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# Tricky question

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Manager
Status: Back to (GMAT) Times Square!!!
Joined: 15 Aug 2011
Posts: 179

Kudos [?]: 87 [0], given: 25

Location: United States (IL)
GMAT 1: 650 Q49 V30
WE: Information Technology (Computer Software)

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23 Nov 2011, 06:28
Just wanted to know whether there is a short-cut way of solving this question? Can anyone please help me with this?

Attachment:

11.jpg [ 61.05 KiB | Viewed 1257 times ]

Thanks,
V.

Kudos [?]: 87 [0], given: 25

Intern
Joined: 20 Aug 2010
Posts: 37

Kudos [?]: 33 [0], given: 0

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23 Nov 2011, 23:45
The area of the figure inside the circle seems to be composed of the half of the circle (because the cords form a right angle and therefore the hypotenuse of the triangle is the diameter of the circle cutting the middle of the circle) + the right triangle in the bottom left of the square. We should find its area.
Area of 1/2 circle = PiR^2/2 = Pi/2 (as radius is 2/2=1)
Area of triangle is 1/2 * 2 (hypotenuse) * 1 (hight to hypotenuse, which is a radius =1) =1
diff = Area of the Square - area of our figure = 2*2 - Pi/2 - 1
Ratio = diff / area of the square

Kudos [?]: 33 [0], given: 0

Intern
Joined: 31 Jul 2011
Posts: 13

Kudos [?]: 4 [0], given: 7

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24 Nov 2011, 01:42
A . Followed the similar approach as Postal

Kudos [?]: 4 [0], given: 7

Re: Tricky question   [#permalink] 24 Nov 2011, 01:42
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