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In the figure above, PQRS is a parallelogram, and P and Q are the midp

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In the figure above, PQRS is a parallelogram, and P and Q are the midp [#permalink]

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New post 26 Sep 2017, 23:21
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In the figure above, PQRS is a parallelogram, and P and Q are the midpoints of adjacent sides of the square as shown. If the area of the shaded region is 2, then the perimeter of PQRS is

(A) 4 + 4√2
(B) 2 + 4√2
(C) 4 + 2√2
(D) 4
(E) 16


[Reveal] Spoiler:
Attachment:
2017-09-27_1015.png
2017-09-27_1015.png [ 5.06 KiB | Viewed 475 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, PQRS is a parallelogram, and P and Q are the midp [#permalink]

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New post 27 Sep 2017, 01:39
Bunuel wrote:
Image
In the figure above, PQRS is a parallelogram, and P and Q are the midpoints of adjacent sides of the square as shown. If the area of the shaded region is 2, then the perimeter of PQRS is

(A) 4 + 4√2
(B) 2 + 4√2
(C) 4 + 2√2
(D) 4
(E) 16


[Reveal] Spoiler:
Attachment:
2017-09-27_1015.png


Let the side of the square be \(2x\),
Area of the shaded region will be \(= \frac{1}{2}*x*x=2\)
or \(x=2\)

Hence \(PQ=SR=\sqrt{2^2+2^2}\) \(= 2\sqrt{2}\)

\(PS=QR=x=2\)

so perimeter of PQRS \(= 2*2\sqrt{2}+2*2 = 4+4\sqrt{2}\)

Option A

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Re: In the figure above, PQRS is a parallelogram, and P and Q are the midp [#permalink]

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New post 01 Oct 2017, 12:14
Bunuel wrote:
Image
In the figure above, PQRS is a parallelogram, and P and Q are the midpoints of adjacent sides of the square as shown. If the area of the shaded region is 2, then the perimeter of PQRS is

(A) 4 + 4√2
(B) 2 + 4√2
(C) 4 + 2√2
(D) 4
(E) 16


[Reveal] Spoiler:
Attachment:
2017-09-27_1015.png


We see that the shaded region is an isosceles right triangle. Since the area of the shaded region is 2, and if we let s = a leg of the isosceles triangle, we have:

s^2/2 = 2

s^2 = 4

s = 2

We see that a leg of the shaded region (the isosceles right triangle) is 2; thus, the hypotenuse of the shaded region, PQ, is 2√2. Since QR and PS are each equal to the length of a leg of the shaded region, QR = PS = 2. Since RS = QP, RS = 2√2.

So, the perimeter of parallelogram PQRS is 2(2) + 2(2√2) = 4 + 4√2.

Answer: A
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Kudos [?]: 906 [0], given: 5

Re: In the figure above, PQRS is a parallelogram, and P and Q are the midp   [#permalink] 01 Oct 2017, 12:14
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In the figure above, PQRS is a parallelogram, and P and Q are the midp

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