Area of Square/Proof 2

Theorem

A square has an area of $L^2$ where $L$ is the length of a side of the square.

Thus we have that the area is a function of the length of the side:

$\forall L \in \R_{\ge 0}: \map \Area L = L^2$

where it is noted that the domain of $L$ is the set of non-negative real numbers.

Proof

Let $\Box ABCD$ be a square whose side $AB$ is of length $L$.

Let $\Box EFGH$ be a square whose side $EF$ is of length $1$.

From the Axioms of Area, the area of $\Box EFGH$ is $1$.

By definition, $AB : EF = L : 1$.

From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\Box ABCD$ to $\Box EFGH$ is the duplicate ratio of the ratio of $AB$ to $EF$.

Thus by definition of duplicate ratio:

$\Box ABCD : \Box EFGH = \left({AB : EF}\right)^2$

That is:

$\dfrac {\Box ABCD} {\Box EFGH} = \left({\dfrac L 1}\right)^2 = L^2$

That is, the area of $\Box ABCD$ has $L^2$ as many units as $\Box EFGH$.

Hence the result.

$\blacksquare$