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ABCD is a square picture frame (see figure). EFGH is a [#permalink]
19 Feb 2012, 22:15

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Question Stats:

100% (01:58) correct
0% (00:00) wrong based on 12 sessions

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 5410 times ]

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

"When the going gets tough, the tough gets going!"

Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]
19 Feb 2012, 22:51

1

This post received KUDOS

Expert's post

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 5401 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).

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

Re: ABCD is a square picture frame (see figure). EFGH is a [#permalink]
10 Oct 2013, 08:02

Expert's post

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).

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

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

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).

ABCD is a square picture frame (see figure). EFGH is a [#permalink]
09 Dec 2014, 16:26

Expert's post

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).

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

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

\(EF^2=18\)

\(EF= 3 \sqrt{2}\) _________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

gmatclubot

Re: ABCD is a square picture frame (see figure). EFGH is a
[#permalink]
11 Apr 2015, 06:54

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