D(-3,5)
what is the perimeter of square abcd?
a(3,4)
+
37 units
cb
4/37 units
28 units
37 units
.
-54-3 -2 -
2
3
4
5
c(-4-1)
b(2,-2)

is the perimeter of square ABCD.

Step-by-step explanation:

Coordinates of square ABCD:

A = (3,4), B = (2,-2), C = (-4-1) , D = (-3,5)

Distance formula:

Distance of AB: A = (3,4), B = (2,-2)

Given that the ABCD is square, then:

AB = BC = CD = DA

Perimeter of the square ABC = AB +BC + CD + DA

is the perimeter of square ABCD.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!