square. If the distance between centers of circle A and В is 4sq2 , what is the area of the shaded region?
(A) 256 – 72п
(B) 256 – 64п
(C) 256 – 32п
(D) 128 – 32п
(E) 64 – 16п
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Joined: 02 Sep 2009
Posts: 34478
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Kudos [?]: 79800 [1] , given: 10022

Re: As the figure above shows, four identical circles are inscri [#permalink]

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17 Feb 2011, 15:19
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Expert's post
banksy wrote:
185. As the figure above shows, four identical circles are inscribed in a square. If the distance between centers of circle A and В is 4sq2 , what is the area of the shaded region?
(A) 256 – 72п
(B) 256 – 64п
(C) 256 – 32п
(D) 128 – 32п
(E) 64 – 16п
Attachment:

untitled.PNG [ 2.74 KiB | Viewed 2790 times ]
As you can see on the diagram above AB is the hypotenuse of the isosceles right triangle with legs equal to 2r (45-45-90 right triangle --> ratio of the sides $$1:1:\sqrt{2}$$). Now, as $$AB=hypotenuse=4\sqrt{2}$$ then $$2r=leg=\frac{4\sqrt{2}}{\sqrt{2}}=4$$ --> $$r=2$$.

Shades region = are of the square - area of the circles = $$(4r)^2-4*\pi{r^2}=64-16\pi$$.

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Kudos [?]: 471 [0], given: 36

Re: As the figure above shows, four identical circles are inscri [#permalink]

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19 Apr 2011, 19:36
AB = 4root(2)

So The diameter of circle = 4

Area of aquare = 4+4 = 8

Area of shaded part = 64 - 4*pi*4 = 64 - 16pi

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Joined: 19 Apr 2011
Posts: 10
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Kudos [?]: 1 [1] , given: 1

Re: As the figure above shows, four identical circles are inscri [#permalink]

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19 Apr 2011, 20:04
1
KUDOS
Distance AB = 4sq2
So each side of triangle = 4
Each side of Square = 8

Area of shaded portion = area of square - area of 4 circles
= (8*8) - 4*pi(2)square
= 64 - 16 pi

Re: As the figure above shows, four identical circles are inscri   [#permalink] 19 Apr 2011, 20:04
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