Show that the area of some shape, in this case a rectangle, equals the sum of the areas of the shapes forming a partition of the shape. 
I've been wondering whether how to show that the area of some shape, in this case a rectangle, equals the sum of the areas of the shapes forming a partition of the shape.

To be more precise:
In the picture below I've partitioned a rectangle $R$ into smaller rectangles and right triangles. Suppose $R$ has side lengths $a,b$ such that the area $A(R)$ equals $a\cdot b$. Can I show that the sum of the areas of the smaller rectangles and right triangles equal $A(R)$ ? 
By a partition $P$ of $R$, I mean a set of shapes not contained in any bigger shape with the exception of $R$. In this case $P = \{1,2,3,4,5,6\}$. Can I show $\sum_{p \in P} A(p) = A(R)$ by induction or how ? (Here $A(p)$ denote the area of $p$ computed as usual for rectangles ($c \cdot d$) and right triangles ($\frac 1 2c \cdot d$) with side lengths $c,d$).
Does the statement holds if I partition any shape into arbitrary shapes like a general region or polygon etc. ?


One possible way you could show this is by applying Green's Theorem.  Let some region $D$ be partitioned into smaller regions $D_i$ with boundaries $C_i$.  Let's find the area of the individual partitions, and sum them up:
$$\sum_i \iint\limits_{D_i}dA = \sum_i\int_{C_i}xdy - ydx$$ 
Note that each partition is endowed with a counterclockwise orientation.  Hence, any interior boundary of any paritition will be canceled out by the boundary of an adjacent partition, and we are left with the following:
$$\int_{C}xdy - ydx = \iint\limits_{D}dA$$ 
Where $C$ is the boundary of the entire region $D$.  We conclude the area of the whole region is equal to the sum of the areas of its partitions.
