Question 241559
I would have to say no.


This is because I found one instance where the areas are equal and the perimeters are not.   That's all that's required to destroy the theory.


Consider that a parallelogram and a trapezoid both are composed of a square in the middle plus two triangles hanging at each end.


I created two squares with 4 inches on each side.


I then attached a triangle to the parallelogram on each end so that the resulting structure formed the parallelogram.


each triangle was a 3/4/5 triangle.


The one on the left had the 3 on top and the one on the right had the 3 on the bottom.


The 5 of each triangle was the sides of the parallelogram.


the other sides of the parallelogram were formed from the 3 and the 5.


A picture at the end will show you what I mean.


The trapezoid was formed from the square with two triangles attached to it, 1 on each end.


1 triangle was a 4/5/sqrt(41) triangle, and the other triangle was a 1/4/sqrt(17) triangle.


The construction was such that the areas of the trapezoid and the parallelogram were equal.


They are equal when b*h of the parallelogram = ((b1+b2)/2)*h of the trapezoid.


Since I made h of the parallelogram equal to h of the trapezoid, this meant that (b1+b2)/2 of the trapezoid equal to b of the parallelogram.


I calculated the areas and they are equal.


b of the parallelogram is equal to 7 (4 + 3)


b1 + b2 of the trapezoid = 4 + 10 = 14


7*4 = 14/2*4 confirming the areas are equal.


The perimeters, however, were not equal.


7+7+5+5 of the parallelogram did not equal to 4+10+sqrt(41)+sqrt(17) of the trapezoid.


Theory doesn't hold.


Perimeters are not equal.


Picture of the resulting structures can be seen <a href = "http://theo.x10hosting.com/problems/241559.html" target = "_blank">HERE !!!!!