Calculating Polygon Area for Dummies

Irregular Polygon Area

If you happen to know a ten- or eleven-year-old, you can probably get the kid to help you calculate the area of a polygon. That is, as long as the polygon's a square, triangle or rectangle. Once the child hits middle-school math, you can probably get an answer for parallelograms, trapezoids, circles, and maybe even an ellipse (but don't ask about ovals...). Of course, the problems become increasingly complex as you add sides to your polygon and tossing in an "irregularity factor" can boggle the imagination of even a college math student. Speaking of which, one apparently asked eHow.com "How to Calculate the Square Foot of an Irregular Polygon" -- and that person got an answer from contributor Peter Flom. The problem with Flom's answer at Sciencing.com? It was wrong.

Peter, as eHow contributors were forced to do, began with an introduction. Sayeth Peter,

"...for irregular polygons, which are made of straight lines of any length, there are no formulas, and you need to be a little creative to find the area. Fortunately, any polygon may be divided into triangles, and there is a simple formula for the area of triangles."

Flom apparently did some quick research and came up with an "answer" from Dr. Math. The answer, as far the Antisocial Network's unofficial geometry buff can tell, is correct. However – and there is a giant "however" here – the answer is not a general solution. In fact, it only works under certain, special conditions.

Here's what Peter tells his readers to do:

Label the vertices... starting with 1 at an arbitrary vertex and continuing clockwise around the polygon.  Draw a line from vertex 1 to vertex 3. This will make one triangle, with vertices 1, 2, and 3. If there are only 4 sides, it will also make a triangle with vertices 1, 3 and 4. If the polygon has more than 4 sides, draw a line from vertex 3 to vertex 5. Continue in this way until you run out of vertices. Compute the area of each triangle.  Add up the areas, and this is the area of the polygon

His "solution" will work for a square or a rectangle. It also works for the figure shown above, which is the figure Dr. Math used in the example. But it will not work for a polygon unless it is consistently convex: even one concavity renders this method useless. That's not to mention that, if you were to add a single additional vertex to the original figure (see red arrow at right) and follow Flom's directions to the letter, you would no longer be calculating the total area of the figure (although Peter might have gotten closer if he had mentioned using the vertex number 1 as a corner of the last triangle in the set)! On top of that, this method forces users to measure the length of each constructed line and somehow measure the height of every triangle to calculate its area (although Dr. Math's example uses trigonometric functions, Peter apparently did not understand them...).

No, while you could calculate the area of an irregular polygon by constructing a large number of triangles among its vertices, it would be horribly inefficient and time-consuming. You definitely could not use Flom's "general" convention of connecting two vertices that are separated by a third – that just plain won't work. There is a surprisingly elegant general formula to calculate the area of an irregular polygon, one that requires the Cartesian coordinates of all the vertices – but Flom's alleged solution just doesn't work.

For being too lazy to test his method or for the intellectual dishonesty of not even trying;- not to mention leaving out all of the drudgery of measurements and calculations – the jackass didn't even provide an example! – Peter Flom is perhaps one of the most deserving recipients of the Dumbass of the Day award in history. Go, Pete...

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