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

Question

https://gmatclub.com/forum/download/file.php?id=16656 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. Picture.PNG

Bunuel · Accepted Answer

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.

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=16657 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. Picture2.PNG

jazzman · Answer

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)

sdas · Answer

Thank you folks. I missed understanding the shaded part and picture area (EFGH). Thanks Bunuel and Jazzman

theGame001 · Answer

Hi,

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

Thanks

theGame001 · Answer

Crystal clear, thanks a lot

coolparthi · Answer

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}

madn800 · Answer

Follow the question stem. Area(EFGH)=area(ABCD)-area(EFGH)------>2*area(EFGH)=area(ABCD). This means that area(EFGH)=1/2*area(ABCD).

nish4every1 · Answer

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 · Answer

Have you read this: abcd-is-a-square-picture-frame-see-figure-efgh-is-a-127823.html#p1276619 ?

PareshGmat · Answer

Refer modified diagram below: Picture.PNG 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}

Lucky2783 · Answer

6^2 -EF^2 = EF^2

EF^2=18

EF= 3 \sqrt{2}

HMC · Answer

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.

EMPOWERgmatRichC · Answer

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:  D

GMAT assassins aren't born, they're made, 
Rich

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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

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