1+1= 12+12=
11+11=
15+15=
23+23=
3+3+3
34+34
12+14
50+50
123+123
21+21
90+90
60+60
34+34
55+55
1+0
2+0
32+32
545+545
54+54
89+89

1+1= 2

12+12= 24

11+11= 22

15+15= 30

23+23= 46

3+3+3=9

34+34 =68

12+14 =26

50+50 =100

123+123 =246

21+21 =42

90+90 =180

60+60 =120

34+34 =68

55+55 =110

1+0 =1

2+0 =2

32+32 =64

545+545 =1090

54+54 =108

89+89=178

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!