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ABCD is a square picture frame (see figure). EFGH is a square inscribe

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ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 19 Feb 2012, 23:15
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ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

A. \(3\sqrt[]{3}\)

B. \(4\sqrt[]{2}\)

C. \(5\sqrt[]{3}\)

D. \(3\sqrt[]{2}\)

E. \(2\sqrt[]{5}\)


Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.

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Attachment:
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Picture.PNG [ 8.57 KiB | Viewed 10014 times ]
[Reveal] Spoiler: OA

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Last edited by Bunuel on 25 Sep 2017, 20:41, edited 3 times in total.
Edited the question and added the figure

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ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 19 Feb 2012, 23:51
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sdas wrote:
Image

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
The attachment Picture.PNG is no longer available



Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.


Image

Look at the figure above, since the area of of the picture frame (shaded region) is equal to the area of EFGH, then the area of EFGH is half of the area of the big square ABCD, which is 6^2=36. Hence, the area of EFGH = 18. The side of the square EFGH is \(\sqrt{18}=3\sqrt{2}\) (since area=side^2).

Hope it's clear.


[Reveal] Spoiler:
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Picture2.PNG
Picture2.PNG [ 8.57 KiB | Viewed 9987 times ]

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 19 Feb 2012, 23:57
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sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG



Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.


Area of ABCD = 36

Area of EFGH = 18 - area of EFGH is equal to area of picture frame - they're each 1/2 of the total area (36)

A=bh
18=x^2 (b & h are equal since it is a square)

Length of EF = sqrt(18) = 3*sqrt(2)

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 20 Feb 2012, 00:05
Thank you folks. I missed understanding the shaded part and picture area (EFGH). Thanks Bunuel and Jazzman
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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 10 Oct 2013, 08:36
Hi,

But where does it say the area of EFGH is half of the area of the big square ABCD?

Thanks

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 10 Oct 2013, 09:02
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theGame001 wrote:
Hi,

But where does it say the area of EFGH is half of the area of the big square ABCD?

Thanks


It's given in the stem: the area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH) --> area(ABCD)-area(EFGH)=area(EFGH) --> area(ABCD)=2*area(EFGH).

Hope it's clear.
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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 10 Oct 2013, 09:14
Crystal clear, thanks a lot

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 13 Nov 2013, 20:22
Also,

Consider area of the picture frame is X.

Area of the Picture frame = Area of ABCD - Area of EFGH (which is equal to the area of the picture frame)
Upon substituting,
X = (6*6) - X
2X = 36
X (area) = 18

Side = \sqrt{18}
= 3\sqrt{2}

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 14 Nov 2013, 18:13
Follow the question stem. Area(EFGH)=area(ABCD)-area(EFGH)------>2*area(EFGH)=area(ABCD). This means that area(EFGH)=1/2*area(ABCD).

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 09 Dec 2014, 17:25
Hi,

How can you infer that the area of EFGH is "Half" of the Area of ABCD?
Im unable to understand that. Please explain

Thank you in advance :)

Bunuel wrote:
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG



Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.


Attachment:
Picture2.PNG
Look at the figure above, since the area of of the picture frame (shaded region) is equal to the area of EFGH, then the area of EFGH is half of the area of the big square ABCD, which is 6^2=36. Hence, the area of EFGH = 18. The side of the square EFGH is \(\sqrt{18}=3\sqrt{2}\) (since area=side^2).

Hope it's clear.

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 09 Dec 2014, 17:26
nish4every1 wrote:
Hi,

How can you infer that the area of EFGH is "Half" of the Area of ABCD?
Im unable to understand that. Please explain

Thank you in advance :)

Bunuel wrote:
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG



Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.


Attachment:
Picture2.PNG
Look at the figure above, since the area of of the picture frame (shaded region) is equal to the area of EFGH, then the area of EFGH is half of the area of the big square ABCD, which is 6^2=36. Hence, the area of EFGH = 18. The side of the square EFGH is \(\sqrt{18}=3\sqrt{2}\) (since area=side^2).

Hope it's clear.


Have you read this: abcd-is-a-square-picture-frame-see-figure-efgh-is-a-127823.html#p1276619 ?
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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 10 Dec 2014, 00:43
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Refer modified diagram below:

Attachment:
Picture.PNG
Picture.PNG [ 16.12 KiB | Viewed 6130 times ]


Given that area of orange region = area of white region in the square

It means that, if the small square EFGH is rotated 45 degree, it will exactly touch midpoints(AE = EB = 3) of square ABCD sides

\(EF = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\)
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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 11 Apr 2015, 07:54
sdas wrote:
ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picure. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?
Attachment:
Picture.PNG



Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.



\(6^2 -EF^2 = EF^2\)

\(EF^2=18\)

\(EF= 3 \sqrt{2}\)
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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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New post 14 Aug 2016, 00:42
I got the answer correct, But I do not see answer options A, B , C , D and E. Looks like they are not provided originally.

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]

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Hello from the GMAT Club BumpBot!

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe   [#permalink] 26 Sep 2017, 03:27
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