find the formula for area of a Trapezium experimentally

He then conjectured the possibility of a proof using that triangle. He asked the audience. If you want to stop here, and go play with some maths, be my guest. Otherwise continue reading…

 

 

So I volunteered the beginnings of a potential solution. I saw straight away that there *must* be a proof using similar triangles, by dropping a perpendicular line from the apex of the triangle, thus forming two pairs of similar right-angled triangles (yellow and small pink,  white and big pink):

He liked the idea, pressed me to take it further, but that was as far as I got. He challenged me to work it out after the session.

Well, after Beth posted her proof on Twitter yesterday, I was reminded of the challenge! So late last night I opened a pad of paper, got out a pen, and started scribbling…

First I need some labels:

Ok So we’re going to work with the two left-hand side Right Angled Triangles:

Now we know the pink and yellowpink triangles are similar, so :

I have an expression for h2, which is important, because that’s the letter I don’t want for my final formula.

So to find the area of the bit that’s important to me, I need to subtract the area of the pink triangle from the larger triangle (I called it yellowpink earlier). I also need to substitute h2 with my new expression:

Now I’ll deal with the other triangles:

Well this is exactly the same method, and will get an almost identical expression, but with b2 and a2 instead of b1 and a1:

So the total area (ie, the area of the trapezium) is these two expressions added together:

Very satisfying!

My original version, which I didn’t simplify until much later on in the proof, and used slightly daft lettering for, is here (click for enlargement):

* edit … until John Golden brilliantly points this out to me! …

There’s a hole in my bucket!

So dropping the perpendicular needs to go. However, the similar triangles can stay. We’ll just use two of them this time. Extend the lines to make a big triangle, and you have two similar (albeit not right-angled) triangles. Therefore the same rules apply, and we only need do everything once: