# Proof ratio squares corr sides similar polygons to their areas..

• Mar 23rd 2007, 08:15 AM
Rich Martin
Proof ratio squares corr sides similar polygons to their areas..
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
• Mar 23rd 2007, 08:40 AM
ThePerfectHacker
Quote:

Originally Posted by Rich Martin
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

I never seen it myself.

It is a fact taught in high-school.
I would imagine this proof being extremely complicated, because polygons can have any form. Convex, Concave.
• Mar 23rd 2007, 09:32 AM
ecMathGeek
Quote:

Originally Posted by Rich Martin
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
• Mar 23rd 2007, 09:34 AM
ecMathGeek
Quote:

Originally Posted by ecMathGeek
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):

This would also have to be proven for my "proof" to hold true.
• Mar 23rd 2007, 09:37 AM
ecMathGeek
Quote:

Originally Posted by Rich Martin
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

I should then add one last note to my "proof" above:

If:
x1 = p*x2

Then:
x1^2 = (p*x2)^2 = p^2*x2^2