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Finding the Area of an Irregular Polygon

Date: 02/23/2008 at 04:13:50
From: dIonAtHan
Subject: finding the area of irregular polygon

What is the formula for finding the area of an irregular polygon?



Date: 03/16/2008 at 21:51:46
From: Doctor Ali
Subject: Re: finding the area of irregular polygon

Hi!

Thanks for writing to Doctor Math.

Let's say the we have an n-gon of A1-A2-A3-...-An.

Let's say that the coordinates of Ai are (Xi, Yi) and we have all 
the Xi's and Yi's. Then the area A is given by the 'ladder':

            |x1   y1|
            |x2   y2|
            |x3   y3|
            |   .   |
            |   .   |
  A = 1/2 * |   .   |
            |   .   |
            |   .   |
            |   .   |
            |xn   yn|
            |x1   y1|    <-- Note that this is the first row,
                             repeated. 

How do you evaluate the ladder?  I'll explain. 

You have to multiply each number of the first column with the number
in the second column of the next row.  Sum the products, and call the
sum S1. 

Then multiply each number of the second column with the number in the
first column of the next row.  Sum the products, and call the sum S2. 

Finally, 

  A = 1/2 * |S1 - S2|

where |x| denotes the absolute value. 

For example, consider a triangle whose vertices are at (x1,y1),
(x2,y2), (x3,y3).  Here's the ladder, 

  |x1  y1|
  |x2  y2|
  |x3  y3|
  |x1  y1|    

here are the sums, 

  S1 = x1*y2 + x2*y3 + x3*y1

  S2 = y1*x2 + y2*x3 + y3*x1

and here is the area:

  A = 1/2 * |S1 - S2|

    = 1/2 * |x1*y2 + x2*y3 + x3*y1 - y1*x2 - y2*x3 - y3*x1|

This theorem comes from "The Surveyor’s Area Formula".  The 
difference is that, in the main formula, the area is calculated by 
the sum of 2x2 determinants.  Here, I tried to simplify the formula 
by defining a SUPER-DETERMINANT--not a real mathematical term, I made 
it myself(!). 

If you would like to learn more about "The Surveyor’s Area Formula", 
take a look at:

  The Surveyor's Area Formula
    http://www.maa.org/pubs/Calc_articles/ma063.pdf 

An easier proof for the formula can be achieved by dividing the 
polygon into triangles.

The are some other interesting methods when we move into three dimensions:

*) In 3-D space, the triangle, made on V1 and V2, which are two 3-D 
vectors, has the area A and:

  A = 0.5 * |V1 x V2|

|V1 x V2| is the area of the PARALLELOGRAM, which is surely twice as 
big as the TRIANGLE.


*) By using the previous theorem, we can calculate the area of a 
triangle in 3-D space by first finding the V1 and V2.  For the 
triangle ABC:

  A:(xA, yA, zA)

  B:(xB, yB, zB)

  C:(xC, yC, zC)

One case is:

  V1 = [xB - xA, yB - yA, zB - zA]
 
  V2 = [xC - xA, yC - yA, zC - zA]

A simpler way to perform the V1 x V2 is to evaluate the following 
3x3 determinant:

  | ^   ^   ^  |
  | i   j   k  |
  |            |
  | x1  y1  z1 |
  |            |
  | x2  y2  z2 |


Please write back if you still have any difficulties.

Bye!

- Doctor Ali, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Polyhedra
College Triangles and Other Polygons
High School Polyhedra
High School Triangles and Other Polygons

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