Relationship between area of similar polygons and their corresponding lengths It is known that ratio of the areas of two similar polygons is equal to the square of the ratio of the corresponding sides. Could one advise me how to prove this assertion? Do we divide the polygon into triangles? Thank you.
 A: It suffices to use the reasoning from a quadrilateral ABCD with AC dividing it into two triangles, namely $a = [\triangle BAC]$ and $b = [\triangle BDC]$.
The similar target will be similarly named as $a’ = [\triangle B’A’C’]$ and $b’ = [\triangle B’D’C’]$
We assume the area and side relation is true for two similar triangles. [For simplicity, I use the side AC. Actually any pair of corresponding sides also works because the image is just k times the original by lengths.]
$a’ = (\dfrac {A’C’}{AC})^2 \times a$
$b’ = (\dfrac {A’C’}{AC})^2 \times b$
$a’ + b’ = (\dfrac {A’C’}{AC})^2 \times (a + b)$
[area of the image] =  $(\dfrac {A’C’}{AC})^2 \times$ [area of the original]
A: The function that scales the plane by a factor of $a$ can be written as $f((x,y))=(ax,ay)$. The Jacobian determinant of this function, which gives the area ratio of the transformed plane to the original, is
$$\begin{vmatrix}a&0\\0&a\end{vmatrix}=a^2$$
So scaling lengths by a factor of $a$ scales areas by a factor of $a^2$.
