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. ?

enter image description here


1 Answer 1


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.

  • $\begingroup$ Thank you for your answer @KajHansen. I was hoping that there was some simple way of proving it, since the problem has origin back to Euclid ? $\endgroup$
    – Shuzheng
    Apr 27, 2014 at 8:15
  • $\begingroup$ Ah, I'm sure there is. One advantage of mine, however, is that it generalizes to include arbitrary partitions that aren't necessarily polygonal. $\endgroup$
    – Kaj Hansen
    Apr 27, 2014 at 9:47

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