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In the figure above, if the area of the larger square region

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In the figure above, if the area of the larger square region [#permalink] New post 21 Jun 2017, 12:22
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In the figure above, if the area of the larger square region is twice the area of the smaller square region, and if a diagonal of the smaller square has a length of 1 foot, then a side of the larger square is how many feet longer than a side of the smaller square?

A) \(\frac{(2-\sqrt{2})}{2}\)

B) \(\sqrt{2-1}\)

C) 1

D) \(\sqrt{2}\)

E) 2
[Reveal] Spoiler: OA

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Re: In the figure above, if the area of the larger square region [#permalink] New post 19 Sep 2017, 09:48
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The smaller square has a diagonal of 1 and since we know the diagonal of a square to be equal to \(l*sqrt(2)\) we get that the side is equal to \(\frac{1}{sqrt(2)}\) and the area equal to \(l^2=\frac{1}{2}\). The area of the larger square is twice the area of the smaller one, thus it is equal to 1 and its side is thus equal to 1 too.
The difference between the two sides is \({1}-frac{1}{sqrt(2)}=frac{sqrt(2)-1}{sqrt(2)}=frac{2-sqrt(2)}{2}\). Thus, the answer is A
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Re: In the figure above, if the area of the larger square region [#permalink] New post 19 Feb 2018, 19:25
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The answer should be (2-sqrt2)/2 instead 2-sqrt2/2.
please update.
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Re: In the figure above, if the area of the larger square region [#permalink] New post 05 Mar 2018, 09:02
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excellent question, checks you ability and flexibility working with geometric concepts as well as fractions and ratios, especially the last part when you get square root 2 - 1 divided by square root 2 and you should multiply both numerator and denominator with \sqrt{2} to get a satisfactory A as an answer.
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Re: In the figure above, if the area of the larger square region   [#permalink] 05 Mar 2018, 09:02
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In the figure above, if the area of the larger square region

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