Solution
For this case we can find the distance
AB= 4
ED= 4
We can use the distance point given by:
[tex]d=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
Now we need to find the distance AE, we have y2= 2, y1= -3 , x2= -3, x1 = -4 and we got:
[tex]AE=\sqrt[]{(2+3)^2+(-3+4)^2}=\sqrt[]{26}[/tex]
Similarly we can find BC and DC.
For BC we have we have y2= 2, y1= -1 , x2= 1, x1 = -1
And for DC we have y2= -1, y1= -3 , x2= 3, x1 = 0
[tex]BC=\sqrt[]{(2+1)^2+(1+1)^2}=\sqrt[]{13}[/tex][tex]DC=\sqrt[]{(-1+3)^2+(3-0)^2}=\sqrt[]{13}[/tex]
Then the perimeter is given by:
[tex]4+4+\sqrt[]{26}+\sqrt[]{13}+\sqrt[]{13}=20.31[/tex]
