Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the coordinates of all points making up a two-dimensional shape and the order in which they are connected, is there any generic formula that will give the area of that shape?

EDIT: The list of points would be a list of all of the vertices.

share|improve this question
    
There are generally infinitely many points in any shape with nonzero area. However, if you restrict yourself to polygons and your points are the corners of the polygon (given in order), then you can use the shoelace formula. –  Henning Makholm Jan 29 '12 at 23:34
    
Given the (algorithm) tag, I have to wonder: are you talking about polygons in particular? Vector calculus gives an area formula given only a parametrization of the boundary, for example. –  anon Jan 29 '12 at 23:34
    
I doubt there is any useful closed-form expression for this. If you partition it into convex parts, and then partition those into triangles, you can then add up the areas of the triangles, and the last part can be made into a closed-form expression. –  Dan Brumleve Jan 29 '12 at 23:35
1  

1 Answer 1

up vote 3 down vote accepted

If you mean arbitrary sets of points in the plane, the answer is probably "no" unless you regard the defintion of Lebesgue measure as such a formula. However, you write about "the order in which they are connected" and that makes me suspect you may have in mind a sequence of points on the boundary with line segments between them as components of the boundary. In that case, maybe the shoelace formula will help you.

And if you mean polygons whose corners are integer points, there is Pick's theorem.

share|improve this answer
    
The Shoelace formula is what I was looking for. Thanks. –  Tyler Crompton Jan 30 '12 at 1:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.