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

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Math Expert
Joined: 02 Sep 2009
Posts: 58396
In the figure above, if the area of the larger square region is twice  [#permalink]

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28 Feb 2019, 00:24
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Difficulty:

45% (medium)

Question Stats:

77% (02:05) correct 23% (01:30) wrong based on 25 sessions

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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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2019-02-28_1121.png [ 2.83 KiB | Viewed 366 times ]

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Joined: 17 May 2018
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Location: India
Re: In the figure above, if the area of the larger square region is twice  [#permalink]

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28 Feb 2019, 01:22
1
Hi Bunuel the figure is missing. Could you please check it. Thank you

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Math Expert
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Posts: 58396
Re: In the figure above, if the area of the larger square region is twice  [#permalink]

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28 Feb 2019, 01:26
1
sumi747 wrote:
Hi Bunuel the figure is missing. Could you please check it. Thank you

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Fixed that. 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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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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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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