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In the figure above, square AFGE is inside square ABCD such [#permalink]
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10 Feb 2012, 06:39
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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 (32\(\sqrt{2}\)) C) 64 (\(\sqrt{2}\)  1)\(^2\) D) 64  16\(\pi\) E) 32  4\(\pi\)
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Re: area of square [#permalink]
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10 Feb 2012, 07:01
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 (32) 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]
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10 Feb 2012, 07:41
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). [108/root(2)]^2 = 32 (32*root(2)).
Answer is B. Hard to explain without a graph and too lazy to make one. Sorry



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In the figure above, square AFGE is inside square ABCD such [#permalink]
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10 Feb 2012, 09:00



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In the figure above, square AFGE is inside square ABCD such [#permalink]
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20 Aug 2014, 15:09
Bunuel wrote: 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 (32\(\sqrt{2}\)) C) 64 (\(\sqrt{2}\)  1)\(^2\) D) 64  16\(\pi\) E) 32  4\(\pi\) No additional drawing is needed. 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 454590 triangle); AG=DC=radius=side=8; Hence \(AG=8\sqrt{2}8=8(\sqrt{2}1)\); Area of a square: \(\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}1))^2}{2}=32*(\sqrt{2}1)^2=32*(32\sqrt{2})\). Answer: B. Bunuel , I think you meant GC in place of AG in the highlighted portion.



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In the figure above, square AFGE is inside square ABCD such [#permalink]
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21 Aug 2014, 00:24
AG \(= 8\sqrt{2}  8\) Attachment:
squ.png [ 7.85 KiB  Viewed 3108 times ]
Area of shaded region \(= \frac{(8\sqrt{2}  8)^2}{2}\) \(= \frac{64 (2  2\sqrt{2} + 1)}{2}\) \(= 32 (3  2\sqrt{2})\) Answer = B
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
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21 Aug 2014, 04:09
maggie27 wrote: Bunuel wrote: 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 (32\(\sqrt{2}\)) C) 64 (\(\sqrt{2}\)  1)\(^2\) D) 64  16\(\pi\) E) 32  4\(\pi\) No additional drawing is needed. 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 454590 triangle); AG=DC=radius=side=8; Hence \(AG=8\sqrt{2}8=8(\sqrt{2}1)\); Area of a square: \(\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}1))^2}{2}=32*(\sqrt{2}1)^2=32*(32\sqrt{2})\). Answer: B. Bunuel , I think you meant GC in place of AG in the highlighted portion. Typo edited. Thank you.
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
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25 Oct 2015, 10:00
we know that ACD will form a 454590 triangle, and, knowing that one side is 8, we can deduce that AC = 8 sqrt(2) CG is 8, and thus, diagonal AG will be 8sqrt(2)8 or 8[sqrt(2)1] again, we have a 454590 triangle. Applying pythagorean theorem, we can see that 2s^2 = 8^2[sqrt(2)1]^2 or s^2 = 32[sqrt(2)1]^2 [sqrt(2)1]*[sqrt(2)1] = sqrt(2)*sqrt(2)  sqrt(2) sqrt(2) +1 = 3 2sqrt(2) +1 = 5  2sqrt(2) now we can rewrite the area of the small square to be 32*[32sqrt(2)], answer choice B.



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Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
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