I can't recover the proof that in similar polygons, the ratio of the squares of corresponding sides equals the ratio of the areas of the polygons themselves. Saw it done for a triangle somewhere once, but how is it derived for just any polygon? Thanks for any ideas.
Rich
Ok, I'm not going to "prove" this, but I'll try demonstrating how a proof might be done:
I know one fact to be true about the area of all polygons (that I have ever seen, and I assume for all polygons period), the area of a polygon is always proportional to some measure on that polygon, n, times some other measure on that polygon, m:
A = k(n*m) where n and m represent some measures (or some sum of measures) on or within that polygon, and k is some constant of proportionality relating the multiplication of these measures to the area.
Since, between similar polygons, the ratio of proportionality, p, between any two similar measures between the two polygons is the same ratio of proportionality between all similar measures between those two similar polygons, the following should hold true (again, this is not a formal proof):
For polygons P1 and P2, if the ratio of proportionality between similar measures is p, then for any two similar measures, x1 and x2:
x1 = p*x2
If the area of each polygon is:
A1 = k(n1*m1)
A2 = k(n2*m2)
Where
n1 = p*n2
m1 = p*m2
Then
A1 = k(n1*m1) = k(p*n2*p*m2) = p^2[k(n2*m2)]
Therefore
A1 = P^2*A2