Combining two different squares: a² + b² = c²

0:00Two squares of the same size can be combined into one, and the thing that does the combining is a diagonal. Now drop the sameness. Here is a square of side three. Its area is nine. And here is a square of side five. Its area is twenty-five. Together they hold thirty-four units of area. The question is whether a single square can hold exactly that much, and whether one cut can still find its side.

0:31It can. And the cut is still a diagonal. It is just no longer the diagonal of a square. Before we build it, two moves that feel right and are not. The first is to add the sides. Try it where the answer is already known. A right triangle with legs of three and four has a longest side of exactly five. Adding the legs gives seven. Seven is not five, and it is not close to five either. The two differ by two whole units.

1:05What adding actually measures is the bent path: out along one leg, then up the other. Of course that is longer. It is the way round. The second wrong move is to take a square root of a sum by rooting the parts. Nine and sixteen make twenty-five, and the root of twenty-five is five. But the root of nine plus the root of sixteen is seven again. Roots do not come apart over a sum. That is not a rule anyone forgot to teach you. It is simply false.

1:41So build it instead. Set the two squares side by side on a single straight line, the smaller one first. Read the baseline. Three, then two, then three: eight in all. That middle stretch of two is the difference between the two sides, and it is the reason this construction is not the old one. Now measure five along the baseline from the left, and mark that point. It sits five from one end and three from the other. Both of the given sides meet there.

2:15Everything that follows comes out of that one point. From the marked point, cut to the far top corner. Look at what that line is the diagonal of. A rectangle three wide and five tall, standing at the far end of the larger square. When the two squares matched, that rectangle was a square and this was its diagonal. It still is a diagonal. The shape it belongs to has just stopped being square.

2:46Now cut the other way, from the same point up to the top corner of the smaller square. Two cuts. Two triangles fall away, and a five-sided piece is left in the middle. And the two triangles are copies of each other. Each has a side of three and a side of five meeting at a right angle. So the two cuts are the same length. Call that length c. We do not yet know what it is.

3:15Three pieces, then. Each triangle covers seven and a half units of area. The middle piece covers nineteen. Seven and a half, and seven and a half, and nineteen. Thirty-four. Which is what we started with, and has to be, because cutting a figure up does not change how much of it there is. Keep hold of that number. It is the only thing that survives the next step unchanged.

3:44Now move the two triangles, and nothing else. The middle piece stays exactly where it is. Watch it, not them. The first triangle swings up to sit on top of the larger square. The second swings up to stand over the smaller one. And the outline that closes is this one. Four sides. Every one of them is a long side of one of the four triangles, so every one of them is c.

4:12Two squares went in. Something with four equal sides has come out. Before asking what that shape is, be sure nothing was smuggled in. The pieces are the same pieces. Seven and a half, seven and a half, nineteen: thirty-four, both before and after. But equal areas alone would not be enough, and it is worth seeing why. Take those same three pieces and lay them against the new outline as they were. They add up to exactly its area, and they do not fit inside it.

4:48Now take only the two triangles. Those do fit inside, and they leave most of the figure bare. A genuine covering needs both halves at once: every piece inside, and the areas adding exactly. This one has both. So the new figure has four equal sides. That does not make it a square. Here is a figure with four sides of five each, and it is not a square. It is a rhombus, leaning over.

5:19A rectangle fails the other way: its diagonals match, and its sides do not. What settles it is the corners. All four corners of the new figure are right angles, and its two diagonals come out equal, which is the same fact counted a second way. So it is a square, and its area is thirty-four. Its side is c. So c squared is thirty-four, which is nine plus twenty-five: the two areas we began with.

5:52Three squared plus five squared. That is the theorem. One step in that argument deserves a harder look than it usually gets. To show a corner is square, you take the whole angle sitting there and subtract the two triangle angles that lean against it. But the whole angle is not the same at every corner. One of these corners sits partway along a straight edge of the old figure. There the angle to subtract from is a straight one.

6:25Another sits at a right-angled corner of the old figure. There it is a right angle you are subtracting from, and the sum comes out differently. A third does not touch the old outline at all. Run the straight-edge version everywhere and you will reach the right answer by a step that is not true. The conclusion survives. The reasoning does not. Now shrink the larger square until it matches the smaller.

6:56The middle stretch of the baseline runs down to nothing. The marked rectangle becomes the whole square again, and the cut becomes its own diagonal. And the result reads: three squared plus three squared, which is twice nine, eighteen. Which is exactly what the equal-squares construction gave. The old result did not get replaced. It got absorbed. That is usually the sign that a generalisation is the right one: the case you already had is sitting inside it, unharmed.

7:33Now it can be used, and it goes both ways. Forwards: legs of three and four give twenty-five, and twenty-five has a whole root. Five. Legs of five and twelve give one hundred and sixty-nine. Thirteen. Backwards it subtracts. A longest side of seventeen with one leg of eight leaves two hundred and twenty-five for the other leg. Fifteen. But whole legs do not promise a whole answer. Five and seven give seventy-four. Eight and twelve give two hundred and eight. Seven and twelve give one hundred and ninety-three.

8:14Not one of those has a whole root. Seventy-four sits between eight and nine, two hundred and eight between fourteen and fifteen, one hundred and ninety-three between thirteen and fourteen. Both ends of each of those, checked separately. The theorem gives you an area every time. Whether that area has a nameable side is a different question, and mostly the answer is no. Two last places it reaches. A rhombus with diagonals of twenty-four and seventy. The diagonals cross at right angles and cut each other in half, so each quarter is a right triangle with legs twelve and thirty-five.

9:00Twelve squared plus thirty-five squared is one thousand three hundred and sixty-nine, and that one does have a whole root. Every side of that rhombus is exactly thirty-seven. And it handles differences as well as sums. Take a square of side five away from a square of side seven and twenty-four is left, which is a square of side something between four and five. One honest note to end on. We used this backwards, to find a leg, and to say a corner was square. Reading it backwards is a second statement, and we have not shown it here.

9:42Baudhayana had the rule long before anyone had a proof of it, stated for exactly this figure: two squares of different sizes, and the cut that combines them. One line, drawn from one point, still does the whole job.