Area of a Polygon

Date: 06/27/97 at 08:36:03
From: Rainer WILHELM
Subject: Area of polygon (3-6 corners)

How do you calculate the area of a polygon (minimum 3, maximum 6 
vertices) in the plane? The polygon is always convex.

I've found a formula, but this formula requires that one know the 
order of the corner points. I have to calculate the area in a computer 
program which only knows the coordinates of the corners, but not their 
order.

Thanks in advance, 
Rainer

Date: 06/27/97 at 11:42:55
From: Doctor Rob
Subject: Re: Area of polygon (3-6 corners)

The program can determine the order. First pick the point with the
smallest y-coordinate. If there is a tie between two, pick the one
with the larger x-coordinate. Call it P(1). Let the last slope be 0.

For each i from 1 to n-1, compute the slopes of the lines from P(i) to
each of the other vertices not already labeled. If there are any with
slope greater than the last slope, pick the point with the smallest 
slope line to be P(i+1). If there are none with greater slope, pick 
the point with the most negative slope to be P(i+1).

The slope from P(i) to P(j) is [y(j)-y(i)]/[x(j)-x(i)].  

(If x(j) = x(i), say the slope is either plus or minus infinity).

When you are done, the vertices will be P(1), P(2), ..., P(n) in 
order. This works because the polygon is convex.

Once you have this information, you can add up the areas of the 
triangles P(1)P(i)P(i+1) for i = 2, ..., n-1.  Compute these by using 
the formula:

                               | 1   x(1)    y(1)  |
A(i) = (1/2)*Absolute value of | 1   x(i)    y(i)  |
                               | 1  x(i+1)  y(i+1) |

This also works because the triangle is convex.

If this doesn't suit your needs, write again, and we'll try again to
solve your problem.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/