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:
\(d = \sqrt{(x2-x1)^2 + (y2-y1)^2}\)
We have then:
For WX:
\(WX = \sqrt{(4-0)^2 + (0+1)^2}
WX = \sqrt{(4)^2 + (1)^2}
WX = \sqrt{16 + 1}
WX = \sqrt{17}\)
For XY:
\(XY = \sqrt{(4-3)^2 + (0+2)^2}
XY = \sqrt{(1)^2 + (2)^2}
XY = \sqrt{1 + 4}
XY = \sqrt{5}\)
Then, since the parallelogram has sides that are parallel, then its distances measure the same.
Thus, the perimeter of the parallelogram is:
\(P = 2WX + 2XY\)
Substituting values:
\(P = 2\sqrt{17} + 2\sqrt{5}\)
Answer:
the perimeter of parallelogram WXYZ is:
\(P = 2\sqrt{17} + 2\sqrt{5}\)
option 2
