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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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02 Apr 2016, 06:12

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

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02 Apr 2016, 09:00

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?

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

Attachments

IMG_20160402_225031.jpg [ 1.41 MiB | Viewed 1446 times ]

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

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03 Apr 2016, 15:08

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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03 Oct 2017, 02:18

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