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## Topic review (newest first)

George,Y
2007-03-05 22:51:14

Actually my solution is a formula already, which I came cross on some site. Using determinants to get the area of a polygon.

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.

MathsIsFun
2007-03-05 21:25:16

I think that is right, George.

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).

George,Y
2007-03-05 19:54:37

Quickest and Absolute Accurate

George,Y
2007-03-05 19:47:28

Suppose you have n points on your polygon. And they are A1,A2,...An counterclockwise sequenced, also you know their positions in Cartesian co-ords.

Let's define A1= (x1,y1),... An= (xn,yn)
And Det(12)=
|x1  y1|
|x2  y2|
=x1y2 - x2y1

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

MathsIsFun
2007-03-05 12:05:01

I have written this up here: Area of Irregular Polygons

MathsIsFun
2006-01-20 09:05:22

I did this once.

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.)

Ricky
2006-01-20 01:21:19

You have to decompose the figure.  For example, if it's a triangle on top of a square, first find the area of the square, then the area of the triangle, and add them together.

kempos
2006-01-20 01:18:53

I would divide your irregular shape into basic shapes, such as rectangles, trapazium, triangles, semicircles etc.

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

johnocarolan
2006-01-20 00:24:31

Hi,

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.