Area of Irregular Polygons


I just thought I would share with you a clever technique I once used to find the area of general polygons.

The polygon could be regular (all angles are equal and all sides are equal) or irregular

regular irregular
Regular Irregular

The Example Polygon

Let's use this polygon as an example:


The first step is to turn each vertex (corner) into a coordinate, like on a graph:

Area Under One Line Segment

Now, for each line segment, work out the area down to the x-axis.

So, how do we calculate each area?

Average the two heights, then multiply by the width

Example: For the shape highlighted above, we take the two heights (the "y" coordinates 2.28 and 4.71) and work out the average height:

(2.28+4.71)/2 = 3.495

Work out the width (the difference between the "x" coordinates 2.66 and 0.72)

2.66-0.72 = 1.94

The area under that line segment is width×height

1.94 × 3.495 = 6.7803

Add Them All Up

Now add them all up!

But the trick is to add when they go forwards (positive width), and subtract when they go backwards (negative width).

If you always go clockwise around the polygon, and always subtract the first "x" coordinate from the second, it works out naturally, like this:

From To      
x y x y Avg Height Width (+/-) Area (+/-)
0.72 2.28 2.66 4.71 3.495 1.94 6.7803
2.66 4.71 5 3.5 4.105 2.34 9.6057
5 3.5 3.63 2.52 3.01 -1.37 -4.1237
3.63 2.52 4 1.6 2.06 0.37 0.7622
4 1.6 1.9 1 1.3 -2.1 -2.7300
1.9 1 0.72 2.28 1.64 -1.18 -1.9352
          Total: 8.3593

And It looks like this:

So that's it! The area is 8.3593

Area of Polygon Tool

Glad you read this far! You are rewarded with a link to the Area of a Polygon Drawing Tool that can do all of this for you. It also accepts manual entry of coordinates.