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In the figure above, each of the four squares has sides of length x.

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In the figure above, each of the four squares has sides of length x. [#permalink]

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New post 11 Nov 2017, 05:54
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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√2
(B) (x√2/2) + x
(C) 2x + √2
(D) x√2 + 2
(E) 2x + x√2

[Reveal] Spoiler:
Attachment:
2017-11-11_1744_002.png
2017-11-11_1744_002.png [ 3.82 KiB | Viewed 380 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 135518 [0], given: 12697

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In the figure above, each of the four squares has sides of length x. [#permalink]

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New post 11 Nov 2017, 07:04
E.

from the fig PQ=x and QR=x. hence PR=X*sqrt(2)

so perimeter = x+x+2*Sqrt(x) = 2x + x*sqrt(2)
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Last edited by Sasindran on 11 Nov 2017, 18:27, edited 2 times in total.

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In the figure above, each of the four squares has sides of length x. [#permalink]

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New post 11 Nov 2017, 16:37
Bunuel wrote:
Image
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√2
(B) (x√2/2) + x
(C) 2x + √2
(D) x√2 + 2
(E) 2x + x√2

[Reveal] Spoiler:
Attachment:
2017-11-11_1744_002.png

If centers of squares with side length x are joined, then

Horizontally, Side PQ =
\(\frac{1}{2}x + \frac{1}{2}x = x\)

Vertically, side QR =
\(\frac{1}{2}x + \frac{1}{2}x = x\)

We have an isosceles right triangle, angle measures are 45-45-90, sides lengths are in ratio \(x : x : x\sqrt{2}\)

Side PR corresponds with
\(x\sqrt{2}\)
Equal sides' length each =\(x\)
PR=\(x\sqrt{2}\)

Perimeter = \((x + x + x\sqrt{2}) =(2x + x\sqrt{2})\)

Answer E

Sasindran , maybe I don't understand your math language, but think you mean: x*sqrt(2), not, as is written, 2*sqrt(x)

Kudos [?]: 401 [0], given: 644

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Re: In the figure above, each of the four squares has sides of length x. [#permalink]

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New post 11 Nov 2017, 18:25
genxer123 wrote:
Bunuel wrote:
Image
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√2
(B) (x√2/2) + x
(C) 2x + √2
(D) x√2 + 2
(E) 2x + x√2

[Reveal] Spoiler:
Attachment:
2017-11-11_1744_002.png

If centers of squares with side length x are joined, then

Horizontally, Side PQ =
\(\frac{1}{2}x + \frac{1}{2}x = x\)

Vertically, side QR =
\(\frac{1}{2}x + \frac{1}{2}x = x\)

We have an isosceles right triangle, angle measures are 45-45-90, sides lengths are in ratio \(x : x : x\sqrt{2}\)

Side PR corresponds with
\(x\sqrt{2}\)
Equal sides' length each =\(x\)
PR=\(x\sqrt{2}\)

Perimeter = \((x + x + x\sqrt{2}) =(2x + x\sqrt{2})\)

Answer E

Sasindran , maybe I don't understand your math language, but think you mean: x*sqrt(2), not, as is written, 2*sqrt(x)

Yeah. Sorry. My bad. Edited the post. Thank you

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Re: In the figure above, each of the four squares has sides of length x.   [#permalink] 11 Nov 2017, 18:25
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In the figure above, each of the four squares has sides of length x.

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