Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

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.

share|improve this answer
    
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 ? –  user111854 Apr 27 at 8:15
    
Ah, I'm sure there is. One advantage of mine, however, is that it generalizes to include arbitrary partitions that aren't necessarily polygonal. –  Kaj Hansen Apr 27 at 9:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.