Area has only two dimensions, whereas you have multiplied four dimensions! You say the area is irregular but you haven't defined the shape, so I'm going to assume you have a 4-sided area (quadrilateral) and you have provided the perimeter measurements. If we split the irregular quadrilateral into two triangles by joining the opposite corners, we may be able to use Heron's formula to find the area of the quadrilateral by adding the areas of the two triangles together. First we need to find the semiperimeter of the two triangles. We'll call the diagonal length x. Triangle 1 has a semiperimeter, S1, of 1/2(149.3+60.8+x) and triangle 2 a semiperimeter, S2, of 1/2(140+60+x). The area of triangle 1, A1=sqrt(S1(S1-149.3)(S1-60.8)(S1-x)) and A2=sqrt(S2(S2-140)(S2-60)(S2-x)). The area of the quadrilateral is A1+A2.
S1=1/2(210.1+x)=105.05+x/2 and S2=1/2(200+x)=100+x/2.
A1=sqrt((105.05+x/2)(x/2-44.25)(x/2+44.25)(105.05-x/2))
A2=sqrt((100+x/2)(x/2-40)(x/2+40)(100-x/2))
We now have to make another assumption, because we don't know what x is and the areas depend on it. The area is almost a rectangle or parallelogram, because 210 and 200 are similar measures, as are 60 and 60.8. Let's assume that triangle 2 is a right-angled triangle, then x=sqrt(140^2+60^2) (Pythagoras), so x=152.315 and x/2=76.158 (approx values). We can now calculate A1 and A2.
A1=4484.92 and A2=4200, therefore the total area is 8684.92 sq ft. Note that A2 is half the area of a rectangle with sides 140 and 60=1/2(60*140)=4200.
If we assume that triangle 1 is right-angled instead, what difference does it make? Its hypotenuse, x=sqrt(149.3^2+60.8^2)=161.21 and A1=4538.72 sq ft. If we calculate A2 using Heron's formula we get A2=4141.80, so total area is 8680.52 sq ft. So the difference between the two areas is less than 4.5 sq ft.
We could average the two estimates of area: 8682.7 sq ft approx.