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# Square ABCD has an area of 9 square inches. Sides AD and BC

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Square ABCD has an area of 9 square inches. Sides AD and BC [#permalink]  31 Jul 2012, 19:11
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79% (01:52) correct 21% (00:30) wrong based on 38 sessions
Square ABCD has an area of 9 square inches. Sides AD and BC are lengthened to x inches each. By how many inches were sides AD and BC lengthened?
Attachment:

Square.png [ 5.13 KiB | Viewed 2124 times ]

(1) The diagonal of the resulting rectangle measures 5 inches.
(2) The resulting rectangle can be cut into three rectangles of equal size.

Can anyone help me vvith B please .. Ive got A right
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Aug 2012, 02:47, edited 1 time in total.
Manager
Joined: 22 Jun 2012
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GMAT 1: 730 Q49 V40
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Kudos [?]: 16 [0], given: 6

Re: Square ABCD has an area of 9 square inches. [#permalink]  31 Jul 2012, 22:01
Edit: never mind I thought it was a rectangle instead of a square, deleted my post to avoid confusion, I keep my explanation of statement 2 since it is mentioned by someone else below.

2) Any rectangle can be divided into 3 rectangles of equal size, insufficient

Last edited by duriangris on 31 Jul 2012, 23:40, edited 1 time in total.
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Re: Square ABCD has an area of 9 square inches. [#permalink]  31 Jul 2012, 22:50
(1) Given that the area of the square is 9, then each side of the square is 3.
The lengthened sides will be of length 3 + x each, and the diagonal of the obtained rectangle being 5,
we can write $$(x+3)^2+3^2=5^2$$, from which $$(x+3)^2=16$$, so $$x = 1$$.
Sufficient.

(2) Obviously, not sufficient, as was already mentioned by "duriangris" in the previous post.

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PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Joined: 22 Jun 2012
Posts: 53
GMAT 1: 730 Q49 V40
Followers: 2

Kudos [?]: 16 [0], given: 6

Re: Square ABCD has an area of 9 square inches. [#permalink]  31 Jul 2012, 23:38
EvaJager wrote:
(1) Given that the area of the square is 9, then each side of the square is 3.
The lengthened sides will be of length 3 + x each, and the diagonal of the obtained rectangle being 5,
we can write $$(x+3)^2+3^2=5^2$$, from which $$(x+3)^2=16$$, so $$x = 1$$.
Sufficient.

(2) Obviously, not sufficient, as was already mentioned by "duriangris" in the previous post.

Gosh! it was a square not a rectangle!!
Re: Square ABCD has an area of 9 square inches.   [#permalink] 31 Jul 2012, 23:38
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