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In the figure above every square is contained (inscribed)

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SVP
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In the figure above every square is contained (inscribed) [#permalink]

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New post 08 Jan 2015, 00:12
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In the figure above every square is contained (inscribed) within a square as shown. What is the length of the side of the smallest square, if the largest square has an area of 16?

A. 4
B. \(2\sqrt{2}\)
C. 2
D. \(\sqrt{2}\)
E. 1
[Reveal] Spoiler: OA

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Re: In the figure above every square is contained (inscribed) [#permalink]

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New post 08 Jan 2015, 01:57
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Answer = C = 2

Area of largest square = 16

So, area of the square inscribed \(= \frac{16}{2} = 8\) (This is by property of the inscribed square. Can be calculated theoretically)

So, area of the smallest square \(= \frac{8}{2} = 4\)

Side of the smallest square \(= \sqrt{4} = 2\)
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Re: In the figure above every square is contained (inscribed) [#permalink]

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New post 08 Jan 2015, 13:26
Hi All,

This is a great "concept" question and can be solved without a lot of complex math. You do need to understand how the squares relate to one another though....

**Note: For organizational purposes, I'm going to refer to the 3 squares as the 'large', 'medium' and 'small' squares.**

To start, we know that the area of the large square is 16, so each of its sides = 4.

If you take the 4 corners of the large square and "fold them in" so that they meet at the center, you will end up with a square with the same size as the medium square. This means that the medium square has the same area as the sum of the 4 white-corner triangles:

medium square = 4 white triangles

medium square + 4 white triangles = large square = area of 16
medium square = 4 white triangles = area of 8

This medium square has an area of 8, so its sides are each 2(root2).

Now we can do the same "math moves" (as above) with the medium square and the small square...

Fold the 4 black triangles of the medium square towards the center and you get a square with the same size as the small square.

small square = 4 black triangles

small square + 4 black triangles = medium triangle = area of 8
small square = 4 black triangles = area of 4

So, the small square has an area of 4, so its sides are each 2.

Final Answer:
[Reveal] Spoiler:
C


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Re: In the figure above every square is contained (inscribed) [#permalink]

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New post 11 Nov 2015, 03:36
Smallest square's side is about half of the largest one, so 4/2=2

Paresh, thanks for concept

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Re: In the figure above every square is contained (inscribed) [#permalink]

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New post 11 Mar 2017, 05:03
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the figure above every square is contained (inscribed)   [#permalink] 11 Mar 2017, 05:03
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