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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? 
Square.png [ 5.13 KiB | Viewed 15951 times ]
(1) The diagonal of the resulting rectangle measures 5 inches.
(2) The resulting rectangle can be cut into three rectangles of equal size.
Attachment:
Square.png [ 5.13 KiB | Viewed 15951 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
duriangris
Updated on: 01 Aug 2012, 00:40
Originally posted by duriangris on 31 Jul 2012, 23:01.
Last edited by duriangris on 01 Aug 2012, 00:40, edited 1 time in total.
Last edited by duriangris on 01 Aug 2012, 00:40, edited 1 time in total.
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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
2) Any rectangle can be divided into 3 rectangles of equal size, insufficient
31 Jul 2012, 23:50
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(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.
Answer A.
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.
Answer A.
