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

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Math Expert
Joined: 02 Sep 2009
Posts: 51229
In the figure above, each of the four squares has sides of length x.  [#permalink]

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05 Jul 2018, 04:11
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Difficulty:

25% (medium)

Question Stats:

95% (01:23) correct 5% (02:46) wrong based on 26 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) $$2x \sqrt{2}$$

(B) $$\frac{x\sqrt2}{2}$$ $$+ x$$

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

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

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

Attachment:

square (1).jpg [ 14.04 KiB | Viewed 373 times ]

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

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05 Jul 2018, 04:44
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) $$2x \sqrt{2}$$

(B) $$\frac{x\sqrt2}{2}$$ $$+ x$$

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

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

(E) $$2x + x \sqrt{2}$$
Attachment:
square (1).jpg

We have, PQ=$$\frac{x}{2}+\frac{x}{2}=x$$
and QR=$$\frac{x}{2}+\frac{x}{2}=x$$
in the right angled triangle $$RP^2=PQ^2+QR^2=x^2+x^2=2x^2$$
So, $$RP=x\sqrt{2}$$

Perimeter of $$∆PQR=PQ+QR+RP=x+x+x\sqrt{2}$$=$$2x + x \sqrt{2}$$

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

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05 Jul 2018, 05:09

Since this triangle is formed by joining the centers of three triangles, the sides
PQ and QR will super-impose into any of the squares. The triangle ∆PQR is half
the square, where the third side(PR) is the diagonal of the square, which is $$x\sqrt{2}$$

Therefore, the perimeter of the triangle(∆PQR) is $$x + x + x\sqrt{2} = 2x + x\sqrt{2}$$ (Option E)

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In the figure above, each of the four squares has sides of length x. &nbs [#permalink] 05 Jul 2018, 05:09
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