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

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In the figure above, if the area of the larger square region is twice  [#permalink]

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New post 28 Feb 2019, 00:24
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A
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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


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Re: In the figure above, if the area of the larger square region is twice  [#permalink]

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New post 28 Feb 2019, 01:22
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Hi Bunuel the figure is missing. Could you please check it. Thank you

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Re: In the figure above, if the area of the larger square region is twice  [#permalink]

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New post 28 Feb 2019, 01:26
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Re: In the figure above, if the area of the larger square region is twice  [#permalink]

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New post 28 Feb 2019, 01:43
Steps:
1.Side of the smaller square= sqrt(1/2)
2.Area of the larger square = 2*1/2=1
3.Side of the larger square = 1
4.Difference between the sides = 1-sqrt(1/2) = {sqrt(2)-1}/sqrt(2)
5.Multiple by sqrt(2)
6. Option A
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Re: In the figure above, if the area of the larger square region is twice  [#permalink]

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New post 28 Feb 2019, 01:46
1
By property of 45-90-45 triangle we get the sides of small triangle= 1/√2
Hence area of small sqr=1/2
Ar of large sqr= 2*1/2=1
Side of large sqr=1
Therefore difference in length of sides for large and small sqr= 1-1/√2 = (√2-1)/√2= √2*(√2-1)/√2*√2= (2-√2)/2
IMO A

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Re: In the figure above, if the area of the larger square region is twice   [#permalink] 28 Feb 2019, 01:46
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In the figure above, if the area of the larger square region is twice

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