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
![l = \sqrt{(a-c)^2 + (b-d)^2}](https://tex.z-dn.net/?f=%20l%20%3D%20%5Csqrt%7B%28a-c%29%5E2%20%2B%20%28b-d%29%5E2%7D)
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)
![EF = \sqrt{ (2-2)^2 + (9-2)^2} = 7 \quad](https://tex.z-dn.net/?f=EF%20%3D%20%5Csqrt%7B%20%282-2%29%5E2%20%2B%20%289-2%29%5E2%7D%20%3D%20%207%20%5Cquad)
![EG = \sqrt{(2-10)^2+(9-2)^2}=\sqrt{8^2+7^2}=\sqrt{113}\approx 10.6](https://tex.z-dn.net/?f=EG%20%3D%20%5Csqrt%7B%282-10%29%5E2%2B%289-2%29%5E2%7D%3D%5Csqrt%7B8%5E2%2B7%5E2%7D%3D%5Csqrt%7B113%7D%5Capprox%2010.6)
![FG=\sqrt{(2-10)^2+(2-2)^2} = 8](https://tex.z-dn.net/?f=FG%3D%5Csqrt%7B%282-10%29%5E2%2B%282-2%29%5E2%7D%20%3D%208)
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.
![EF = |9-2|=7](https://tex.z-dn.net/?f=EF%20%3D%20%7C9-2%7C%3D7)
![DF = |2-10|=8](https://tex.z-dn.net/?f=DF%20%3D%20%7C2-10%7C%3D8)
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)
![MN = 9-1 = 8](https://tex.z-dn.net/?f=MN%20%3D%209-1%20%3D%208%20)
![NO = 8 -2 = 6](https://tex.z-dn.net/?f=NO%20%3D%208%20-2%20%3D%206%20)
![OP=9-1 = 8](https://tex.z-dn.net/?f=OP%3D9-1%20%3D%208)
![PM=|2-8|=6](https://tex.z-dn.net/?f=PM%3D%7C2-8%7C%3D6)
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