Bunuel wrote:

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

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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