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

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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]
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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]
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]
1
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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]
Crystal clear, thanks a lot
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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]
1
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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]
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]
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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Re: ABCD is a square picture frame (see figure). EFGH is a square inscribe [#permalink]
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 square inscribe [#permalink]
3
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Refer modified diagram below:

Attachment:

Picture.PNG [ 16.12 KiB | Viewed 25017 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]
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]
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]
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)

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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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