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Bunuel
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Looks scary if you're inclined to hate geometry like myself, but quite simple when you remember rules and take good notes.

\(x^2+x^2=PR^2\), so \(2^2+2^2=8\)

From there, we know that \(PR=\sqrt{8}\). That can be simplified to \(2\sqrt{2}\).

Since all angles of triangle PRS = 60, we know that we have an equilateral triangle and all sides are a the same. We're looking for the perimeter of the whole quadrilateral, so side RS and SP will contribute to it. Those are both equal to PR, so we'll take 2 of those, or \(2*2\sqrt{2}\), or more simply, \(4\sqrt{2}\).

We're given that \(x=2\), and 2 of those contribute to the perimeter, so the final perimeter will be \(4+4\sqrt{2}\).

Answer choice B.
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Bunuel

If QR = 2, then what is the perimeter of quadrilateral PQRS in the figure above?

A. \(6\sqrt{2} + 4\)

B. \(4\sqrt{2} + 4\)

C. \(4\sqrt{2} + 2\)

D. \(2\sqrt{2} + 2\)

E. \(3\sqrt{2}\)

Attachment:
2018-02-21_1019_003.png
Answer must be B as x=2 and triangle QRS is isosceles then 3rd side is 2root 2 so perimeter is sum of all the side 2 sides with 2root 2 and 2 sides with 2 length so total is 4root2+ 4
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Bunuel

If QR = 2, then what is the perimeter of quadrilateral PQRS in the figure above?

A. \(6\sqrt{2} + 4\)

B. \(4\sqrt{2} + 4\)

C. \(4\sqrt{2} + 2\)

D. \(2\sqrt{2} + 2\)

E. \(3\sqrt{2}\)

Attachment:
2018-02-21_1019_003.png

Since triangle PQR is an isosceles right triangle, side PR = 2√2. Since triangle PRS is an equilateral triangle, PR = RS = PS = 2√2.

So the perimeter of quadrilateral PQRS is 2 + 2 + 2√2 + 2√2 = 4 + 4√2.

Answer: B
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