A sector's area is its angle's share of the whole

0:00Draw a circle, mark two points on it, and join each of them to the centre. You have fenced off a piece of the disc. An arc, the two radii through its ends, and everything between them. That piece is a sector. The angle those two radii make at the centre is the only thing that decides how much of the disc you have taken. Not where the sector points, and not which two points you chose.

0:28Just the angle. Once you believe that sentence the formula is decided; the rest of this is why you should believe it. Start with the easiest case. Draw a diameter straight across; it cuts the disc into two pieces, and each is a sector of one hundred and eighty degrees. Now reflect the disc in that diameter. The circle lands exactly on itself, because a circle reflected in any of its diameters is unchanged.

1:02So the top piece lands exactly on the bottom piece. They match, and together they are the whole disc, so each is half of it. Half of pi r squared. Nothing there was about circles in particular - it was about a motion that changed no area. Now two diameters at right angles. Four pieces, each a sector of ninety degrees. Turn the disc a quarter of the way round.

1:31Again the circle lands on itself, and now each quarter lands on the next one. First onto second, second onto third, third onto fourth, fourth back onto first. All four match. Four matching pieces filling the disc, so each is a quarter. A quarter of pi r squared. Same argument, different motion. Reflection gave the half; a quarter-turn gives the quarter. Two answers so far, and both look like fractions of the disc.

2:05One half, and one quarter. Now rewrite them, deliberately awkwardly. One half is one hundred and eighty over three hundred and sixty. One quarter is ninety over three hundred and sixty. That looks like a step backwards - two clean fractions turned into two ugly ones. But look at what is now sitting in each numerator. One hundred and eighty is the angle of the half; ninety is the angle of the quarter.

2:36The number on top is no longer a coincidence of the case - it is the thing the case was about. So write it for any angle at all. A sector of theta degrees has area pi r squared, times theta over three hundred and sixty. The whole disc, times the angle's share of a whole turn. But be careful about what just happened. Two cases were examined, a pattern was spotted, and the pattern was declared to hold everywhere.

3:10That is not a proof. A formula with the angle squared in it also agrees at ninety and at one hundred and eighty, and parts company everywhere else. The formula is right - but so far the only reason offered is that it fits two cases. Here is the argument that closes the gap. Cut a whole turn into n equal angles about the centre; you get n sectors. Turn the disc through one of those angles and every sector lands on the next, so all have the same area.

3:45They do not overlap, and together they are the whole disc, so each is pi r squared over n. Take m of them side by side: m times pi r squared over n, and the angle taken is m out of n of a turn. Which is the formula. It holds for every angle that is a whole-number fraction of a turn - and any other angle is trapped between two of those.

4:13The two things doing the work are that turning changes no area, and that pieces add. Now put this beside something you already have. An arc's length is two pi r, times theta over three hundred and sixty. A sector's area is pi r squared, times theta over three hundred and sixty. The same fraction twice - what differs is only which whole is being shared. They are not two results; they are one argument applied to two things.

4:46Divide one by the other and the angle cancels: a sector's area over its arc's length is half the radius, whatever the angle. Which carries a warning. On a circle of radius two that is one - so area and length come out as the same number, at every angle you try. Test the two formulas against each other there, and you cannot tell them apart. Two radii do not make one sector.

5:19They make two - the small one between them, and the large one going the other way round. Their angles add to three hundred and sixty, so their areas add to the whole disc. That is a free check on every question of this kind. Radius ten, pi taken as three point one four, a chord at a right angle to the centre. The small sector is seventy eight point five, the large one two hundred and thirty five point five.

5:50Add them: three hundred and fourteen - which is pi r squared with that pi. If those had not matched, one answer was wrong, and you would know before anybody told you. A sector is not the only piece you can cut off a disc. Draw the chord across the arc's two ends and take the piece between chord and arc - that is a segment, and it is not a sector.

6:17What separates them is exactly the triangle on the two radii and the chord. So a segment is its sector, minus that triangle. Radius fifteen, sixty degrees, pi as three point one four and root three as one point seven three. The sector is one hundred and seventeen point seven five. The triangle is equilateral, so ninety seven point three one. Subtract: the segment is twenty point four four. And at a straight angle that triangle collapses to nothing, so there the segment and the sector are the same piece.

6:58Here is a question that gets written down wrongly very often. In a circle of radius r, a chord makes sixty degrees at the centre; find the smaller segment. Sector minus triangle is r squared, times the bracket pi over six minus root three over four. About nought point zero nine of r squared. Now the version that goes wrong. Put the pi outside the bracket instead, multiplying both terms, and evaluate that.

7:31Minus nought point eight four of r squared. Negative - and that is the point. You do not need to know the right answer to know this one is wrong, because an area cannot be negative. Checking the sign before you trust an answer costs two seconds and catches this every time. Two places this turns up wearing a disguise. A clock's minute hand, seven centimetres long: what does it sweep in ten minutes?

8:02Ten minutes is a sixth of an hour, so the hand turns a sixth of three hundred and sixty - sixty degrees. With pi as twenty two over seven, that is about twenty five point six seven square centimetres. The same number as a plain sixty degree sector of radius seven, because that is what it is. Second: two wiper blades, each twenty eight centimetres long, each sweeping one hundred and twenty degrees.

8:32One clears about eight hundred and twenty one square centimetres, so two clear about one thousand six hundred and forty three. But only because the sweeps do not overlap - and that clause is the whole licence to double. One last thing, where a tempting picture goes wrong. Draw a regular polygon inside a circle and ask what share of the disc it takes. A triangle takes about nought point four one, a square about nought point six four, a hexagon about nought point eight three.

9:09And the hexagon's share is exactly twice the triangle's - not roughly, exactly. So there ought to be a picture. The hexagon splits from its centre into six equilateral triangles, and the inscribed triangle should be three of them. Here is the trap: highlight three alternate ones and the areas do match. But look at the shape you have highlighted - it is a pinwheel. Its pieces meet only at the centre, and between them they reach all six corners, where the inscribed triangle reaches three.

9:45Equal area, different region, which is exactly why the wrong picture is so easy to draw. Join the centre to the triangle's own corners instead, and those three pieces really do fill it - each the size of one of the six.