It is currently 20 Oct 2017, 16:56

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 18 Feb 2011
Posts: 41

Kudos [?]: 51 [2], given: 26

GPA: 3.91
In the figure above, square AFGE is inside square ABCD such [#permalink]

### Show Tags

10 Feb 2012, 06:39
2
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

65% (02:29) correct 35% (02:03) wrong based on 174 sessions

### HideShow timer Statistics

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

Kudos [?]: 51 [2], given: 26

Director
Status: Enjoying the GMAT journey....
Joined: 26 Aug 2011
Posts: 713

Kudos [?]: 584 [2], given: 264

Location: India
GMAT 1: 620 Q49 V24

### Show Tags

10 Feb 2012, 07:01
2
KUDOS
1
This post was
BOOKMARKED
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.

A WAY TO INCREASE FROM QUANT 35-40 TO 47 : http://gmatclub.com/forum/a-way-to-increase-from-q35-40-to-q-138750.html

Q 47/48 To Q 50 + http://gmatclub.com/forum/the-final-climb-quest-for-q-50-from-q47-129441.html#p1064367

Three good RC strategies http://gmatclub.com/forum/three-different-strategies-for-attacking-rc-127287.html

Kudos [?]: 584 [2], given: 264

Manager
Joined: 31 Jan 2012
Posts: 74

Kudos [?]: 26 [0], given: 2

### Show Tags

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

Kudos [?]: 26 [0], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129034 [0], given: 12187

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

### Show Tags

10 Feb 2012, 09:00
Expert's post
2
This post was
BOOKMARKED
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: $$\frac{diagonal^2}{2}=\frac{(8(\sqrt{2}-1))^2}{2}=32*(\sqrt{2}-1)^2=32*(3-2\sqrt{2})$$.

_________________

Kudos [?]: 129034 [0], given: 12187

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16636

Kudos [?]: 273 [0], given: 0

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

### Show Tags

15 Jul 2014, 14:13
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Manager
Joined: 25 Apr 2014
Posts: 143

Kudos [?]: 73 [0], given: 1474

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

### Show Tags

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.

Kudos [?]: 73 [0], given: 1474

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1853

Kudos [?]: 2626 [0], given: 193

Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
In the figure above, square AFGE is inside square ABCD such [#permalink]

### Show Tags

21 Aug 2014, 00:24
AG $$= 8\sqrt{2} - 8$$

Attachment:

squ.png [ 7.85 KiB | Viewed 2157 times ]

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

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

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

_________________

Kindly press "+1 Kudos" to appreciate

Kudos [?]: 2626 [0], given: 193

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129034 [0], given: 12187

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

### Show Tags

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.
_________________

Kudos [?]: 129034 [0], given: 12187

CEO
Joined: 17 Jul 2014
Posts: 2604

Kudos [?]: 395 [0], given: 184

Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: In the figure above, square AFGE is inside square ABCD such [#permalink]

### Show Tags

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.

Kudos [?]: 395 [0], given: 184

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16636

Kudos [?]: 273 [0], given: 0

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

### Show Tags

29 Nov 2016, 00:07
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Re: In the figure above, square AFGE is inside square ABCD such   [#permalink] 29 Nov 2016, 00:07
Display posts from previous: Sort by