[ Trapezoid Week continues! ]
Yesterday I set out a diagram, showing off one example of a trapezoid, with which I mean to show one way to get the formula for a trapezoid’s area. The approach being used here is to find two triangles so that the difference in area between the two is the area of the trapezoid. This can often be a convenient way of finding the area of something: find simple shapes to work with so that the area we want is the sum or the difference of these easy areas. Later on I mean to do this area as the sum of simple shapes.
For now, though, I have the trapezoid set up so its area will be the difference of two triangle areas. The area of a triangle is a simple enough formula: it’s one-half the length of the base times the height. We’ll see much of that formula.
In the diagram I put the longer base, connecting points A and B, of length b1, at the bottom, and the shorter base, connecting points D and E, of length b2, on top. And I extended the two non-parallel legs so that they eventually come together at a point, labelled C. The trapezoid is the figure bounded by the points A, B, E, and D, or to put it succinctly, ABED. The distance between the line AB and the line DE is the altitude a. And the perpendicular distance between line AB and the point C is the first height, h1. The perpendicular height between the line DE and the point C is the second height, h2. That’s all background.
There are two triangles in the figure. The bigger one is triangle ΔABC, and its area is (1/2) * b1 * h1, half its base times its height. The smaller triangle is triangle ΔDEC, and its area is (1/2) * b2 * h2, half its base times its height. The trapezoid ABED is the part of triangle ΔABC not already in the triangle ΔDEC; its area has to be the difference between them. That is, the trapezoid has area (1/2) * b1 * h1 – (1/2) * b2 * h2. This is perfectly true, although it requires we figure out whta h1 and h2 are, which might be work. What we’ll do is try to get rid of h1 and h2.
Here’s one piece that lets us get rid of h2 at least. (It’ll return, but, we’ll get rid of it for good shortly after). The height h1 is the distance between the parallel bases, which we were calling a, plus the distance between the shorter base DE and the point C, h2. So we can either replace h1 with a + h2, or we can replace h2 with h1 – a. Either way will work, but, I want to use the second expression. By replacing h2 where it appears in the area for the trapezoid, we get the equation:
That second part of the above line we can expand, according to the distributive property. According to the distributive property, for any three numbers, s * (x + y) is the same number as s * x + s * y. So we can change the way we write the equation — though not its value, or its truth — into this:
Now we have the area as the sum of three terms, the middle one negative. This is about where I got stuck in class, by the way, as I couldn’t think of how to get rid of the h1 there. At best I could shuffle between h1 and h2, without eliminating both together.
The insight that I needed, and that came to me after class, was to realize that the triangles ΔABC and ΔDEC are similar. That has a precise mathematical definition: it means the triangles have the same interior angles. If they differ at all, it’s only in their size. If I were to show a picture of ΔABC by itself, and to show a zoomed-in picture of ΔDEC by itself, there’d be no way of telling them apart, other than the labels.
Why this is important is because of a neat property regarding similar triangles. If you have two similar triangles, then, the ratios of corresponding pieces of the two triangles will be equal. That is, for example, the ratio of the length of AB to the length of AC will be the same as the length of DE to the length of DC. More immediately usefully, the ratio of the length of AB, b1, to the height of ΔABC, h1, is equal to the ratio of the length of DE, b2, to the height of ΔDEC, h2.
Put in a more familiar equation form, h1 / b1 = h2 / b2. Or we can rewrite that in a couple of ways; the one I am going to use is: b2 = b1 * h2 / h1.
Now I’ll put this expression for b2 in to the equation just where I got stuck, and before long, both the h2 and the h1 terms will melt away. Starting with the expression where I got stuck:
Replacing b2 as I intended gives me this:
In the middle term, we have a quantity which is divided by h1 only to be multiplied by h1 again. As long as h1 isn’t equal to zero, this division and multiplication come out to the same thing as multiplying the rest of the quantity by 1, which is just leaving the quantity alone. And we can be fairly sure that whatever h1 is, it isn’t equal to zero. So:
Now I’m going to call on the distributive property again, combining the first and second terms again:
That’s a lot of work not to have got rid of h1 or h2 … except, that we do know something about h1 minus h2. That difference is the distance between the two parallel bases, that is, a. So the area of the trapezoid is:
And — you may have seen this coming — with yet another round of the distributive property we get:
So we’ve gotten exactly the formula we should have gotten.
This is one way of showing the formula’s true. It’s not the best. I’m not sure this is even the best way of doing it by the difference-between-triangles, particularly as it seems to need to remember this bit about similar triangles. It’s not an obscure property of similar triangles, but it isn’t as obvious as the area for the triangles. The proof also does make one hard-to-remove assumption: we’re supposing that the legs AE and BE, if we keep drawing those lines out, eventually come together in a point. That is, the proof is not going to work on parallelograms or on rectangles, which at least last time around I had wanted to include as trapezoids.
So, how could this be done better?