Exact perimeters given coordinates are ugly, the sum of a bunch of square roots. A calculator approximation isn't so bad, especially when the choices are so far apart; we can just keep one decimal place.
There's only one rule in play, that the length of a segment with endpoints (a,b) and (c,d) is given by the Pythagorean Theorem as
[tex] l = \sqrt{(a-c)^2 + (b-d)^2}[/tex]
It's going to be too boring for me to do more than one or two, so hopefully you'll learn how and do the rest yourself.
E(2,9),F(2,2),G(10,2)
[tex]EF = \sqrt{ (2-2)^2 + (9-2)^2} = 7 \quad[/tex]
[tex]EG = \sqrt{(2-10)^2+(9-2)^2}=\sqrt{8^2+7^2}=\sqrt{113}\approx 10.6[/tex]
[tex]FG=\sqrt{(2-10)^2+(2-2)^2} = 8[/tex]
Of course there's a shortcut when one of the coordinates between the endpoints is the same; then we can just skip the square root.
[tex]EF = |9-2|=7[/tex]
[tex]DF = |2-10|=8[/tex]
So an approximate perimeter of 7+8+10.6=25.6
First choice
One more, a quadrilateral
M(1,8), N(9,8), O(9,2), P(1,2)
[tex]MN = 9-1 = 8 [/tex]
[tex]NO = 8 -2 = 6 [/tex]
[tex]OP=9-1 = 8[/tex]
[tex]PM=|2-8|=6[/tex]
No square roots needed for that one, which is apparently a rectangle, perimeter 8+6+8+6=28
Second choice
I leave the rest for you
