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# ABCD is a square, and EFGH is a square, each vertex of which is on a s

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Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4489
ABCD is a square, and EFGH is a square, each vertex of which is on a s  [#permalink]

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27 May 2016, 16:05
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57% (02:05) correct 43% (02:41) wrong based on 99 sessions

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Square inside square.png [ 4.88 KiB | Viewed 2379 times ]

ABCD is a square, and EFGH is a square, each vertex of which is on a side of ABCD. What is the ratio of the area of square EFGH to the area of square ABCD?

Statement #1: AE:AB = 4:7

Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24

Geometry is a truly beautiful subject! This question is one of a set of ten practice DS questions about geometry. To see the others, as well as the OE for this particular question, see:
GMAT Data Sufficiency Geometry Practice Questions

Mike

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Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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Joined: 02 Aug 2009
Posts: 7108
Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s  [#permalink]

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27 May 2016, 18:53
mikemcgarry wrote:
Attachment:
Square inside square.png

ABCD is a square, and EFGH is a square, each vertex of which is on a side of ABCD. What is the ratio of the area of square EFGH to the area of square ABCD?

Statement #1: AE:AB = 4:7

Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24

Geometry is a truly beautiful subject! This question is one of a set of ten practice DS questions about geometry. To see the others, as well as the OE for this particular question, see:
GMAT Data Sufficiency Geometry Practice Questions

Mike

we have two squares...
the one inside, EFGH, will have least area when its vertex is in center of the sides.....
and will become max as it moves closer to the vertex of the bigger square ABCD..

Second point is that all the triangles formed on the vertex of ABCD will be similar...

lets see the statement

Statement #1: AE:AB = 4:7
let the common ratio be x....
so AE = 4x and AB = 7x, so BE = 3x...

side of ABCD = 7x and side of EFGH = HYP of triangle whose sides are 3x and 4x = $$\sqrt{(3x)^2+(4x)^2}$$..
ratio =$$\frac{7x*7x}{\sqrt{(3x)^2+(4x)^2}^2}$$..
variable x will get cancelled out and we will have a numeric value...
Suff

Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24
all triangles are similar so their Area = 0.24*Area of EFGH*4.....
Let area of square EFGH = x
Area of ABCD = Area of 4 triangles + area of EFGH = 0.24*x*4 + x...

ratio =$$\frac{0.24*x*4 + x}{x}$$ = 1.96/1
Suff

D
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Joined: 28 Dec 2015
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Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s  [#permalink]

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05 Jul 2016, 02:13
chetan2u wrote:
mikemcgarry wrote:
Attachment:
Square inside square.png

ABCD is a square, and EFGH is a square, each vertex of which is on a side of ABCD. What is the ratio of the area of square EFGH to the area of square ABCD?

Statement #1: AE:AB = 4:7

Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24

Geometry is a truly beautiful subject! This question is one of a set of ten practice DS questions about geometry. To see the others, as well as the OE for this particular question, see:
GMAT Data Sufficiency Geometry Practice Questions

Mike

we have two squares...
the one inside, EFGH, will have least area when its vertex is in center of the sides.....
and will become max as it moves closer to the vertex of the bigger square ABCD..

Second point is that all the triangles formed on the vertex of ABCD will be similar...

lets see the statement

Statement #1: AE:AB = 4:7
let the common ratio be x....
so AE = 4x and AB = 7x, so BE = 3x...

side of ABCD = 7x and side of EFGH = HYP of triangle whose sides are 3x and 4x = $$\sqrt{(3x)^2+(4x)^2}$$..
ratio =$$\frac{7x*7x}{\sqrt{(3x)^2+(4x)^2}^2}$$..
variable x will get cancelled out and we will have a numeric value...
Suff

Statement #2: The ratio of the area of triangle AHE to the area of square EFGH is 0.24
all triangles are similar so their Area = 0.24*Area of EFGH*4.....
Let area of square EFGH = x
Area of ABCD = Area of 4 triangles + area of EFGH = 0.24*x*4 + x...

ratio =$$\frac{0.24*x*4 + x}{x}$$ = 1.96/1
Suff

D

Dear sir,

I don't get the first part...
AE/AB=4/7
AE/AE+EB=4/7
AE/EB=4/3

So,AE=4x and EB=3x
But then how EBF or any other triangle is a 3-4-5 triangle??
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4489
Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s  [#permalink]

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05 Jul 2016, 09:31
1
Ashishsteag wrote:
Dear sir,

I don't get the first part...
AE/AB=4/7
AE/AE+EB=4/7
AE/EB=4/3

So,AE=4x and EB=3x
But then how EBF or any other triangle is a 3-4-5 triangle??

Dear Ashishsteag

This is Mike McGarry, the author of this question. I'm happy to respond.

So I believe you get that, for some unknown x, AE = 4x and EB = 3x.

You see, the entire figure has four-fold symmetry. We could rotate the entire shape by 90 degrees and it would be the same. The four triangles, {AEH, BFE, CGF, and DHG} have to be entirely congruent. Thus,

AE = BF = CG = DH = 4x
AH = BE = CF = GD = 3x

Thus, all of them are 3-4-5 triangles.

Does this make sense?
Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Intern
Joined: 28 Dec 2015
Posts: 39
Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s  [#permalink]

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05 Jul 2016, 21:30
mikemcgarry wrote:
Ashishsteag wrote:
Dear sir,

I don't get the first part...
AE/AB=4/7
AE/AE+EB=4/7
AE/EB=4/3

So,AE=4x and EB=3x
But then how EBF or any other triangle is a 3-4-5 triangle??

Dear Ashishsteag

This is Mike McGarry, the author of this question. I'm happy to respond.

So I believe you get that, for some unknown x, AE = 4x and EB = 3x.

You see, the entire figure has four-fold symmetry. We could rotate the entire shape by 90 degrees and it would be the same. The four triangles, {AEH, BFE, CGF, and DHG} have to be entirely congruent. Thus,

AE = BF = CG = DH = 4x
AH = BE = CF = GD = 3x

Thus, all of them are 3-4-5 triangles.

Does this make sense?
Mike

Thanx a lot mike....:D
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Joined: 09 Sep 2013
Posts: 9197
Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s  [#permalink]

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12 Jul 2018, 11:03
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Re: ABCD is a square, and EFGH is a square, each vertex of which is on a s &nbs [#permalink] 12 Jul 2018, 11:03
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