A segment as what is left when the triangle is taken away

0:00A circle of radius twenty one, and a chord across it opening a hundred and twenty degrees at the centre. The piece caught between that chord and the arc above it measures about two hundred and seventy one. Now halve the angle: sixty degrees, same circle, same kind of piece. If you expect about a hundred and thirty five, you are about to be wrong by a factor of three.

0:28It measures about forty. Halving the angle did not halve the piece. It cut it to less than a seventh of what it was. So whatever argument gave us the wedge cannot be run again here. Be precise about what failed, because the wedge behaved perfectly: four hundred and sixty two at a hundred and twenty degrees, two hundred and thirty one at sixty. Exactly half, as promised. The piece we want sits inside that wedge. And between the two of them there is one other thing: a triangle, standing on the chord, with its point at the centre.

1:10Watch that triangle as the angle halves. It does not halve. It does not change at all. That is the whole failure. One of the two pieces scales with the angle and the other does not. The picture that rescues this is almost too simple to be worth drawing. Take the arc, and join each of its two ends to the centre. You now have a wedge, and inside it, two pieces.

1:39One is the piece past the chord: the arc on one side, the chord on the other. The other is a triangle - two radii and the chord, three straight edges and no curve anywhere. If a curve has crept into your triangle you are subtracting the wrong region, and that is the commonest way to get this wrong. The claim is that those two pieces are the wedge. Not roughly - exactly.

2:08Exactly is a strong word, and two things have to be true for it: nothing counted twice, and nothing missed. So take every point of the wedge and ask which side of the chord it lies on. Past it, and the point belongs to the piece; on the centre's side, to the triangle. Not one point answers both. At nine angles, from fifteen degrees to a straight one, the count provably on both sides comes out at nought every time.

2:41And what is left over is the chord itself: a line, which has length and no width, and which measured against the wedge it lies across comes to about two parts in ten thousand even at its worst. So the piece plus the triangle is the wedge, nothing overlapping and nothing left over, and therefore the piece is the wedge minus the triangle. No approximation entered anywhere. The only thing approximate in what follows is whatever number you decide to use for pi.

3:18So the answer is a subtraction, and its two halves behave completely differently. Double the angle and the wedge doubles: six angles tried, and it doubles at every one of them. Double the angle and neither the piece past the chord nor the triangle doubles, at any of those six. The easy half is the wedge: its share of the turn, times the whole disc. The hard half is the triangle. Its size depends on the angle, but not in proportion to it.

3:52In fact it climbs to the right angle and falls away again, which is behaviour no proportional quantity can have. That is why this is the first area here that needs trigonometry: not because circles are hard, but because that triangle is not a fixed fraction of anything. Come back to those two triangles, because something about them is easy to miss. The triangle at sixty degrees and the triangle at a hundred and twenty have the same area. Not roughly. The same.

4:29Nor is it confined to those two. Every pair of angles that adds to a straight one gives two triangles of identical size: thirty and a hundred and fifty, forty five and a hundred and thirty five, seventy five and a hundred and five. But they are emphatically not the same triangle. At sixty degrees it is tall and narrow; at a hundred and twenty, short and wide. The chords differ and the heights differ.

5:01Same area, different shape - which matters, because in a moment we build the triangle out of its chord and its height, and the area alone cannot tell those two apart at all. Now one all the way through, with the numbers we opened on. Radius twenty one, the chord opening a hundred and twenty degrees at the centre, pi taken as twenty two sevenths. The whole disc first: twenty two sevenths times twenty one squared. Twenty one squared is four hundred and forty one, seven goes into that sixty three times, so this is twenty two sixty threes. One thousand three hundred and eighty six.

5:48The wedge is its share of the turn, and a hundred and twenty over three hundred and sixty is a third. A third of one thousand three hundred and eighty six is four hundred and sixty two. Easy half done. Now the triangle. The triangle has two sides of twenty one with a hundred and twenty degrees between them, and no obvious base and height. So drop a perpendicular from the centre onto the chord, and call where it lands the foot.

6:21That splits the triangle into two right angled ones, and those two are congruent: they share the perpendicular, their long sides are both radii, and each has a right angle at the foot. Now read off two things, because that one congruence gives you both and you need both. First, the two halves of the chord are equal, so the foot is the middle of it. Second, the two halves of the angle at the centre are equal, so the perpendicular has bisected the angle.

6:55The foot is not the middle because it looks like it. It is the middle because of that congruence, and the same congruence hands you the half angle. And the half angle is what makes a hundred and twenty degree problem workable, because sixty is an angle you have a ratio for. Take one of the two right triangles. Hypotenuse twenty one, and the angle at the centre sixty degrees, half of the hundred and twenty.

7:26The side along the perpendicular is adjacent to that angle, so it is twenty one times the cosine of sixty. Cosine of sixty is a half, so the height is ten and a half. The side along the chord is opposite, so it is twenty one times the sine of sixty. Sine of sixty is root three over two, so that half is twenty one root three over two, and the whole chord is twenty one root three.

7:57Notice the pattern under those two. The chord is twice the radius times the sine of the half angle, and the height is the radius times its cosine. That is general, and it is what turns a problem about an arc into one about a right triangle. Half the base times the height. Half of twenty one root three, times ten and a half. Twenty one root three over two, times twenty one over two, is four hundred and forty one root three over four.

8:32So the piece we want is four hundred and sixty two minus four hundred and forty one root three over four. Stop there, and look at it rather than reaching for a calculator. Four hundred and sixty two is twenty one twenty twos, and four hundred and forty one is twenty one twenty ones. So twenty one over four comes out of both, leaving eighty eight minus twenty one root three. That is the answer, and it is exact.

9:04If you are handed no value for root three, that is where you stop. Reaching for one point seven three two is not finishing the question, it is answering a different one - and it turns this into about two hundred and seventy one, which is the number we opened on. Here is a check that costs one line and catches a swapped ratio instantly. Push the angle out to a straight one, so the chord becomes a diameter.

9:36The triangle's apex is now on the line its base sits on. It has flattened completely and measures nothing. And the piece past the chord is a semicircle, plainly half the disc. So at a straight angle the formula has to return the wedge with nothing taken off it, and it does - not by luck, but because the height is the radius times the cosine of the half angle, and the cosine of ninety is nought.

10:06Every chord cuts the disc into two pieces, and the larger one is where nearly everybody slips. The reflex side has its own wedge, the big one, so surely the big piece is that big wedge minus the triangle? It is not. It is the big wedge plus the triangle, and the picture shows why. The triangle lies on the far side of the chord from the small piece - which is the same side as the large one. It is part of the large piece, so you add it.

10:42Take a circle of radius fifteen, a chord opening sixty degrees, pi as three point one four and root three as one point seven three. The small wedge is a hundred and seventeen point seven five and the triangle is ninety seven point three one two five, so the small piece is twenty point four three seven five. Off the disc's seven hundred and six point five, that leaves six hundred and eighty six point zero six two five.

11:15Now the other way. The large wedge is three hundred degrees' worth, five hundred and eighty eight point seven five, and adding the triangle back gives exactly that again. Subtract it instead and you are nearly two hundred short, at four hundred and ninety one point four three seven five. Two more, both quick, and both making a point. Radius ten, chord opening a right angle, pi as three point one four.

11:48The wedge is a quarter of the disc, seventy eight point five, and the triangle needs no trigonometry at all: with a right angle at the centre the two radii are the base and the height, so half of ten times ten is fifty. The piece is twenty eight point five, and no ratio table was opened. Sixty degrees does the same favour differently: there the chord equals the radius, so the triangle is equilateral and reads straight off.

12:23Finish with one that puts the whole thing to work. A round table cover, radius twenty eight, carrying six identical designs round its edge. The making costs thirty five cents a square centimetre, and root three is to be taken as one point seven. Note that number. The worked example used one point seven three two, and the last digits of an answer depend on which you were handed. Six equal chords round the circle means six equal angles at the centre, and six equal angles filling a full turn means sixty degrees each.

13:04So each design is the piece past a chord opening sixty degrees on a radius of twenty eight: one subtraction, done six times. The wedge is a sixth of the disc, a sixth of two thousand four hundred and sixty four: four hundred and ten and two thirds. Sixty degrees, so the triangle is equilateral on a side of twenty eight, a hundred and ninety six root three, which at one point seven is three hundred and thirty three point two.

13:38One design is seventy seven point four seven, six of them four hundred and sixty four point eight, and the bill a hundred and sixty two point six eight. Now the mistake worth naming. Read those six designs as six wedges instead of six pieces past chords and you get two thousand four hundred and sixty four, which is the entire cover: more than five times the real figure, and a bill of eight hundred and sixty two point four zero.

14:12That is not being slightly generous. That is quoting for decorating the whole table. The wedge is the piece plus the triangle, and all of this comes down to knowing which of the three you were asked for.