The basic part is that the area of triangle ABO equates one half of the determinant of coords of AB.

But the principle of this method is the same as that of yours.

]]>By rearranging how my procedure works: (x2-x1)(y2+y1)/2 + (x3-x2)(y3+y2)/2 + ... + (x1-xn)(y1+yn)/2

You do come up with 1/2 [ Det(12) + Det(23) + ... Det(n1)]

My first term expands to x2y2 + x2y1 - x1y2 - x1y1. The x2y2 is cancelled by the next term's -x2y2, so the only functional part is x2y1-x1y2 which is the determinant (well, the negative of it, but that can be handled when the final answer is achieved).

]]>Let's define A[sub]1[/sub]= (x[sub]1[/sub],y[sub]1[/sub]),... A[sub]n[/sub]= (x[sub]n[/sub],y[sub]n[/sub])

And Det(12)=

|x[sub]1[/sub] y[sub]1[/sub]|

|x[sub]2[/sub] y[sub]2[/sub]|

=x[sub]1[/sub]y[sub]2[/sub] - x[sub]2[/sub]y[sub]1[/sub]

The Area of the Polygon=1/2 [ Det(12)+Det(23)+Det(34)+...+Det(n-1 n)+Det(n1) ]

ie (1,0) (0,3) (-1/3,0) (0,-2)

Det 3 1 2/3 2

Area=(3+1+2/3+2)/2

Comments Invited!

]]>Take the area from each line segment down to the x-axis, then add them up! Go clockwise, and if the line goes forwards the area is positive, if backwards negative. (Or the other way around, it doesn't matter, because if the result is a negative area just make it positive.)

]]>Forgot to add that you find their areas and add them together :-(

]]>Can anybody help me out with this -

I want to be able to calculate the are of irregular polygons - using the easiest and quickest but most accurate way - anybody any ideas?

examples are something like 5 or 6 sides (maybe more up to maximum of 10 sides), some other areas might have circular boundaries instead of straight lines.

Thanks in advance.

J

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