The first thing we must do for this case is to find the distance between points.
For this, we use the formula of distance between points:
[tex] d = \sqrt{(x2-x1)^2 + (y2-y1)^2} [/tex]
We have then:
For WX:
[tex] WX = \sqrt{(4-0)^2 + (0+1)^2}
WX = \sqrt{(4)^2 + (1)^2}
WX = \sqrt{16 + 1}
WX = \sqrt{17} [/tex]
For XY:
[tex] XY = \sqrt{(4-3)^2 + (0+2)^2}
XY = \sqrt{(1)^2 + (2)^2}
XY = \sqrt{1 + 4}
XY = \sqrt{5} [/tex]
Then, since the parallelogram has sides that are parallel, then its distances measure the same.
Thus, the perimeter of the parallelogram is:
[tex] P = 2WX + 2XY [/tex]
Substituting values:
[tex] P = 2\sqrt{17} + 2\sqrt{5} [/tex]
Answer:
the perimeter of parallelogram WXYZ is:
[tex] P = 2\sqrt{17} + 2\sqrt{5} [/tex]
option 2
