# Video: Using a Sector to Form a Cone

This video explains how to calculate the radius and the measure of the central angle of the sector required to construct a right circular cone of specified height and base radius.

14:53

### Video Transcript

In this video, we’re gonna look at how you can make a cone from a sector of a circle. But first I’d like to tell you about a lesson; I want to talk on volumes of cylinders and cones. To start the lesson, we’ll recap how to calculate the volume of a cylinder. First you need to work out the area or the base, which is a circle with a radius of 𝑟. So that’s 𝜋𝑟 squared. Then you drag the base up through space to the top of the cylinder and this gives you the volume of the cylinder. Since the height of the cylinder is ℎ, then we have to multiply the area of the base by ℎ because that’s the length through which we drag the base in the third dimension.

So far so good, I then suggested that we think about a right circular cone that sits neatly inside that cylinder — the same base radius, 𝑟, and the same height, ℎ. We were going to work out the volume of that cone. Obviously, it’s smaller than the volume of the cylinder that it sits in because we’ve got all this wasted space around the outside here that we’ve cut away. But exactly how much smaller would it be? Now one student suggested that it could be considered as a set of right-angled triangles of height ℎ and base 𝑟, all connected to the axis here in the middle of the cylinder and then swivelled round to make a circle, so rotated around the axis like this making like a little decoration thing here. Obviously these- there will be loads and loads of these triangles, so you wouldn’t have lots of gaps here; they’ll be much, much closer together, so it would approximate to the cone.

They then said that we could consider the cylinder to be a series of rectangles arranged in much the same way and again rotated round here with lots and lots of them really, really packed in together to suggest a cylinder. Then they said if we lay the two on top of each other, for every rectangle that we’ve got there will be a corresponding triangle and each triangle is exactly half of one of the rectangles. So if we throw away this half of that rectangle, if we do the same for each of those shapes, they thought that maybe the volume of the cone will be half the volume of the cylinder. We decided we would try a little experiment.

Now I happen to have some thin card a cylinder with width of eight centimetres and a height of twelve point five centimetres and a bag of rice in my desk. Well teachers do! So we needed to make a cone exactly the right size to fit inside the cylinder, so they had to be of the same base size and the same height. We could fill the cone of rice and pour it into the cylinder and see how many times we needed to do this before the cylinder was full. If the student was right, we need to do that two times, and- but if the student was wrong we might have to do it more or fewer times than this. So there was the cylinder, there was the cone, there was me putting the cone in the cylinder, and there was the cone sitting neatly inside the cylinder with the same width, the same radius, and the same height. So I filled up the cone once and poured it in. And when I’ve done that, the cylinder wasn’t quite half full; it look like the student might be wrong. So I topped up the cone with rice again and poured all of that into the cylinder. And after this second cone full of rice, it was just over half full. And with three cones full of rice, the cylinder was exactly full. So our conclusion was that the volume of the cylinder was three times greater than the cone with the same base radius and height.

Now we got another video called “deriving the volume of a cone” which explains in more detail why the volume of the cylinder will be three times bigger than the volume of the cone, not just twice as big. But just quickly if you think about it, that triangle that we’ve got from the count if we genuinely did stack up loads of them, they wouldn’t go round in a circle to make a cone; they would actually go in front of each other and make a triangular prism like this. In order to get a cone, those triangles would actually need to be 3D wedges and not 2D triangles so that wedge would be infinitesimally thin here on the axis, but a bit thicker out here on the circumference of the cone. And we could organise them in a big circle round here and that would in fact make a cone. And likewise the rectangles wouldn’t be two-dimensional rectangles; they would be three-dimensional wedges like this so they would have a little bit of depth to them here and they will be infinitesimally thin in the middle here on the axis. And again we’d stack them in a big circular pattern like that to make our cylinder.

So if we cut the rectangular wedge in two through here like this and then we threw away the top one, that would leave us with all of these little triangular wedges down here to make the cone. Now if we consider the height of the green wedge — that’s gonna make the cone — we can see that the longest bit here is actually the infina- infinitesimally thin section. But for the other bit — the bit that we’re throwing away — the long section here is the bit that is a lot thicker. And I think you can see therefore that in fact this bit here is gonna be bigger than this bit here. So that’s why the volume of the cylinder is not twice the volume of the cone; it’s more than twice that volume. So that neatly brings us onto the point of this video: how do you go about making a cone that fits neatly inside a cylinder with the same base radius and the same height?

So to make our cone, we cut out our sector and then these two exposed radii we drag them towards each other. So we drag that one in; we drag that one in. And as we do that, the top pops up and we get a cone And we can see that this our claim for the way around the outside here makes up the base of the cone; so that’s this distance — the circumference of all the way around here, around the edge. Now also this radius here, let’s call it “s” for the radius of the sector. And that they- these two sides remember came together like this to join up to- it’s like a slit down the side of our cone there. And so that length there must be 𝑠 as well. So the slopy length of the cone, so maybe this length as well, this length as well; all those are the slopy lengths of the cone and are the same length as the radius of the sector that we-we created the cone from.

so I suppose the crucial question now is if we want to make a specific cone with a base radius of 𝑟 and a height of ℎ, what should the radius of this sector be? and what should this angle, the measure of this angle round here, be; let’s call it 𝜃 in order to create that particular cone? So remember then as we form this cone and we pull these sides together like this, the radius of the original sector is now making up the length of the slopy side of that cone. Now what we’re gonna do is drop a perpendicular down from the tip of the cone to the centre of the base of the cone. And then the radius remember we said was 𝑟 and the height was ℎ; so that’s this distance here. Now because we said we’ve dropped a perpendicular, we know that this angle here is ninety degrees. So that perpendicular is at ninety degrees to the base of the cone. So what we’ve got is a right-angled triangle. The hypotenuse is this length 𝑠 from the radius of the sector and then the two of the sides of the radius of the base of the cone and the height of the cone. Now when we’ve got a right-angled triangle like this, we start thinking Pythagorean theorem. And the Pythagorean theorem tells us that in a right-angled triangle, the square of the longest side is equal to the sum of the squares of the other sides. So in this case, 𝑠 squared is equal to ℎ squared plus 𝑟 squared. Now if we take the square root of both sides of that equation, then we see that the radius of the sector that’s needed to create that cone is equal to the square root of ℎ squared plus 𝑟 squared. Now we know ℎ squared is the height of the cone and we know what 𝑟 squared is; that’s the radius that we want for our cone. So we can easily work this out if we need to.

So now to work out the size of the angle at the centre of the sector that we’re trying to create, we need to consider some proportions. If we think about this distance here this, this arc length here, which of course is the circumference of the base of the cone and if we compare that to the entire circumference of that circle for the sector, those two measurements are in the same proportion as this angle here — the measure of this angle here — compared to the measure of the entire angle; in other words, three hundred and sixty degrees all the way round. Now we also because we’ve said that that we want the base radius of the cone to be 𝑟, we know that the circumference of that cone at the base of the cone is gonna be two times 𝜋 times 𝑟. So we’ve got the arc length; we can calculate that. We’ve just seen how to calculate the-the radius here of the sector. So that means the circumference of the whole circle would be two times 𝜋 times that radius, two 𝜋𝑠. And we obviously we know three hundred and sixty cause that’s a complete turn; that’s the measure of the angle of a complete turn. So knowing those three things is gonna enable us to calculate the value of 𝜃 here.

Okay, let’s put all this into action then and do the calculations for the cone that I used in the lesson that I’ve just told you about. So the cylinder that we were trying to fit the cone into was twelve point five centimetres high and eight centimetres wide. That means the base radius here was four centimetres, half of eight centimetres. So now we can use a bit of our Pythagorean theorem. By constructing this right-angled triangle here, the height is twelve point five centimetres; the radius as we said was four centimetres. We want to know the length of this slope side of the cone. So that we can work that, that will be the radius of the circle that we’re gonna use to make the sector. So just to make that a little bit clearer, let’s pull that out and do a new diagram. So here’s our right-angled triangle with those things on there. We can say that 𝑠 squared, that’s the longest side, is equal to twelve point five squared plus four squared, which when we evaluate those tells us that 𝑠 squared is a hundred and seventy-two point two five. But remember we don’t want 𝑠 squared; we want 𝑠, the radius of the circle for the sector. So we need to take square roots of both sides.

And since we’re gonna be drawing this, I’m not gonna get more accurate than about to the nearest millimetre. So one decimal place in terms of centimetres should give us a reasonable answer; so thirteen point one centimetres was the radius that we needed. Now in fact by an amazing coincidence when I went to open up my compasses, the largest radius that I could possibly get was just exactly thirteen point one centimetres. So that was a nice happy coincidence; I was able to just about draw this circle that I needed to So now let’s gather the information we need to work out the size of the angle, the measure of that angle at the centre of the sector that we’re gonna create. Well the base of the cone is the circumference here in orange. So the radius of that is four. And the circumference is two times 𝜋 times that radius of four, which gives us eight 𝜋. Now actually I’m gonna leave that in that format because if I started introducing as decimal places, I’m gonna introduce rounding errors into my later calculations. So we just gonna use the value eight 𝜋 in our subsequent calculations.

So now let’s think about the circumference of the whole circle from which my sector is gonna be cut. Now remember, we’ve just calculated the value of 𝑠, thirteen point one centimetres. Although we do have a completely accurate version of that here, the square root of one hundred and seventy-two point two five; that’s the value I’m gonna use in my calculation. So plugging those values in, the circumference of this circle is two times 𝜋 times the radius of square root of a hundred and seventy-two point two five, so again just leaving that in its nice format — not so nice format; perhaps I should say two root a hundred and seventy-two point two five times 𝜋 centimetres. So to calculate the size of the angle at the centre of our sector, we’re gonna need this value and we’re gonna need this value.

Now remember we’ve said these things are in the same proportion. The ratio of circumference of the base of the cone to circumference of sector of circle is the same as the ratio of 𝜃, the measure of the angle of central sector, to three hundred and sixty degrees of full turn. And bringing across those values that we’ve just calculated, we can see that eight 𝜋 over two root a hundred and seventy-two point two five 𝜋 is equal to 𝜃 divided by three hundred and sixty. Now if I divide the top and the bottom of the left-hand side by two, I’ve got eight divided by two is four; two divided by two is one. And I’m gonna divide them both by 𝜋. So 𝜋 divided by 𝜋 is one and 𝜋 divided by 𝜋 is also one; that simplifies to this. And if I now multiply both sides by three hundred and sixty like this, then over on the right-hand side I’ve got three hundred and sixty on the top; I’ve got three hundred and sixty on the bottom. Those two are gonna cancel cancel out, leaving me with 𝜃.

And well here’s the final answer; I’m not gonna be able to measure that accurately using my protractor. So in fact the value I would use to measure out the angle is probably gonna be to one decimal place at most, a hundred and nine point seven degrees. So let’s have a look at what I actually did then. I’ve got an angle of a hundred and nine point seven on a semicircle in this case of radius thirteen point one centimetres. I then cut it out and stuck those two radii together to make that cone, which then fitted neatly inside the cylinder.