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Re: Square ABCD has an area of 9 square inches. [#permalink]

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31 Jul 2012, 23:01

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 01 Aug 2012, 00:40, edited 1 time in total.

Re: Square ABCD has an area of 9 square inches. [#permalink]

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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.
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Re: Square ABCD has an area of 9 square inches. [#permalink]

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01 Aug 2012, 00:38

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.

Re: Square ABCD has an area of 9 square inches. Sides AD and BC [#permalink]

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07 Jan 2016, 21:04

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Re: Square ABCD has an area of 9 square inches. Sides AD and BC [#permalink]

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30 Jul 2016, 11:34

venmic wrote:

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

(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

(1)-let side will be stretched x inches then side BC=3+x so by pythagoras theorem (3+x)^2+3^2=5^2 thus we can find x suff... (2) let BC be extended 6 inches then we have identical three 3*3 sized rectangle But if BC extended 12 inches then also we get identical three 3*5 sized rectangle

Re: Square ABCD has an area of 9 square inches. Sides AD and BC [#permalink]

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31 Dec 2016, 03:42

If the square has an area of 9 square inches, it must have sides of 3 inches each. Therefore, sides AD and BC have lengths of 3 inches each. These sides are lengthened to x inches, while the other two remain at 3 inches. This gives us a rectangle with two opposite sides of length x and two opposite sides of length 3. Then we are asked by how much the two lengthened sides were extended. In other words, what is the value of x – 3? In order to answer this, we need to find the value of x itself.

(1) SUFFICIENT: If the resulting rectangle has a diagonal of 5 inches, we end up with the following: We can now see that we have a 3-4-5 right triangle, since we have a leg of 3 and a hypotenuse (the diagonal) of 5. The missing leg (in this case, x) must equal 4. Therefore, the two sides were each extended by 4 – 3 = 1 inch.

(2) INSUFFICIENT: It will be possible, no matter what the value of x, to divide the resulting rectangle into three smaller rectangles of equal size. For example, if x = 4, then the area of the rectangle is 12 and we can have three rectangles with an area of 4 each. If x = 5, then the area of the rectangle is 15 and we can have three rectangles with an area of 5 each. So it is not possible to know the value of x from this statement. The correct answer is A.

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