Question 1012063
In the figure, ABCD is NOT a square and it is NOT a rectangle. It's simply a quadrilateral. If you look at the information given that AC = 12 and BD = 9, this also tells us we don't have a rectangle (if we don't have a rectangle, it's certainly not a square). The diagonals have to be equal in order to have a rectangle.



However, PQRS is a parallelogram. This is true no matter where you place A,B,C or D. The proof of that is <a href = "http://mathforlove.com/2012/02/midpoints-of-a-quadrilateral-form-a-parallelogram/">here</a>. We'll use the fact that the opposite sides of any parallelogram are congruent. 


Connect A to C. This segment is 12 units long. Once you have segment AC drawn, you have triangle ABC. The segment PQ is parallel to AC and it's half as long. This is <a href = "http://www.regentsprep.org/regents/math/geometry/gp10/midlinel.htm">midsegment property of triangles</a>.



So,


PQ = (1/2)*(AC)
PQ = (1/2)*12
PQ = 6


So PQ is 6 units long. So is SR (opposite side is congruent). 



Using the same idea for BD and PS, we get


PS = (1/2)*(BD)
PS = (1/2)*9
PS = 4.5


The opposite sides of the parallelogram are congruent, so PS = RQ


Now we know all 4 sides of the parallelogram PQRS and they are


PQ = 6
SR = 6
PS = 4.5
RQ = 4.5


Add up the four sides: 

PQ+SR+PS+RQ = 6+6+4.5+4.5 = 21


This is no coincidence that the perimeter of PQRS is equal to AC + BD. So in the future, you can use this shortcut to make things faster.