Any polygon is a pile of triangles

0:00Here is a quadrilateral. Not a square, not a rectangle, not a rhombus, not a trapezium. Four different sides, no two of them parallel, and nothing marked on it at all. So: what would you ask to have measured? Most people reach for the sides. Hold that thought - we will find this shape's area without knowing one of them. Join two opposite corners. One straight cut, and the shape falls into two triangles.

0:32That is the whole idea. We know a triangle: half the base, times the height. So if a shape can be cut into triangles that do not overlap, its area is the pieces, added up. Cut from the other corner instead and you get two different triangles, of different sizes. They still add to the same total. The pieces were never the point; the shape was. Now a shape with some numbers on it - long, thin, and just as nameless as the first.

1:06The diagonal across it is twenty-two. From the corner above it, drop a perpendicular onto that diagonal: three. From the corner below it, drop another. That one is three as well. That is everything. Three numbers. The upper triangle stands on twenty-two with a height of three, so it holds thirty-three. The lower one does the same. Together, sixty-six. And look at what we were never given. Not one of the four sides, not a single angle. Three numbers were enough.

1:43There is a tidier way to see that arithmetic. Half of twenty-two times three, plus half of twenty-two times three again. Both halves stand on the same diagonal, so take it out: half of twenty-two, times three plus three. Half the diagonal, times the sum of the two drops. That is the whole formula, and it never needed a name for the shape. Uneven drops are no trouble. Go back to the first quadrilateral.

2:15Its diagonal is fifteen; the drops onto it are five and seven. Half of fifteen, times twelve, is ninety - exactly what its four corners say it is. Read it across the other diagonal instead and you get three different numbers, and the same ninety. A pentagon, then. Five sides. Pick any corner and join it to the corners it is not already attached to. Two cuts, three triangles. Any corner works, and each gives a different three.

2:50A hexagon takes four. A twelve-sided shape takes ten. Always two fewer triangles than there are sides. And here is the claim it all rests on: every polygon can be cut into triangles. So the triangle formula, plus a pair of scissors, handles every straight-sided shape there is. That claim is true - and it is being asserted to you, not shown. Worth knowing which is which. Because the instruction usually given in its place is false. Join a diagonal, people say.

3:28Here is a quadrilateral with one corner pushed inward, pointing back into the shape. Its area is twenty-five. Both diagonals are whole numbers long, five and ten, so either could have been measured. The first lies inside and cuts two triangles totalling twenty-five. The second leaves the shape almost at once and cuts nothing. Use it anyway and you will confidently report seventy-one. It gets worse. Drive a thin spike into the side of a square, and a cut can leave the shape and come back. The middle of that cut is comfortably inside; the cut is still useless.

4:11A shape can always be cut into triangles. Which cuts work is a separate question, and you have to look. Back to the four sides you wanted to measure. A rectangle four by three, so twelve. Take two opposite corners and lean it over. Every side keeps its length exactly: four, three, four, three. But it is now nine point six. Four sides settle nothing. A four-sided frame flexes, and its area moves the whole time while every side stays put. What changed was the diagonal, from five to something shorter.

4:54So: four sides and a diagonal is five numbers. A diagonal and two drops is three. Fewer measurements is what makes one cut better than another. Not fewer triangles. Fewer numbers you have to go and find. Which brings us to the cheapest shape of all. A regular hexagon. How many lengths would you need? Join the centre to all six corners. Six triangles - and because the hexagon is regular, every one is equilateral, on the hexagon's own side.

5:30One length. That is the whole cost. Take that small triangle as the unit. The hexagon is six of them. Now cut it differently. Join two opposite corners: that cut is twice the side and runs straight through the centre. Above it sits a trapezium of three. From the centre, cut down to the corner below: an equilateral triangle of one, and a rhombus of two. Three, one and two. They total six, and the ratio is counted rather than calculated.

6:06Take a cut that skips a single corner instead and it is shorter, misses the centre, and leaves nothing that ratio would recognise. Cutting a shape up is not the only way to use triangles. A rectangle, eighteen by ten, holding a hundred and eighty, with a shaded region inside it reaching from one bottom corner to the other. Two triangles are left over at the top. The left one has legs of six and ten, so thirty. The right one has eight and ten, so forty.

6:43A hundred and eighty, less thirty, less forty, leaves a hundred and ten. You could cut the shaded region into two triangles and add them instead. Same answer. Subtracting is a decomposition too, and here it is the shorter road. You were handed more lengths than you needed; part of choosing a cut is deciding which numbers it uses. A rectangle, twelve across and seven up, so eighty-four. Two triangles inside it: one on the whole top edge, one on the whole bottom edge, their tips meeting at a point in the middle.

7:23What fraction of the rectangle is shaded? The point is drawn well off centre, so you might expect the answer to depend on where it sits. The two pieces hold twenty-four and eighteen - not equal at all. But twenty-four and eighteen make forty-two, and forty-two is half of eighty-four. Move the point. Anywhere. The two pieces change every time; the total never moves. Both triangles stand on a base of twelve, and their two heights always add up to seven, because between them they span the rectangle. Half of twelve times seven is forty-two, whatever the split.

8:07One more, running the same trick backwards. Three equal squares set in an L, and a straight cut from the bottom left corner up to the top. That cut crosses the middle line exactly at its midpoint. The red region stands on a side two squares long, held one square away, so it holds exactly one square's worth of area. The blue sliver has half a side for its base and a whole side for its height, so it is a quarter of a square.

8:40If the red region is forty-nine, a square is forty-nine and the blue one is twelve point two five. And if red and blue together make a hundred and eighty, then since they come to a quarter more than a square, each square is a hundred and forty-four. Which is twelve squared - the sign that we read the picture right. Last one, and it is open. Given a quadrilateral, construct another one with exactly half its area.

9:11One that works: draw a diagonal, mark the midpoints of two of the sides, and join them in. The shape you get is half. It works because a line from a corner to the midpoint of the opposite side cuts a triangle into two equal halves - not congruent halves, just equal ones. Do it on both triangles and you have halved the lot. Another that works: join the midpoints of all four sides. Half as well, and a completely different shape.

9:45And one that does not, although people offer it: join the midpoints of two opposite sides. On this shape, which holds eleven, it splits five and six. Which leaves the shortcuts. A parallelogram is base times height. A rhombus is half the product of its diagonals - but only because its diagonals cross square on. A trapezium is half the sum of the parallel sides, times the gap. Every one of those is a pile of triangles somebody has already added up for you. They are worth having. They are never necessary.

10:25One cut, one formula, and every straight-sided shape in the world.