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ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]
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Updated on: 25 Sep 2017, 20:41
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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. Attachment:
Picture.PNG [ 8.57 KiB  Viewed 13400 times ]
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Originally posted by sdas on 19 Feb 2012, 23:15.
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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19 Feb 2012, 23:51
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: 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. 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. Attachment:
Picture2.PNG [ 8.57 KiB  Viewed 13362 times ]
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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]
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19 Feb 2012, 23:57
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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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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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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10 Oct 2013, 09:02



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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]
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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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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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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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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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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: abcdisasquarepictureframeseefigureefghisa127823.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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10 Dec 2014, 00:43
Refer modified diagram below: Attachment:
Picture.PNG [ 16.12 KiB  Viewed 8806 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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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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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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20 Mar 2018, 11:34
Hi All, In this prompt, we're told that a square is inscribed in another square. With the rest of the info in the prompt, we know the following: 1) (area of small square) = (area of "frame") 2) (area of small square) + (area of "frame") = (area of big square) = 36 With these two formulas, we know that the small square and the frame are EQUAL and that they add up to 36. So, they must both = 18 Area of a square = S^2 so…. S^2 = 18 S = 3(root2) Final Answer: GMAT assassins aren't born, they're made, Rich
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