Video: Answering Questions About Using a Sector to Form a Cone | Nagwa Video: Answering Questions About Using a Sector to Form a Cone | Nagwa

# Video: Answering Questions About Using a Sector to Form a Cone

Here, we are talking through questions about the radius and central angle of a sector and the height, slope length, and base radius of the right circular cone it creates.

14:53

### Video Transcript

In this video, we’ll recap the basic facts about forming right circular cones from sectors of circles and then go on to answer a few questions about them. To form a right circular cone from a sector, just cut out a sector and pull the exposed radii together like this. The centre will pop up and you’ll have a cone.

When you do this, the radius of the original sector here ends up here on our cone. So the radius of the original sectors the same as the length of the slopy side of the cone. Also, the arc all away around the outside of the original sector makes up the base of the cone. So the arc length for the original sector is the same as the circumference of the base of the cone.

If you want to form a cone with a specific height ℎ and a base radius of say 𝑟, you can calculate the radius needed for the sector — let’s call that 𝑠 — and the measure of the angle — let’s call that 𝜃 — at the centre of the sector.

Now remember that the length of the side of the radius 𝑠 is the same as the sloped side of the cone. So we can write that in here. And we’ve got, with the height, the radius, and sloped side, we’ve got a right-angled triangle so we can use the Pythagorean theorem, which says that the square of the longest side in a right-angled triangle is equal to the sum of the squares of the other two sides.

And then by taking square roots of both sides of that, we’ve worked out that the radius of the sector that we need to draw is the square root of the height of the cone squared plus the radius of the base of the cone squared. Now also, the ratio between the measure of the angle at the centre of the sector here and a full turn of three hundred and sixty degrees here is the same as the ratio of the arc length here to the complete circumference of that sector’s circle here.

And we can write that in this way, so 𝜃 divided by three hundred and sixty is the same as the arc length on the sector divided by the circumference of the whole of the sector circle. But remember, the arc length is the same as the circumference of the base of the cone; it was that arc that was used to make the base of the cone so that is two times 𝜋 times 𝑟.

And the circumference of the sector’s circle, that has a radius of 𝑠, so that is equal to two times 𝜋 times 𝑠. So if we write those in, we see that we can actually cancel. If I divide top and bottom by two 𝜋, okay? we just get 𝜃 over three hundred and sixty is equal to 𝑟 over 𝑠. And if I just multiply both sides by three hundred and sixty, I can cancel those out in the left-hand side.

So if we know ℎ and 𝑟, the height of the cone that we’re looking for and its radius, we can easily calculate the-the radius of the sector that we need to create and the angle at the centre of that sector that we need to use in order to create that cone. Now we’re gonna use this information in a couple of questions.

Right, the first question then is the radius of the base of a cone is fifteen centimetres, so this measurement here is fifteen centimetres, and the length of its sloped side is twenty-five centimetres, so that’s this dimension here. Find the measure of the angle at the centre of the sector which was used to form the cone; so that’s this here. Now we can just plug this into the formula that we’ve just created.

𝜃 is equal to three hundred and sixty times the radius of the base of the cone, so that’s fifteen, divided by 𝑠, the length of the sloped side which we’re given in the question, which is twenty-five. And when we work that out, we get an answer of two hundred and sixteen degrees.

So if you can remember that formula, this was a pretty simple calculation to make because we were given lots of the relevant information in the question. Let’s pretend though that we couldn’t remember that formula, and let’s try and work this out more from first principles by looking at what we know about the situation.

We know that the ratio of the angle at the centre to a whole turn is the same as the ratio of the length of this arc to the whole circumference of that sector’s circle. And we can write that as 𝜃 over three hundred and sixty, so this angle divided by three hundred and sixty, is the same as the sector arc length divided by the circumference of that full sector’s circle.

Now we also know that the arc length of the sector is the same as the circumference of the base of the cone because that’s what’s used to form the cone and we know the radius of the base of the cone, so we can work that out. This length here is two times 𝜋 times the radius; that’s two times 𝜋 times fifteen, which is equal to thirty 𝜋.

We also know, remember, that-that the length of the sloped side of the cone is the same as the length of the radius of the sector that was used to create it because remember these two radii come together to form that slopy side. So that means that the radius of this sector is twenty-five centimetres. And again we can use that radius to work out the circumference of the whole of that sector’s circle.

That’s two times 𝜋 times twenty-five, which is the same as fifty 𝜋. So let’s plug those numbers into the calculation that we created on the other side. And we’ve got thirty 𝜋 divided by fifty 𝜋 is equal to 𝜃 over three hundred and sixty, so we can do a bit of cancelling and we can multiply both sides by three hundred and sixty; they will cancel out on that side. We’ve now got an expression for 𝜃 here, which when we work out we get the same answer as before.

Now obviously that second method was quite a lot longer than the first method. If you can remember the formula and you just plug the numbers in, you’ll get the answers very very quickly. But actually, doing the question like this really forces you to understand where the numbers are coming from, understand the logic of the situation. And then if you have a bit of a panic in the exam and everything goes horribly wrongly, you forget everything, at least you will definitely be able to recreate the method by understanding the situation because you had been through it so many times before.

In this second question, we’re asked- we were told that the radius of the base of a cone is five centimetres, the measure of the angle at the centre of the sector that was used to form the cone is a hundred and fifty degrees, and we’ve got to find the height of the cone. So this is kind of what I call a backwards question, where we-we don’t know the radius and height of the original cone, we’re not working back to the dimensions of the sector, but we’re given other bits of information. We’ve gotta try and work out the height of the cone. But never mind! We can still work out the answer.

So the first obvious thing that we can work out is the circumference of the base of the cone, so we’ve got the radius here of five, so the circumference of that circle is two times 𝜋 times the radius. And that is ten 𝜋 centimetres. I’m gonna leave it in that format; I’m not gonna put it into decimal places because I want as accurate as answer as possible.

Now remember that that circumference there is the same as the arc length of the sector that we were looking at. Now we don’t seem to be very much closer to where we want to be at the moment. Remember, we’re trying to work out the height of the cone so that’s the- this distance here. Now we do also know that the sloped length of the cone is the same as the radius of the sector here. And if we knew what that slope length was, let’s call that 𝑠, we would then be able to use the Pythagorean theorem to work out what ℎ is.

Right then, so we can work out then a- an expression for the circumference of this big sector’s circle over here because that’s two times 𝜋 times that radius. So that’s two 𝜋𝑠 centimetres. No that’s an 𝑠 and not a five. Now we also know that the ratio of the angle at the centre to a full turn is the same as the ratio of the length of the arc here compared to the full circumference of this sector’s circle, so we can write down that equation.

So 𝜃, which in fact is a hundred and fifty degrees, over three hundred and sixty is the same as the sector arc length over the sector circumference- a sector’s circle’s circumference. So let’s plug those values in then. We know that the sector arc length is the same as the circumference of the base of the cone, so that’s ten 𝜋, and we know that the circumference of the sector’s circle is two 𝜋𝑠 so we can write that in as well. Now we can do a bit of cancelling; I can divide the top and the bottom by 𝜋; I can divide the top and the bottom by two; and I can divide the top and the bottom over here by thirty, giving me five over twelve.

I can then see that five over twelve is the same as five over 𝑠, so clearly twelve must be 𝑠. So now I’ve calculated the length of the sloped side, so I can cross that out and write twelve. So this length is twelve centimetres. And now I can use Pythagorean theorem to work out the value of ℎ.

So that means that the length of the hypotenuse squared, twelve squared, is equal to the sum of the squares of the other sides, ℎ squared plus five squared. And evaluating twelve squared and five squared, I’ve got this. I can now subtract twenty-five from both sides of that equation, leaving a hundred and nineteen is equal to ℎ squared. Now if I take square roots of both sides and I work out the square root of a hundred and nineteen, ℎ is ten point nine o eight seven one two one one dot dot dot dot dot centimetres. But of course, remember our question asked us to give the answer correct to one decimal place. So if we look at that first decimal place, ten point nine, the next digit is only a zero so we’re not gonna round that up. So the answer is the height of the cone is ten point nine centimetres correct to one decimal place.

So this final question then: a sector is formed into a cone. The radius of the circle from which the sector was taken is twel- is ten centimetres. The measure of the angle at the centre of the sector is two hundred degrees. Find the radius 𝑟 and height ℎ of the cone that was formed, giving your answers correct to one decimal place.

Well the formulae that we worked out earlier, we’re kind of going in the other direction. They were assuming that we knew ℎ and 𝑟 and we can work back to the other values. So we’re sort of working backwards in this question, but let’s just look at the situation and think, “What do we know?” Well, we know that the circumference of the base of the cone is equal to the length of the arc of the sector, and we know that the radius of the sector is equal to the sloped length, the sloped side length, of the cone.

So first, let’s look at the length of that arc. We know that this angle here is two hundred degrees, and we know that the ratio of that angles there compared to three hundred and sixty degrees is the same as the ratio of this arc length here compared to the circumference of the whole of that circle. Now we’ve got the radius of that circle, so we can work out the circumference.

And that’s two times 𝜋 times ten, which is twenty 𝜋 centimetres. Again let’s keep these- let’s not use decimal places at moment. Let’s keep these numbers as accurate as we can as we’re going through the calculation. And as we’ve just said, let’s write this as an equation. So two hundred over three hundred and sixty, the comparison, the ratio of these two angles is the same as the ratio of the arc length to the sector’s circle’s circumference. And remember, the arc length is the same as this value here, the circumference of the base of the cone.

So we can cancel the left-hand side of that down, doing it with division, so two hundred over three hundred and sixty is the same as five over nine. We know that the circumference of the sector’s circle is twenty 𝜋, so the arc length, remember, is this value here: two 𝜋𝑟. So we can put two 𝜋𝑟 into here.

And we can cancel that down. So if I divide the top by 𝜋 and the bottom by 𝜋 on this side, and I could divide the top by two and the bottom by two. Now I’ve got an expression- an equation just involving 𝑟. So I’m now gonna multiply both sides by ten, meaning that the tens cancel on the right-hand side and I’m left with 𝑟 being equal to fifty over nine. Now obviously the question is looking for an answer correct to one decimal place, so that’s going to be five point six. But in the future calculations, I’m still gonna use this most accurate version of the answer to do my calculation so that I don’t create any rounding errors.

So I can replace 𝑟 here with fifty over nine. I also know, remember, that the radius of this sector was ten and that that length is the same as the sloped side here, so that’s ten centimetres over here. And now we know that that’s our perpendicular, so that’s a right-angled triangle so we can use a bit of Pythagorean theorem to work out value of ℎ.

Remember, the square of the longest side is equal to the sum of the squares of the other sides. Now using the accurate value fifty over nine, I can expand those, then take away two thousand five hundred over eighty-one from both sides, and then take square roots of both sides of that equation, giving me my final answer of ℎ is eight point three to one decimal place.

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