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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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Difficulty:

55% (hard)

Question Stats:

62% (03:23) correct
38% (02:07) wrong based on 74 sessions

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\(\pi\)

Re: area of square [#permalink]
10 Feb 2012, 06:01

1

This post received KUDOS

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.. _________________

Fire the final bullet only when you are constantly hitting the Bull's eye, till then KEEP PRACTICING.

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

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\(\pi\)

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

Re: In the figure above, square AFGE is inside square ABCD such [#permalink]
15 Jul 2014, 13:13

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In the figure above, square AFGE is inside square ABCD such [#permalink]
20 Aug 2014, 14:09

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\(\pi\)

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.

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

Expert's post

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\(\pi\)

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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