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

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Joined: 02 Sep 2009
Posts: 51188
In the figure above, each of the four squares has sides of length x.  [#permalink]

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11 Nov 2017, 05:54
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Difficulty:

15% (low)

Question Stats:

90% (01:23) correct 10% (02:02) wrong based on 52 sessions

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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

Attachment:

2017-11-11_1744_002.png [ 3.82 KiB | Viewed 1056 times ]

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

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Updated on: 11 Nov 2017, 18:27
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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Originally posted by Sasindran on 11 Nov 2017, 07:04.
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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11 Nov 2017, 16:37
Bunuel wrote:

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

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

Sasindran , maybe I don't understand your math language, but think you mean: x*sqrt(2), not, as is written, 2*sqrt(x)
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Joined: 17 Oct 2016
Posts: 318
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Re: In the figure above, each of the four squares has sides of length x.  [#permalink]

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11 Nov 2017, 18:25
genxer123 wrote:
Bunuel wrote:

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

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

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

Posted from my mobile device
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Re: In the figure above, each of the four squares has sides of length x. &nbs [#permalink] 11 Nov 2017, 18:25
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