SOLUTION: If a rectangle and a trapezoid have equal areas, then they must have the same perimeters. A) True B) False

Question 241559: If a rectangle and a trapezoid have equal areas, then they must have the same
perimeters.
A) True
B) False
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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 HERE !!!!!

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