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!
