The perimeter of the parallelogram is [tex]\sqrt[2]{29}+ \sqrt[6]{6}[/tex]
Calculation of the perimeter of the parallelogram:
Since the coordinates are (2,10), (4,2), (7,8) and (-1,4).
Now for determining the perimeter first we have to calculate the distance of AB and BC
So,
For AB, the distance is
[tex]= \sqrt{(2-7)^2 + (10-8)^2}\\\\ = \sqrt{29}[/tex]
Now
For BC, the distance is
[tex]= \sqrt{4-7)^2 + (2-8)^2}\\\\ = \sqrt{54}\\\\= 3\sqrt{6}[/tex]
Now the perimeter is
= 2(AB + BC)
= [tex]2(\sqrt29 + 3\sqrt 6)[/tex]
= [tex]\sqrt[2]{29}+ \sqrt[6]{6}[/tex]
Therefore, The perimeter of the parallelogram is [tex]\sqrt[2]{29}+ \sqrt[6]{6}[/tex]
Learn more about perimeter here: https://brainly.com/question/17615672
