Heron's formula: area from the three sides alone

0:00You already have a formula for the area of a triangle. Half the base, times the height. It always works, and it has one inconvenience: a height is not something a triangle comes with. It has to be found - somebody drops a perpendicular, or reaches for the theorem of Baudhayana and Pythagoras, before the multiplication can even start. So here is a fair question. Suppose all you are handed is the three side lengths.

0:33No height, no angle, no drawing. Is there a formula that turns those three numbers straight into the area? Before hunting for such a formula, ask whether one can exist at all. Take three sticks of fixed length and pin them at the corners. There is exactly one triangle you can build. You can flip it over, but you cannot flex it - a triangle has nowhere to hinge. Now take four sticks and pin those.

1:03That figure moves. Push a corner and it leans, and it keeps leaning, through shape after shape, without one side changing length. So four lengths do not decide a four-sided figure. Three lengths decide a triangle completely, and if they decide the shape, they have already decided the area. That is worth measuring rather than asserting. Here is a rhombus with all four sides five units long. Standing up straight its area is twenty-five.

1:37Lean it and the area drops to twenty; lean it further and it is fifteen. Same four sides, three different areas. Sweep sixteen thousand seven hundred and eighty-six four-sided figures on a grid and sort them by side length. Of the four hundred and nineteen sets of four sides, two hundred and sixteen hold more than one area between them. Do the same with six thousand seven hundred and sixty-eight triangles, and of their hundred and seventy-two sets of three sides, the number holding more than one area is zero.

2:18So the area is already decided by a, b and c. The only question left is which expression in them it is. The answer needs one piece of bookkeeping first. Add the three sides, then halve the total. That half-total is written s, and it is called the semi-perimeter. Half. Not the whole perimeter, half of it. That confusion is the commonest way this formula is got wrong, and a nasty one, because the wrong answer still looks like an answer.

2:53And now the formula itself. The area is the square root of s, times s minus a, times s minus b, times s minus c. Do not use it yet. Read it. Four factors sit under one root sign, three of them subtractions. Write s out in full and s minus a becomes b plus c minus a, all halved. So that bracket is positive exactly when b plus c is bigger than a - when the other two sides are long enough to reach across the first.

3:30The formula is carrying the condition for a triangle to exist inside its own brackets. Feed in two, three and nine. The semi-perimeter is seven, the three brackets are five, four and minus two, and the product goes negative. There is no square root of a negative, so there is no area. The formula has refused. Refusing is only worth something if it cannot be talked out of. Two negative brackets would multiply back to a positive, handing you a confident number for lengths that make no triangle.

4:11So can two go negative at once? If s minus a and s minus b were both negative, a and b would each beat s, so a plus b would beat twice s - which is a plus b plus c. That forces c below zero, and no side is. Across all six hundred and eighty triples of whole numbers from one to fifteen, the most brackets that ever go negative together is one.

4:38Three hundred and seventy-two of those close up, fifty-six fall exactly flat, two hundred and fifty-two cannot close at all, and the sign of the product gets every one right. Two, three and five is the flat case: the last bracket is zero, so the area is zero. A formula that strange deserves testing. Start with the triangle you know best: equilateral, every side a. The perimeter is three a, so s is three a over two.

5:13Subtract a and you get a over two - and all three sides being a, that is every bracket. Multiply: three a over two, times a over two, three times over. That is three, a to the fourth, over sixteen. Take the root and the area is root three over four, times a squared. Now the old route. The height splits the base into halves of a over two, so h squared is three a squared over four.

5:43Half of a times that height gives the same number back. Second test, one degree harder. Two equal sides of length a, and a base of two b. The perimeter is two a plus two b, so s is simply a plus b. The four brackets come out a plus b, then b, then b, then a minus b. Multiply in pairs: the middle two give b squared, the outer two give a squared minus b squared.

6:14So the area is b times the root of a squared minus b squared. And the height route: the base halves into b and b, so the height is that same root, and half of two b times it is the same thing. Across a hundred and five different isosceles shapes, not one disagreement. Third test, and this one hides a bonus. Sides three, four and five. The perimeter is twelve, so s is six, and the brackets are three, two and one.

6:48Six times three times two times one is thirty-six, whose root is six. Six square units. Now look at those three numbers again. Nine plus sixteen is twenty-five - three squared plus four squared is five squared, so the triangle has a right angle, and nobody put it there, it was found. With that right angle, three is a base and four is its own height, and half of three times four is six.

7:18The same six. Three tests, three agreements. That is not a proof, and it is better to see why than be told. Here is a different formula, and it is not Heron's. It agrees with Heron on every equilateral, every isosceles and every right-angled triangle - all three tests just run, and every relative of them. Fifteen equilateral cases, a hundred and sixty-nine isosceles, four right-angled, and it matches every one exactly.

7:55Then hand it sides four, five and six. Heron gives nine point nine two two. This one gives eleven point zero seven nine. A positive number, a believable number, and wrong. Checks catch carelessness; they do not establish a general truth. A proof does exist - it runs on the theorem of Baudhayana and Pythagoras together with the difference of two squares, used over and over - and it is a job for later.

8:28Two more formulas give the same area, and both are borrowed from circles. Draw the circle through all three corners - the circumcircle - and call its radius capital R. Then the area is a times b times c, divided by four R. Now draw the circle that fits inside, touching all three sides - the incircle, radius small r - and the area is r times the semi-perimeter. There is s again, turning up in a second formula.

9:01Both were put to seventeen triangles with whole-number sides: the inner centre sits the same distance from all three sides, each touching point lands strictly inside its own side, and both formulas return the area exactly. But notice the cost - each wants a radius, and a radius is something extra to go and find. So which route do you take? Heron, when three sides are what you were given.

9:34Sides eight and eleven with perimeter thirty-two: the third is thirteen, s is sixteen, the area is eight root thirty. But handed an area and asked for a length, Heron is the long way round. Base ten, area sixty, two equal sides: the height has to be twelve, and twelve with five gives thirteen. And an area of fifty-four alongside one side of twelve fixes no triangle at all - seven genuinely different ones have both.

10:07It is the right angle that pins it, and then the legs are twelve and nine, the third side fifteen, the perimeter thirty-six. Three sides fix an area. An area, on its own, fixes nothing.