So proving the area formula for a trapezium is turning into something I do when I get bored. Here is my third proof. This one requires two versions annoyingly, as I mapped out all the different types of trapezium I could think of (are there more??)

Type 1 has a base that exceeds the top in width, both to the left and right. Type 2 and type 3 differ only in that type 2 has a top right vertex in line with the bottom left vertex, whereas type 3 does not. Type 4 has a base and top completely unaligned, and type 5 has two right angles.

I needed a slightly different proof for type 1. Or perhaps I’m just missing something so that I can combine these two proofs nicely without writing a completely different one? Seems likely.

Anyway, it’s a simple premise: enclose the shape in the smallest possible rectangle, and show that the area of the trapezium is equal to the area of the rectangle minus the remaining triangle(s). If you play with the algebra a little, you end up with the formula we’re all familiar with. I think with the added arrows it’s pretty clear where everything comes from, albeit a little messy.

### Like this:

Like Loading...

*Related*

Hi!

Lovely stuff, clever trick in proof 1 when you factorise h/2 to leave b+b in the bracket…

How are the two proofs different to your eyes though?

I felt like I needed to separate the two cases as the labelling doesn’t work otherwise. Case 1 has 3 elements at the top, whereas the other cases have only 2