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A square ABCD is drawn and point E is marked on AB such that AE=AB/3

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A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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A square ABCD is drawn and point E is marked on AB such that AE=AB/3 . Similarly points F, G and H are marked on the sides of the square such that BF=BC/3 , CG=CD/3 and DH=DA/3. If the points E, F, G, and H are connected to make a parallelogram, what is the ratio of the area of square ABCD to the area of parallelogram EFGH?

A. 81/16

B. 9/4

C. 9/5

D. 5/4

E. 4/5
[Reveal] Spoiler: OA

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Re: A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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New post 02 Apr 2016, 07:31
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shasadou wrote:
A square ABCD is drawn and point E is marked on AB such that AE=AB/3 . Similarly points F, G and H are marked on the sides of the square such that BF=BC/3 , CG=CD/3 and DH=DA/3. If the points E, F, G, and H are connected to make a parallelogram, what is the ratio of the area of square ABCD to the area of parallelogram EFGH?

A. 81/16

B. 9/4

C. 9/5

D. 5/4

E. 4/5


HI,

Two ways to find the answer
lets take each side of ABCD as 3

1)what you have as a parallelogram inside is also a square EFGH..
what is each side of the square EFGH ?
Take TRiangle EHA..
its sides are \(AE=\frac{1}{3} *3=1\) and \(AH = 3-\frac{1}{3}*3= 2\)
so HYP or side EH = \(\sqrt{1^2+2^2}\)=\(\sqrt{5}\)
so area of EFGH= \(\sqrt{5}\)*\(\sqrt{5}\)=5
and area of ABCD=3*3=9
ans = 9/5..

2) when we know
Take TRiangle EHA..
its sides are AE= \(\frac{1}{3} *3=1\)and \(AH = 3-\frac{1}{3}*3= 2\)
the area of this triangle = 1/2 * 1*2 =1..
there are 4 such triangles at each vertices A,B,C,D..
so Area =1*4..
area of ABCD =9..
area of EFGH= area of ABCD - Area of 4 triangles= 9-4=5
ans=\(\frac{9}{5}\)
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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New post 02 Apr 2016, 10:00
chetan2u wrote:
shasadou wrote:
A square ABCD is drawn and point E is marked on AB such that AE=AB/3 . Similarly points F, G and H are marked on the sides of the square such that BF=BC/3 , CG=CD/3 and DH=DA/3. If the points E, F, G, and H are connected to make a parallelogram, what is the ratio of the area of square ABCD to the area of parallelogram EFGH?

A. 81/16

B. 9/4

C. 9/5

D. 5/4

E. 4/5


HI,

Two ways to find the answer
lets take each side of ABCD as 3

1)what you have as a parallelogram inside is also a square EFGH..
what is each side of the square EFGH ?
Take TRiangle EHA..
its sides are \(AE=\frac{1}{3} *3=1\) and \(AH = 3-\frac{1}{3}*3= 2\)
so HYP or side EH = \(\sqrt{1^2+2^2}\)=\(\sqrt{5}\)
so area of EFGH= \(\sqrt{5}\)*\(\sqrt{5}\)=5
and area of ABCD=3*3=9
ans = 9/5..

2) when we know
Take TRiangle EHA..
its sides are AE= \(\frac{1}{3} *3=1\)and \(AH = 3-\frac{1}{3}*3= 2\)
the area of this triangle = 1/2 * 1*2 =1..
there are 4 such triangles at each vertices A,B,C,D..
so Area =1*4..
area of ABCD =9..
area of EFGH= area of ABCD - Area of 4 triangles= 9-4=5
ans=\(\frac{9}{5}\)


Hi Chetan,

Could you please explain how the parallelogram is also a square?

Thanks.

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Re: A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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New post 02 Apr 2016, 10:21
KS15 wrote:
chetan2u wrote:
shasadou wrote:
A square ABCD is drawn and point E is marked on AB such that AE=AB/3 . Similarly points F, G and H are marked on the sides of the square such that BF=BC/3 , CG=CD/3 and DH=DA/3. If the points E, F, G, and H are connected to make a parallelogram, what is the ratio of the area of square ABCD to the area of parallelogram EFGH?

A. 81/16

B. 9/4

C. 9/5

D. 5/4

E. 4/5


HI,

Two ways to find the answer
lets take each side of ABCD as 3

1)what you have as a parallelogram inside is also a square EFGH..
what is each side of the square EFGH ?
Take TRiangle EHA..
its sides are \(AE=\frac{1}{3} *3=1\) and \(AH = 3-\frac{1}{3}*3= 2\)
so HYP or side EH = \(\sqrt{1^2+2^2}\)=\(\sqrt{5}\)
so area of EFGH= \(\sqrt{5}\)*\(\sqrt{5}\)=5
and area of ABCD=3*3=9
ans = 9/5..

2) when we know
Take TRiangle EHA..
its sides are AE= \(\frac{1}{3} *3=1\)and \(AH = 3-\frac{1}{3}*3= 2\)
the area of this triangle = 1/2 * 1*2 =1..
there are 4 such triangles at each vertices A,B,C,D..
so Area =1*4..
area of ABCD =9..
area of EFGH= area of ABCD - Area of 4 triangles= 9-4=5
ans=\(\frac{9}{5}\)


Hi Chetan,

Could you please explain how the parallelogram is also a square?

Thanks.


Hi
See the attached image
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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New post 02 Apr 2016, 11:23
Agreed-I could deduce until here and got the correct answer-but it can also be a rhombus right?

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Re: A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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New post 02 Apr 2016, 11:33
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KS15 wrote:
Agreed-I could deduce until here and got the correct answer-but it can also be a rhombus right?


Hi,
every Square is a RHOMBUS but it is not vice versa..
here all sides are equal as \(\sqrt{5}\) and all angles are 90 degree, so it is a square, You may call it a rhombus with all angles 90 degree..
something like all Squares are also rectangle and vice versa is not true..
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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New post 03 Apr 2016, 16:08
chetan2u wrote:
KS15 wrote:
Agreed-I could deduce until here and got the correct answer-but it can also be a rhombus right?


Hi,
every Square is a RHOMBUS but it is not vice versa..
here all sides are equal as \(\sqrt{5}\) and all angles are 90 degree, so it is a square, You may call it a rhombus with all angles 90 degree..
something like all Squares are also rectangle and vice versa is not true..



Hey I did get the Correct answer but i used this approach =>
as we can see all the lengths are equal => it is definitely a rhombus
Then i saw that each of the diagonals are of same length ..
As far as i can recall from m 10 grade studies a Rhombus with equal diagonals is a square .
so i took it as a square .
Another approach (THIS IS A LAME ONE ) =>if we take it as a rhombus => area = 1/2 *3x*3x => 9x^2/2 => Hence the ratio will be => 9x^2/2 /9x^2 => 1/2
But 1/2 is not in the options ..
hence It must be a square ..

Anyway all i have to ask is is my understanding that a rhombus with equal diagonals is a square correct ..
And Also why are the diagonals both equal to 3x (side of bigger square ) and not √10 x each...
Any theory related to this will be helpful..
Regards
S.C.S.A
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Re: A square ABCD is drawn and point E is marked on AB such that AE=AB/3 [#permalink]

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Re: A square ABCD is drawn and point E is marked on AB such that AE=AB/3   [#permalink] 03 Oct 2017, 03:18
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