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In the figure above [#permalink]
Expert's post 00:00

Question Stats: 70% (00:32) correct 30% (01:43) wrong based on 10 sessions
Attachment: #GREpracticequestion In the figure above, each of the four squares.jpg [ 14.04 KiB | Viewed 105 times ]

In the figure above, each of the four squares has sides of length x. If ∆PQR is formed by joining the centers of three of the squares, what is the perimeter of ∆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

_________________ Director  Joined: 07 Jan 2018
Posts: 604
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Kudos [?]: 545  , given: 88

Re: In the figure above [#permalink]
1
KUDOS
Given,
each of the square are of side x imagine a point $$a$$ between P and Q such that the point is intersected by a side of the square and point $$b$$ between Q and R such that it is also intersected by a side of the square

Now distance between P to a = distance between a to Q. If a side of the square has length $$x$$ then distance between p to a is half of it hence $$\frac{x}{2}$$ similarly distance between a to Q is also $$\frac{x}{2}$$ therefore total distance between P to Q is $$\frac{x}{2} * 2 =x$$
Similar conclusion can be made for the distance Q and R going through point $$b$$ and hence distance between Q and R is also $$x$$
This triangle is an Isosceles right triangle therefore sides are in the ratio $$x: x: x\sqrt{2}$$
Hence, perimeter = $$x + x + x\sqrt{2}$$
option E
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes Re: In the figure above   [#permalink] 23 Mar 2018, 08:10
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