Area and Perimeter of Similar Figures

The Area of a Square Postulate (this postulate actually has no universal name) states that the area of a square is equal to the product of the side length with itself.

Suppose that the area of a rectangle with length $l$ and width $w$ is $A$. Construct a square $S$ of side length $l+w$. Then, we may figure out the area of the square in two ways: $$[S]=(l+w)^2$$ $$[S]=l^2+w^2+2A$$ Equating these two expressions, we get that $A=lw$. Thus, the area of a rectangle is equal to the product of its length and its width.

Now suppose that we have a rectangle $R_1$ of length $l$ and width $w$ and a similarity transformation with scale factor $k$.

A similarity transformation maps shapes to similar shapes, so the rectangle will be mapped to a similar rectangle $R_2$.

Moreover, since the scale factor is $k$, the length of $R_2$ is $kl$, and the width of $R_2$ is $kw$.

Now, we have the following two equations: $$[R_1]=lw$$ $$[R_2]=(kl)(kw)=k^2lw=k^2[R_1]$$ Thus, we have proven that for rectangles, a similarity transformation scales areas by a factor equal to the square of the scale factor.

Now, given any shape, we may approximate it by a set of rectangles. We may make the rectangles thinner and thinner, and as we do so, the rectangles become better and better approximations of the area of the shape. In fact, in the limiting case, this approximation becomes exact (I won't prove this here); this is the foundation of integral calculus.

Now, the similarity transformation with scale factor $k$ scales up each rectangle's area by a factor of $k^2$. Thus, it also scales the total area by $k^2$. By the properties of limits, the new limiting case is also $k^2$ times the original limiting case.

Thus, we have proved that for any closed two-dimensional figure and any similarity transformation, the area of the figure is scaled by a factor equal to the scale factor squared.