Question

Can someone help me the answer choices are still the same for the rest of the questions

Answer 1

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

$l = \sqrt{(a-c)^2 + (b-d)^2}$

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)

$EF = \sqrt{ (2-2)^2 + (9-2)^2} = 7 \quad$

$EG = \sqrt{(2-10)^2+(9-2)^2}=\sqrt{8^2+7^2}=\sqrt{113}\approx 10.6$

$FG=\sqrt{(2-10)^2+(2-2)^2} = 8$

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.

$EF = |9-2|=7$

$DF = |2-10|=8$

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)

$MN = 9-1 = 8$

$NO = 8 -2 = 6$

$OP=9-1 = 8$

$PM=|2-8|=6$

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