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

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

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19 Feb 2012, 22:15
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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?
Attachment:

Picture.PNG [ 8.57 KiB | Viewed 8073 times ]

Hi, this question is annoying me, as I believe I am unable to read it straight. Pls help.
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Last edited by Bunuel on 19 Feb 2012, 22:43, edited 1 time in total.
Edited the question and added the figure
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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19 Feb 2012, 22:51
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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:
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.

Attachment:

Picture2.PNG [ 8.57 KiB | Viewed 8055 times ]
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 [#permalink]

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19 Feb 2012, 22:57
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 [#permalink]

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19 Feb 2012, 23:05
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 [#permalink]

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10 Oct 2013, 07:36
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 [#permalink]

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10 Oct 2013, 08:02
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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 [#permalink]

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10 Oct 2013, 08:14
Crystal clear, thanks a lot
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Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]

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13 Nov 2013, 19:22
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 [#permalink]

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14 Nov 2013, 17:13
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 [#permalink]

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09 Dec 2014, 16:25
Hi,

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

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

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09 Dec 2014, 16:26
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

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 [#permalink]

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09 Dec 2014, 23:43
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Refer modified diagram below:

Attachment:

Picture.PNG [ 16.12 KiB | Viewed 4316 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 [#permalink]

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11 Apr 2015, 06:54
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 [#permalink]

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

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13 Aug 2016, 23:42
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.
Re: ABCD is a square picture frame (see figure). EFGH is a   [#permalink] 13 Aug 2016, 23:42
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