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# In the figure above, square AFGE is inside square ABCD such

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In the figure above, square AFGE is inside square ABCD such [#permalink]  10 Feb 2012, 05:39
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In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = s^2)?

A) 32 (1-\sqrt{2})
B) 32 (3-2\sqrt{2})
C) 64 (\sqrt{2} - 1)^2
D) 64 - 16\pi
E) 32 - 4\pi
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Re: area of square [#permalink]  10 Feb 2012, 06:01
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nafishasan60 wrote:
In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = )?

A) 32 (1-)
B) 32 (3-2)
C) 64 ( - 1)
D) 64 - 16
E) 32 - 4

we can find diagonal AC = 8 \sqrt{2}
so AG = AC -CG( radius of the arc) = 8\sqrt{2} - 8
now AE = AG/\sqrt{2} = 8\sqrt{2} - 8 / \sqrt{2}
IMO B..
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Re: area of square [#permalink]  10 Feb 2012, 06:41
Took me 10 mins... but I got it... Surprising it wasn't super difficult. Since C is the center of the Circle, the length of GC = 8. Since AEFG touches the circle with it's corner, we know the angle of GC is 45%. When you spilt a square in half diagonally it's going to be 45 degrees. If the angle GC is 45% you know the triangle form with would be a 1:1:root(2). Since GC is the hypotenuse the other 2 will be 8/root(2). The length of FG = 10 (height of the square) - 8/root(2) (height of the triangle). [10-8/root(2)]^2 = 32 (3-2*root(2)).

Answer is B. Hard to explain without a graph and too lazy to make one. Sorry
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]  10 Feb 2012, 08:00
Expert's post
nafishasan60 wrote:

In the figure above, square AFGE is inside square ABCD such that G is on arc BD, which is centered at C. If DC=8, what is the area of square AFGE (area of square of side s = s^2)?

A) 32 (1-\sqrt{2})
B) 32 (3-2\sqrt{2})
C) 64 (\sqrt{2} - 1)^2
D) 64 - 16\pi
E) 32 - 4\pi

The length of the diagonal AG is AC-AG;

AC is a diagonal of a square with a side of 8, hence it equal to 8\sqrt{2} (hypotenuse of 45-45-90 triangle);

Hence AG=8\sqrt{2}-8=8(\sqrt{2}-1);

Area of a square: \frac{diagonal^2}{2}=\frac{(8(\sqrt{2}-1))^2}{2}=32*(\sqrt{2}-1)^2=32*(3-2\sqrt{2}).

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Re: In the figure above, square AFGE is inside square ABCD such   [#permalink] 10 Feb 2012, 08:00
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