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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]

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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 (3-2$$\sqrt{2}$$)
C) 64 ($$\sqrt{2}$$ - 1)$$^2$$
D) 64 - 16$$\pi$$
E) 32 - 4$$\pi$$
[Reveal] Spoiler: OA
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10 Feb 2012, 07:01
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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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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). [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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In the figure above, square AFGE is inside square ABCD such [#permalink]

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10 Feb 2012, 09:00
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-GC;

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]

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15 Jul 2014, 14:13
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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 (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: $$\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}-1))^2}{2}=32*(\sqrt{2}-1)^2=32*(3-2\sqrt{2})$$.

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 1267 times ]

Area of shaded region $$= \frac{(8\sqrt{2} - 8)^2}{2}$$

$$= \frac{64 (2 - 2\sqrt{2} + 1)}{2}$$

$$= 32 (3 - 2\sqrt{2})$$

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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 (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: $$\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}-1))^2}{2}=32*(\sqrt{2}-1)^2=32*(3-2\sqrt{2})$$.

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 45-45-90 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 45-45-90 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*[3-2sqrt(2)], answer choice B.
Re: In the figure above, square AFGE is inside square ABCD such   [#permalink] 25 Oct 2015, 10:00
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