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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] New post 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
[Reveal] Spoiler: OA
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Re: area of square [#permalink] New post 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] New post 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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In the figure above, square AFGE is inside square ABCD such [#permalink] New post 10 Feb 2012, 08:00
Expert's post
nafishasan60 wrote:
Image

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

No additional drawing is needed.

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);

GC=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*(3-2\sqrt{2}).

Answer: B.
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Re: In the figure above, square AFGE is inside square ABCD such [#permalink] New post 15 Jul 2014, 13:13
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In the figure above, square AFGE is inside square ABCD such [#permalink] New post 20 Aug 2014, 14:09
Bunuel wrote:
nafishasan60 wrote:
Image

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


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 45-45-90 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*(3-2\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] New post 20 Aug 2014, 23:24
AG = 8\sqrt{2} - 8

Attachment:
squ.png
squ.png [ 7.85 KiB | Viewed 116 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] New post 21 Aug 2014, 03:09
Expert's post
maggie27 wrote:
Bunuel wrote:
nafishasan60 wrote:
Image

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


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 45-45-90 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*(3-2\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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the figure above, square AFGE is inside square ABCD such   [#permalink] 21 Aug 2014, 03:09
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