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__Preliminary __

__ __**
****Area of a triangle**

From linear algebra, we know that the magnitude of the cross product of two vectors is the area of the parallelogram they determine: let a(x

_{a }, y_{a}), b(x_{b }, y_{b}), c(x_{c }, y_{c}) are three vertices of a triangle T, see Figure 10, let us denote this area as A(T)

Fig.10

2*A(T) = (x _{A}y_{B}
- y_{A}x_{B} )
(1)*

where A=b-a, B=c-a, so we have:

* *
2*A(T) = 2A(a, b, c) =x _{a}y_{b}
- y_{a}x_{b}+y_{a}x_{c} - x_{a}y_{c}+x_{b}y_{c}
- x_{c}y_{b}
(2)*

__ __**Area of a convex quadrilateral**

The area of a convex quadrilateral Q=(a, b, c, d) can be written in two ways, depending on the two different triangulations (see Fig.11)

Fig.11

*A(Q) = A(a, b, c) + A(a, c, d) = A(d, a,
b) + A(d, b, c) *

2*A(Q) *= *x _{a}y_{b}
- y_{a}x_{b}+y_{a}x_{c} - x_{a}y_{c}+x_{b}y_{c}
- x_{c}y_{b}*

*
+x _{a}y_{c}
- y_{a}x_{c}+y_{a}x_{d} - x_{a}y_{d}+x_{c}y_{d}
- x_{d}y_{c} *

Note that the terms y

_{a}x_{c}- x_{a}y_{c}apear in A(a, b, c)and in A(a, c, d) with opposite signs and so cancel, that means the diagonal ac cancel. Thus the exact expression of A(t) isof the triangulation.independent

__
__**Area of a nonconvex quadrilateral**

Fig.12

See Figure 12. even though the diagonal ac is external to Q, the algebraic expression is

the diagonal chosen, the equation:still independent of

*A(Q) = A(a, b, c) + A(a, c, d)*

Because note that (a.c.d) is a clockwise path, so the cross product for the area will be negative, i.e. A(a, c, d)is , and is therefore subtracted from the surrounding triangle (a, b, c).

*
*

__ __**
Area from an arbitrary Center**

Let us generalize the method of summing the areas of the triangles in a triangulation to summing areas based on an arbitrary, external, point p. Let T=(a, b, c) be a triangle with the vertices oriented counterclockwise, p is any points in plane

*A(T) = A(p, a, b) + A(p, b, c) + A(p, c, a)*

It is easy to know that A(p, a, b) is negative because clockwise. and A(T) is

(see Fig.13)independent of point P

Fig.13

So we may generalize the result for arbitrary polygons.

__
__**Lemma:** let a polygon (convex or
simple) P have vertices v_{0},..., v_{n-1}
labeled counterclockwise, let p be any point in plane, then

*A(P) =
A(p,*v_{0}, v_{1}*) + A(p,*v_{1},
v_{2}*)+ ...+ A(p,*v_{n-2},
v_{n-1}*) + A(p,*v_{n-1}, v_{0}*)*

*
*

where v

_{i}=( x_{i}, y_{i}), it can be seen as vectors from the origin. Here the method is known as the.polar formula

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