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Tricky question [#permalink] New post 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?

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Re: Tricky question [#permalink] New post 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
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Re: Tricky question [#permalink] New post 24 Nov 2011, 01:42
A . Followed the similar approach as Postal
Re: Tricky question   [#permalink] 24 Nov 2011, 01:42
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