Area of a rhombus from its diagonals

0:00Here is a rhombus. Four equal sides, leaning over. Its opposite sides are parallel, both pairs. So a rhombus is a parallelogram that happens to have all four sides the same. Which means we already know its area. Base times height. This side is twelve point five, the distance across to the side opposite is twelve, and that gives one hundred and fifty. No new formula needed. And yet there is a second formula for a rhombus, and everybody learns it. Half the product of the diagonals.

0:39So the question here is not how to use it. It is: what is that second formula actually about? Because being a rhombus was already covered. A rhombus comes with two properties, and it matters enormously that we keep them apart. The first: all four sides are equal. The second is about the diagonals, and it is really two facts wearing one coat. They cut each other exactly in half. And they cross at a right angle.

1:12Two separate things. Cutting in half. Crossing squarely. Hold on to that. The long diagonal is twenty, the short one fifteen. Because they bisect each other, the halves are ten and seven point five. Because they meet squarely, those two halves stand at a right angle - so the side of the rhombus closes off a right-angled triangle, and it comes to twelve point five. That is a three, four, five triangle scaled up by two and a half.

1:46Now cut the rhombus along one diagonal. Along the short one, the fifteen. Two triangles fall out, and they are both isosceles - two equal sides each, because all four sides of the rhombus were equal. They stand on the same base: that diagonal of fifteen. And their heights are the two halves of the other diagonal. Ten above, ten below. So each one holds half of fifteen times ten. Seventy five. And seventy five plus seventy five is one hundred and fifty, which is what the whole shape holds.

2:25That is already an answer. But the way this pair gets rebuilt is where the famous formula comes from. Take one of those isosceles triangles on its own. Base fifteen, height ten. Drop the perpendicular from the apex to the base. It lands in the middle, and it splits the triangle into two right-angled halves. Swing each half outward about the midpoint of its slanted side. What you get is a rectangle. Nothing was added and nothing thrown away, so it holds what the triangle held.

3:03Look at its measurements. It is seven point five wide - half of the base. And it is ten tall - the full height. Half the base, all of the height. Remember that, because it is about to explain the word half. Do the same to the other triangle. Same base, same height, so the same rectangle: seven point five by ten. Now set one on top of the other.

3:31The stack is seven point five wide and twenty tall. And now read those two numbers back into the original rhombus. Twenty is the whole of the long diagonal. Seven point five is half of the short one - the one we cut along. So the rhombus is now a rectangle whose sides are one whole diagonal and half of the other. Multiply: one hundred and fifty. There is the formula. Half the product of the diagonals.

4:03And now the thing that is worth stopping for. Where did the half come from? Not from the algebra. Nobody halved anything at the end. It arrived much earlier, when each triangle became a rectangle only half as wide as its base. The half lives in the width. Which tells you what goes wrong when this is misremembered. Suppose you take the rectangle's sides to be the two diagonals, twenty and fifteen.

4:35Then you get three hundred. The shape holds one hundred and fifty. You have doubled it. That is the single most common slip in the whole derivation, and it is not an arithmetic slip. It is forgetting which of the two diagonals got halved. There is a second route to the same formula, and it is shorter, and it tells us far more. Go back to the two triangles. Both stand on the same base - one whole diagonal. Their heights are the two pieces of the other diagonal.

5:10So the area is half the base times the first height, plus half the base times the second. Collect the common half and the common base: half the base times the sum of the heights. And those two heights add up to the whole of the other diagonal. Half of one diagonal times the other. The same formula, with no rectangle anywhere. Now look at what that argument used. Only that the heights lie along one straight line, at a right angle to the base.

5:43It never used the four equal sides. It never used the bisection either - the pieces did not have to be equal, only to add up. If the argument never used those things, the formula should survive without them. So let us check. Here is a kite. Its diagonals still cross squarely. But one of them is not cut in half - the crossing sits at the middle of the short diagonal and nowhere near the middle of the long one. And its sides are certainly not all equal.

6:20The diagonals are sixteen and twelve, so half their product is ninety six. Now measure the kite itself, from its four corners. Ninety six. Take away the bisection, take away the equal sides, and it holds. So try taking away the other one instead. Keep the bisection, and lose the right angle. A rectangle. Its diagonals cut each other dead centre, but they cross at a lazy angle, not a square one.

6:53Sixteen by twelve, so both diagonals are twenty. Half their product says two hundred. The rectangle holds one hundred and ninety two. Out by eight. Here is a quadrilateral with no name at all, diagonals thirteen and ten. The formula says sixty five. It holds sixty three. So there it is. The bisection can go and the formula survives. The right angle goes and it collapses. Half the product of the diagonals is a fact about any quadrilateral whose diagonals cross squarely. The rhombus is just the most famous shape that does.

7:37One more way of seeing it, needing no cutting at all. Draw a rectangle around the rhombus with its sides parallel to the diagonals. It touches all four corners. That box is fifteen wide and twenty tall - the two diagonals exactly - so it holds three hundred. Now look at what is left over. Four triangles in the corners of the box, outside the rhombus. Each is a copy of one of the four inside. Thirty seven and a half each.

8:10Four in, four out. The rhombus is exactly half of its box, and half of three hundred is one hundred and fifty. Same answer - and this time the half is something you can see. The two formulae have to agree, so we can play them against each other. From the diagonals the area is one hundred and fifty, and the side is twelve point five. So twelve point five times the height is one hundred and fifty, and the height is twelve.

8:44A length nobody measured. It came out of making the two formulae agree. Try a flatter one. Diagonals twenty four and ten, halves twelve and five, so the side is thirteen and the area is one hundred and twenty. The height is one hundred and twenty over thirteen, a little over nine. And notice: its long diagonal, twenty four, beats its side. But its short diagonal is ten, and the side is thirteen.

9:15So a diagonal is not always longer than a side. Squash a rhombus far enough and the short one gets as small as you please. Finally, run the whole construction backwards. Start with a rectangle, eight by three, holding twenty four. The height was one whole diagonal and the width half the other. So read it in reverse: one diagonal is eight, the other is twice three, six. Half-diagonals four and three, so the side is five, and half of eight times six is twenty four. The rectangle we started from.

9:55So keep the formula, by all means. Half the product of the diagonals. Just remember what it is a statement about. Not four equal sides. Not diagonals cutting each other in half. Only this: two diagonals meeting at a right angle. Everything else was scenery.