At the Don Steward event last weekend, Don demonstrated how the formula for a trapezium can be derived in different ways. He asked the participants if they could figure out one of them, and my fantastic student Beth came up with this beauty:

What a star!

Following on from that, Don pointed out that a trapezium is in fact just the bottom of a larger triangle:

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 h_{2}, 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 h_{2} 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 b_{2} and a_{2} instead of b_{1} and a_{1}:

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:

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Love this. Especially the celebration of multiple approaches. Here’s what I had in geogebra on it.

http://tube.geogebra.org/student/m105530

Helped my students to redraw it as a transformed rectangle.

Not sure if you need to split triangle into left and right sides.

By similar triangles (h2-h)/a =h2/b since perpendicular height from base will be in proportion to the sides and in particular the base

Rearranging h2/a – h2/b = h/a

Take out h2 from LHS gives h2 (1/a -1/b) = h/a

Putting 1/a -1/b over a common denominator gives h2 (b – a)/ab = h/a

So h2 = b h / (b-a)

Area of trapezium = whole triangle less top triangle = b h2 / 2 – a (h2 – h) / 2

= (b – a) h2 / 2 + a h / 2

Substitute formula for h2

Area =( (b-a) . b h / (b-a) ) / 2 + a h / 2

Cancelling (b-a) gives

Area = b h / 2 + a h /2 = (a+b) h / 2

I assumed you could do it without, I wanted to follow my initial trail of thought. Thanks for offering another one!

Clearly an easy way to derive the formula for a trapezium’s area is to label the vertices of the trapezium A, B, C and D where AB is the base say and DC is parallel to AB. Draw a line from A to the opposite corner (C). The line AC splits the trapezium into two triangles ABC (with base length b and height h) and DCA (an upside down triangle with base a and height h). Area of ABC = bh/2, area of DCA is ah/2. Adding gives (a+b)h/2

Yes, but that wasn’t the point of the exercise. That’s the method we teach in schools.

Linear variation give you another way to see the formula for h2 (and, as noted above, you don’t need to cut into left and right sides): if it takes a linear relationship h distance to go from b length to a, then how long does it take to go from a to 0? Simple slope formula gives you ah/(b-a).

Assume you’ve seen another approach where you make a paralellogram out of your original quadrilateral and a copy that has been reflected through a line parallel to the two parallel sides.

Interesting

A nice alternative if you want;

Start with your trapezium; have another copy rotated 180degrees.

Push it together.

The top and bottom are now of length a+b and the height is the same

Shape is a parallelogram so area base x height

Which is really (a+b) x height

Obviously the parallelogram is formed by two identical trapezia so multiply by 1/2

(I tend to teach area of parallelograms first then area of triangles linked to that) and then area of trapezium linked back to parallelograms.

Love that! I’ve seen it before, but what I’m really enjoying about the comments on this post is the love of alternative proofs for things 🙂

Hi. Another approach is to use integration. Rotate the triangle 90deg clockwise so that its apex is at the origin and the perpendicular height is along the positive x-axis. Integrate y from x=h2 – h to x=h2 (where y=b, the short parallel side when x=h2-h and y=a, the long parallel side when x=h2). Substitute t=x-(h2-h) noting that dt/dx=1. Easy to see (with a diagram!) that y=b+t*(b-a)/h. At x=h2-h we get t=0 and at x=h2 we get t=h. So integrate b+t*(b-a)/h to get [bt + (a-b)*t^2/(2h)] from t=0 to t=h. This = bh + (a-b)*h^2/2h = bh + (a-b)*h/2 = ((a+b)/2)*h

Separately I took a look at Don Stewards powerpoints you kindly posted recently on this site. Very interesting. I particularly liked the problem in the 2nd powerpoint on cutting up lengths of cardboard into three to get the average length of two pieces to equal the length of the 3rd.

Back to the trapezium ‘though it is interesting to draw an analogy between calculating the area of a trapezium and summing an arithmetic series – average of first term and last term times the number of terms.

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