Brahmagupta's formula, and Heron's as the case where a side vanishes

0:00Three side lengths fix a triangle completely, and Heron's formula turns them straight into its area. So here is the obvious next question. Four sides. A four-sided figure - the short word for it is a 4-gon, and a quadrilateral is the same thing under a longer name. Hand me four lengths. Is there a formula that turns those into the area? There is one, it is Brahmagupta's, and it is worth being suspicious of before you meet it.

0:34Because the answer to the question exactly as I asked it is no. Here is why. Take four rods of length three and pin them at the corners. Stand it up square and the area is nine. Lean it, and the area is seven point two. Lean it further and it is five point four. Nothing cut, nothing stretched - every side is still three - and the area has fallen by nearly half.

1:03Four lengths do not fix a four-sided figure, and that is not one awkward example. Sweep seven thousand two hundred and twenty-eight four-sided figures whose corners and sides are all whole numbers, sort them by their four side lengths, and fifty-nine of the hundred and ninety-eight sets of sides hold more than one area between them. One set holds nine different areas. So if Brahmagupta's formula works, it is buying its answer from somewhere.

1:39What would be enough? One angle of the figure would do it. So would the length of a diagonal, or the angle at which the two diagonals cross. Any of those pins the hinge, and a pinned hinge has one area. But there is a fourth kind of answer, and it is the one the formula uses. Not a measurement added - a property imposed. A condition on the figure itself that leaves it no room to flex.

2:10The condition is one word. Cyclic. All four corners lie on a single circle. Draw a circle, mark four points on it, join them in order: that is a cyclic 4-gon, and its four sides are enough again. That is not obvious, so it is worth testing. One thousand one hundred of the figures in the sweep have all four corners on a circle. Grouped by their side lengths they give a hundred and eight sets, and the number of sets where two of them came out with different areas is zero.

2:45Not merely the four lengths - even the order they run round the figure makes no difference. So here it is. Add the four sides, halve the total, and call it s - the same semi-perimeter as before. The area is the square root of s minus a, times s minus b, times s minus c, times s minus d. Now put it underneath Heron's formula, and the family resemblance is impossible to miss.

3:17Heron has an s out in front and three brackets after it. Brahmagupta has four brackets and no s in front. That is the whole difference between them. It is not a coincidence, and we will come back to it. Before using it, read it, the same way we read Heron's. Each bracket, s minus a, is the other three sides minus this one, all halved. So a bracket is positive exactly when its own side is shorter than the other three together - the condition for four lengths to close up at all.

3:55Feed in two, three, four and twenty. The semi-perimeter is fourteen and a half, and the last bracket is minus five and a half. The product goes negative and the formula refuses. And it cannot be fooled, for the same reason as before: across all seven hundred and fifteen quadruples of whole numbers from one to ten, the most brackets that ever go negative together is one. First check, on the easiest figure there is.

4:28A rectangle. Its four corners do lie on a circle, so the formula is allowed near it. The sides are a, b, a, b, so s is a plus b. Now subtract each side in turn: s minus a is b, s minus b is a, and then b and a again. Four brackets: b, a, b, a. Multiply them and you get a squared, b squared, whose root is a b - what a rectangle's area was long before anyone had heard of Brahmagupta.

5:03Second check, and harder. An isosceles trapezium: two parallel sides, and two equal slants. Write the top as two a and the bottom as two b, with slants c. Doubling those labels keeps the height clean, and in a moment it will cost us something. The perimeter is two a plus two b plus two c, so s is a plus b plus c. The four brackets come out c plus b minus a, then c plus a minus b, then a plus b, twice.

5:36The first pair multiply to c squared minus, b minus a, squared. So the area is a plus b, times the root of that. And that root is the height: the overhang is b minus a, so Baudhayana and Pythagoras hand you exactly that number. Which makes the formula read a plus b, times h - half the sum of the parallel sides, times the height, walking back out. Now a number, and the trap.

6:09Parallel sides six and fourteen, both slants five. In the doubled letters a is three, b is seven, c is five, so s is fifteen. The brackets are s minus six, s minus fourteen, and s minus five twice: nine, one, ten and ten. The product is nine hundred and the area is thirty. The old way agrees: overhang four, height three, and half of six plus fourteen times three is thirty.

6:40Now the trap. Forget the doubling and reach for s minus a and s minus b - twelve and eight. The product comes to nine thousand six hundred, the area to nearly ninety-eight. But that trapezium fits inside a rectangle fourteen by three, which is forty-two. More than twice the box the figure lives in, so it is not an answer. A word about the shape of what is going on.

7:10A square is a rectangle with the two sides made equal, so the square's area formula is the rectangle's with b set to a. The square is the special case; the rectangle is the generalisation. An isosceles right-angled triangle is a right-angled one with the legs made equal, and the theorem collapses to the hypotenuse being root two times a leg. Generalising does not mean getting vaguer. It means dropping a condition, so the result covers more.

7:47So: is Heron's formula the special case here? Shrink the fourth side. Let d go to zero, so two corners slide together and the 4-gon becomes a triangle. Every triangle is cyclic - three points not in a line lie on exactly one circle - so the formula ought to survive the trip. With d at zero the semi-perimeter is the triangle's own, and the fourth bracket, s minus zero, is simply s.

8:19Four brackets become s, s minus a, s minus b, s minus c. That is Heron's formula exactly, on all one thousand seven hundred and twenty-eight triples tested. But notice what just happened. A figure with a side of length zero is not a four-sided figure, so this is a limit, not a line of a proof. It is legitimate because the area moves continuously as the side shrinks - through six settings the gap closes to under one part in a hundred thousand, and is never once actually zero.

8:58It explains the resemblance; it does not prove Heron. One last thing, which everything on the board already makes available and almost nobody says. Feed three, three, three, three into Brahmagupta. s is six, all four brackets are three, the product is eighty-one, the area is nine. Nine - the square, the first and largest of the three hinge positions. That is not a coincidence. Hand the formula a figure that is not cyclic and it still returns a number, and the number is wrong.

9:36Across six thousand one hundred and twenty-eight non-cyclic figures in the sweep it lands above the true area every single time, and below it not once. And no figure anywhere in the sweep beats it. The one thousand one hundred that reach it exactly are precisely the one thousand one hundred whose corners lie on a circle. So Brahmagupta's formula is not only the area of the cyclic figure - it is the most area those four lengths can enclose.