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WHAT IS THE AREA OF AN S-SIDED IRREGULAR POLYGON?
It is well known that any oblique triangle with specified vertices A,B,C has an area given by half the absolute value of the cross-product between two of its side-vectors. Alternatively one can use the Hero formula which requires only the side lengths to find the triangle area. In most cases the vector product approach is the simpler way to find the area. For example, consider a triangle with vertices at A[0,0], B[0,2], and C[1,1] In this case the side-vector from A to B is BA =2j and the side-vector from A to C is CA =i+j. So evaluating |( BA x CA |/2 we find it equals 1 for the area of the isosceles triangle under consideration. To check this answer we can compare things with the Heron Formula-
beingthesemi perimeter
a b c Area ss a s b s c with s
For the present case we have s=1+sqrt(2) and Area=sqrt[(sqrt(2)+1)(sqrt(2)-1)]=1, which checks. It is our purpose here to extend the area calculations for a triangle to any larger S sided closed polygon of irregular side-length lying in the x-y plane.
In looking at S-sided polygons, one notices that they can always be broken up into S- contiguous triangles. So a rectangle can be represented as two triangles, a pentagon by 3 triangles, a hexagon by 4 triangles, and so on. That is, the number of triangles T=S-2. One also notices that the number of lines L required to form these sub-triangles is three less than the number of sides, so L=S-3. We summarize these geometric observations in the following figure-
Note that the rule we are using in drawing the red lines is always to connect those vertices separated from each other by two vertex points and that this continuous to hold when only red line polygons remain.
Let us next consider a specific area in the form of a pentagonal polygon which has straight lines connecting its five vertices located at A[0,0], B[1,0[, C[2,2] , D[1,3], and E[-1,2]. To determine the area of this polygon we begin with the following figure-
The figure is drawn to scale with [x,y] representing the Cartesian coordinates of the vertices. Here S=5, T=3 and L=2. That is, we need to use only two division lines to form three sub-triangles. One possible configuration is as shown. We next write down the seven vectors defining all S+L=7 straight lines. These vectors are-
BA = i , CB = i +2 j , DC =- i + j , ED =-2 i - j , AE = i -2 j , CE =3 i , and CA =2 i +2 j
The vectors are indicated in bold lettering with the first letter referring to the head of the vector. Here i and j are unit length base vectors in the x and y direction , respectively. To find area 1 in the figure we write-
Area 1 =| BA x CA |/2=|2 k |/2=
Next we find that area 3 is given by-
Area 3 =|-CA x - CE |/2=|6k|/2=
and area 2 by-
Area 2 =|-ED x CE|/2=|3k|/2=3/
This is a very easy problem to analyze when making use of its eight-fold symmetry. Following our rule of connecting the vertices n with n+2 we get 8 isosceles triangles and a central regular octagon. Next connecting every second vertex of the octagon we get 4 more triangles plus a central square which can be divided into two equal triangles as shown. The results are 8+4+2=14 isosceles triangles and 8+4+1=13 red lines needed to construct these. The total number of sides of the star is 16. Note that there are other ways to divide up the star into sub-triangles , but these would violate the condition that T=S- and L=S-3. Fortunately here we don’t need to run through an evaluation of the areas of each of the 14 triangles in order to find the star area. The much simpler approach is to note that all 16 vertices are given in cylindrical coordinates by-
)] (
1 2 [cos(^2
n and
n r
So we need to only look at the area of the single triangle with vertex points at a[0,0], b[3,0] and C[cos(π/8),sin(π/8)] and multiply the result by 16 to find the area of the star. The area of the triangle is simply –
Areatriangle=|(3i) x [( i cos(π/8)+ j sin(π/8)]|/2=(3/2)sin(π/8)
so that the area of the star becomes 24sin(π/8)=8.306809… This example clearly shows that the evaluation of polygon areas should always be looked at from a standpoint of symmetry.
As a last example, look at the following thirteen sides polygon and its eleven sub- triangles –
Here we see that all but the one triangle C-D-E have an area of 1 each. The area of C-D- E however is-
AreaC-D-E=|(-2 i + j ) x (- i - j )|/2=3/
Thus the total area of the 13 sided irregular polygon becomes 23/2.
