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# In the figure above, each of the four squares

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Joined: 22 Aug 2016
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In the figure above, each of the four squares [#permalink]  06 Jan 2017, 07:52
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Question Stats:

50% (01:50) correct 50% (00:00) wrong based on 8 sessions
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In the figure above, each of the four squares has sides of length x. If $$\triangle$$$$PQR$$ is formed by joining the centers of three of the squares, what is the perimeter of $$\triangle$$$$PQR$$ in terms of x ?

A. $$2x \sqrt{2}$$

B. $$\frac {x{sqrt2}}{2}+ x$$

C. $$2x + \sqrt{2}$$

D. $$x \sqrt{2} + 2$$

E. $$2x + x \sqrt{2}$$
[Reveal] Spoiler: OA
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Joined: 07 Jun 2014
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GRE 1: Q167 V156
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Kudos [?]: 1897 [1] , given: 397

Re: In the figure above, each of the four squares [#permalink]  06 Jan 2017, 08:18
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Expert's post
Explanation.

P is the center of square and so is Q. From P to edge the square of is $$\frac{x}{2}$$. And from edge of square to Q is $$\frac{x}{2}$$ as well.

So PQ = x
Similarly QR is also equal to x.

Again P to edge of the square is half of the length of diagonal $$\frac{1}{2}\sqrt{2}x$$. Again from corner to R is also $$\frac{1}{2}\sqrt{2}x$$.

PR =$$2 \times \frac{1}{2}\sqrt{2}x$$= $$\sqrt{2}x$$.

Hence perimeter = PQ + QR + PR = $$x + x + \sqrt{2}x$$ = $$2x + \sqrt{2}x$$

Hence option E is correct.
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Sandy
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Re: In the figure above, each of the four squares   [#permalink] 06 Jan 2017, 08:18
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