The isosceles right triangle: why its hypotenuse must be a√2

0:00Here is a right triangle with two sides of length one. The third side has no obvious length at all, and a ruler will not settle it. You cannot look this one up and you cannot pace it out. It has to be reasoned to. And the way in is going to look like a detour, because we are not going to go after the length. We are going after an AREA instead, and the length will fall out of it.

0:27That move - finding a length through an area - is the whole idea, and it is worth watching closely. First, the answer everybody reaches for. One and one make two. But look at what two measures. It is the length of the bent path, out along one side and up the other. The third side is the straight path, and a straight path is shorter than a bent one.

0:54So two is too big. It is an upper limit, not the answer. The next guess is usually about one and a half, and that is a reasonable guess. One and a half, multiplied by itself, is two point two five. Hold on to that number; in a moment it will matter. Guessing is fine. Stopping at the guess is not. The side we are chasing has a name. It is the hypotenuse.

1:23And the definition is worth getting exactly right, because it is not about how the drawing is tilted. The hypotenuse is the side FACING the right angle. The other two are the ones that meet at it. Turn the triangle round and the right angle moves to a different corner, but the answer does not change. This matters when you are handed a figure. In one drawing the marked sides are forty and forty-one, and the forty-one is the hypotenuse.

1:56In another they are twenty-seven and forty-five, and the forty-five is the hypotenuse. The missing sides are nine and thirty-six. Hunt for the slanted-looking one and you will misread half the figures you meet. Now back to our triangle, and the first move of the detour. Take a second copy of it and turn it over. The two together make a square, with sides of one all the way round.

2:26So our triangle is exactly half of a unit square, and the side we want is that square's diagonal. That is worth saying on its own. The hypotenuse of this triangle and the diagonal of that square are the same segment. Which means anything we learn about one, we have learned about the other. And there is something we already know about a square's diagonal. Build a new square standing on that diagonal, using it as one side.

2:59This is the construction from before, and its one fact is the fact we need. The square on the diagonal has exactly twice the area of the square you started with. Not roughly twice. Exactly, and it was proved by cutting both squares into copies of one triangle and counting them. Our first square has area one, so the new square has area two. And notice what has just happened. We have the area of a square whose side we cannot measure.

3:32We know the area without knowing the length. That is the detour paying for itself. So write that area down twice, in two different ways. The first way is the one we just got. It is two square units. The second way is the ordinary way you write any square's area. Side times side. Call the side c, because it is the hypotenuse we are after. Then the area is c times c.

4:03Two ways of writing the same area, so they are equal. c times c is two. And that is the answer, in the only form it has. The hypotenuse is the length whose square is two. It gets written with a mark in front of the two, and that mark means exactly that: the length whose square is the number under it. Nothing in that argument used the number one.

4:30So put a letter in its place. Two equal sides, each of length a, with a right angle between them. The square on a leg has area a times a. The square on the hypotenuse has twice that. And it also has area c times c, because that is what a square's area is. So c squared is two a squared. That is the relation, and it holds for every isosceles right triangle there is.

5:00Notice the order of the argument. This did not come from a general theorem about right triangles. It came from one construction about squares, and the general theorem is still ahead of us. Now the slip that this relation invites, because it catches almost everyone once. c squared is two a squared. So c is two a. Cancel the squares and go home. That is wrong, and it is worth seeing exactly how wrong.

5:32If c were two a, then c squared would be two a, times two a, which is four a squared. Four a squared, not two a squared. The doubling has happened twice over instead of once. Undoing a square reaches everything inside it, not just the part you were looking at. Keep the relation in the form that is true, and read the length off the square at the end.

5:58Put it to work. Two equal sides of twelve. c squared is two times twelve times twelve, which is two hundred and eighty-eight. Two hundred and eighty-eight is not a whole number times itself, so there is no exact whole answer to write. That is not a failure. It is the answer, and you can pin it down as tightly as you like. Sixteen times sixteen is two hundred and fifty-six, which is under.

6:28Seventeen times seventeen is two hundred and eighty-nine, which is over. So the hypotenuse is between sixteen and seventeen, and both halves of that were checked. Notice how far that is from the wrong answer of twenty-four. The relation runs the other way just as well. Suppose you are given the hypotenuse and asked for the equal sides. If c squared is seventy-two, then two a squared is seventy-two, so a squared is thirty-six.

7:02And thirty-six IS a whole number times itself. Each equal side is exactly six. Check it forwards: two times six times six is seventy-two. It closes. Now a hypotenuse of ten. A hundred is two a squared, so a squared is fifty. Fifty is not a whole number times itself, so we bound it instead. Seven sevens are forty-nine and eight eights are sixty-four, so each side is between seven and eight.

7:36And a bound can always be tightened, by exactly the same method. Seven point nought, squared, is forty-nine. Seven point one, squared, is fifty point four one. Fifty sits between them, so the side is between seven point nought and seven point one. Two squarings, and the answer is pinned to a tenth. Two more would pin it to a hundredth. Here is the same work on five triangles at once, with equal sides three, four, six, eight and nine.

8:10The squares of their hypotenuses are eighteen, thirty-two, seventy-two, one hundred and twenty-eight, and one hundred and sixty-two. And to one decimal place: between four point two and four point three, five point six and five point seven, eight point four and eight point five, eleven point three and eleven point four, twelve point seven and twelve point eight. One last thing, and it is free. A square of side five. How long is its diagonal?

8:46You have already done it. The diagonal is the hypotenuse of an isosceles right triangle with sides of five, so its square is two times five times five, which is fifty. The same fifty as a moment ago, arriving from a question that sounded completely different. And a warning to close on. The relation is about the RIGHT ANGLE, not about the two sides being equal. An isosceles triangle with two sides of five and a third side of six is a perfectly good triangle, and thirty-six is not fifty.

9:27Equal sides are not enough. It is the right angle between them that does the work.