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Area of an Irregular Polygon


Date: 03/10/97 at 16:09:13
From: Ron Smith
Subject: Area of irregular polygon

I guess I've been out of school way too long.  I have searched your 
database and I see where you have explained to several people that you 
can't find the area of an irregular polygon with only the perimeter 
values. However, you did not give a formula for finding the area if 
you do have the angles.

I have the lengths of all the sides and some of the angles (I know I 
am going to need to know all of the angles) for a patio I am going to 
build.  They are (clockwise):

22'(90deg), 4'(90deg), 10'(90deg), 25'(?deg), 25'(?deg), 21'(90deg),  
4'(90deg), 4'(90deg), 14'(?deg), 25'(?deg), 12'(90deg).

I would really like some software that I could plug these values into 
to get the area of the polygon. Since I am a software engineer, I 
could write the program if you give me the formula.

Thanks.


Date: 03/11/97 at 12:03:13
From: Doctor Ken
Subject: Re: Area of irregular polygon

Hi Ron-

What I'm going to tell you about is more of an algorithm and not 
really a formula.

Let's say you're given a bunch of side lengths and angles, in order. 
Call the side lengths S1, .., Sn, and call the angles A1, .., An.  For 
the angles, let's not think about the interior angles, but rather the 
exterior angles (each of which is 180 minus the interior angle at that 
corner).  So these side lengths and angles are sort of a set of 
directions for tracing out the polygon: "Walk S1 meters forward, then 
turn A1 degrees to the left, then walk S2 meters forward, then turn A2 
degrees to the left, ...."  Note that to specify an n-gon, we only 
need n-1 sides and n-2 angles, since the last step is to just go back 
to the starting point.

I've drawn a diagram that may be helpful here:

        

Here I'm using S1, .., Sn to be both labels of segments, and also the 
length of those segments.  The same goes for A1, .., An.

The way we'll find the area is to add up the areas of a bunch of 
triangles. If it's an n-gon, it can be divided into n-2 triangles and 
these are the triangles whose areas we'll sum. The general rule we'll 
use to find all the areas is this: the area of a triangle whose sides 
are the vectors (a,b) and (c,d) is 1/2 (ad-bc). This is just 1/2 the 
determinant of the matrix whose rows are (a,b) and (c,d).

First we'll find the area of the first triangle (triangle 1). The two 
vectors we'll use are the sides labeled S1 and S2. If you're not 
familiar with vector notation, here's what it means: if you put the 
origin of a coordinate system at one end of the line segment, then the 
vector is the coordinates of the other point. You can see that the 
vector of the side S1 is (S1,0). The vector of the side S2 is a 
little trickier. To find it, we remember a little polar coordinate 
math, and we get the vector to be (S2 * Cos(A1), S2 * Sin(A1) ).  
So the area of triangle 1 is 1/2 (S1*S2*Sin(A1) - 0*S2*Cos(A1)), 
i.e., 1/2*S1*S2*Sin(A1).

Now we'll find the area of the second triangle, and in doing so, we'll 
see the general pattern for how to iteratively find the area of the 
entire polygonal region.

To find the area of triangle 2, we again need to know a couple of 
vectors that are its sides. The vectors we'll use are the segment 
shared between triangle 1 and 2 (let's call that the current_base), 
and the segment S3.  

So let's find the vector notation for the base. Actually, we've 
already done most of the work in the previous section of the problem.  
If we add together the vectors S1 and S2, we'll get the current_base.  
So the current_base is (S1 + S2*Cos(A1), S2*Sin(A1)).  

Now let's find the vector notation for segment S3. Using polar 
coordinate geometry again, we see that it's:
 
(S3*Cos(A1+A2), S3*Sin(A1+A2))

So the area is 1/2 the determinant of the matrix of these two vectors:

(S1 + S2*Cos(A1)) * (S3*Sin(A1+A2)) - (S2*Sin(A1)) * (S3*Cos(A1+A2))
---------------------------------------------------------------------
                                   2

Let's review the general procedure. For an n-gon, we want to sum the 
areas of n-2 triangles. To find the area of each triangle, we pick 
two sides and write them in vector notation, then take 1/2 the 
determinant of the matrix formed by these two vectors. In general, 
one vector will be the vector sum of the two vectors from the previous 
triangle. This will look something like this:

current_base = (current_base[0] + prev_side[0], current_base[1] + 
prev_side[1])

The other vector will be the "vectorization" of one of the sides, 
which will look something like this:

partial_angle_total += A[i]

current_side = (S[i] * Cos(partial_angle_total), 
                S[i] * Sin(partial_angle_total)

area = Absolute_value(.5 * (current_side[0] * current_base[1] -
                              current_side[1] * current_base[0]) );
prev_side = current_side

I hope this is relatively clear.  For a brush-up on (or first-time 
experience with) polar coordinates or determinants, you might want to 
check out some problems in our archives. 

-Doctor Ken,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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