Lesson video

Hello there, and thanks for joining me today.

My name is Dr.

Rowlandson, and I'll be guiding you through today's lesson.

Let's get started.

Welcome to today's lesson from the unit of 2D and 3D shape for surface area and volume, including pyramids, spheres, and cones.

This lesson is called, "Forming equations involving complex shape calculations." And by the end of today's lesson, we'll be able to form and solve equations involving complex shapes and solids.

Here are some previous keywords that will be useful during today's lesson.

So you might want to pause the video while you remind yourself what these words mean, and then press play when you're ready to continue.

This lesson is broken into two parts, and we're going to start by forming and solving the equations from solids.

Unknown lengths on a shape can be represented by algebra.

For example, here we have a cuboid where we know two of the lengths where 3 and a 10, but what we don't know is the height.

So, we could use a letter to express the height such as x.

Now, using this, we can express other measures algebraically as well.

For example, we can't actually work out what the volume is for this cuboid because we don't have all the information required.

But we could write an algebraic expression using what we have here and the letter we've used.

We could do 10 times x times 3 to get 30x, as an expression for the volume.

And we can't work out what the surface area is, but we could do a similar thing.

We could do the normal calculations that we do to find the surface area, but include the x as one of the numbers in there and simplify it to get 26x + 60.

That means if we do find out what the height is, we can substitute it into these expressions to work out the volume or surface area.

Or if we know what either the volume or surface area is, we could then find the value of x.

That is, equations can be formed by equating expressions with relevant known measures and then solved.

Let's take a look at an example.

First cuboid here.

If the volume is 150 cubic units, then we could work out the value of x by creating an equation with that.

We could write 150 is equal to 30x, 150 is the volume and 30x is the expression that we wrote for the volume.

And then we could solve that equation by dividing both sides by 30 to get 5 = x or x = 5.

Here we have a composite solid this time, which is constructed from two cuboids.

All the lengths are given in centimetres, but we can see that we have some lengths as numbers, but we have one length expressed as x.

We're told it's volume though.

The volume is 162 centimetres cubed, and we need to find the value of x.

Well, Laura's gonna help us with this, and Laura says, "We could break this complex problem down into smaller steps." Let's consider what those steps would be.

We could write an expression for the area of the cross-section.

We can't find the area if we don't know all the measurements we need, but we can write an expression.

And we can use that to write an expression for the volume.

And then we could form an equation using our expression for the volume and the actual volume and then solve it.

So let's do that.

Area of the cross-section.

Well, it's in composite shape in itself.

It's like an L-shape, but we can break it into two rectangles.

We can do x times 3 for one rectangle, and 8 times 3 for the other rectangle, and add them together and we get 3x + 24.

And then to get the volume as an expression, we can multiply that by 4, which is its depth, and we'd get 12x + 96.

And then we can create an equation.

Our expression is 12x + 96, the volume actually is 162, if we equate those, we can then subtract 96 from both sides, divide both sides by 12, and we get 5.

5 for the value of x.

So let's check what we've learned.

Here we have a composite solid which is constructed from cuboids.

Write down an expression for the area of the cross-section.

Pause the video while you do that, and press play when you're ready for an answer.

Our answer is 4x + 30.

So, now we have that the area of the cross-section can be express as 4x + 30, write down expression for its volume.

Pause the video while you do that, and press play when you're ready for answer.

The answer is 14x + 105, and we get that by multiplying the 4x + 30 by 3.

5.

So, now I'm gonna tell you that the volume actually is 189 centimetres cubed.

Could you use that and the expression we found earlier to find the value of x? Pause the video while you do that, and press play when you're ready for an answer.

The answer is x = 6.

We get it by equating 14x + 105 with 189, and then rearrange it to solve.

Let's take a look again at the composite solid that we saw earlier, which is constructed from cuboids.

But let's imagine for a second that we haven't previously worked out the of x as being 5.

5.

And in rather than knowing the volume, I'm gonna tell you this time that the surface area is 213 centimetres squared, and let's use that information to find the value of x this time.

Sam's gonna help us with this.

Sam says, "We could calculate the areas of faces that have two known lengths.

And we could write expressions for the areas of faces that have an unknown length.

And then we could write an expression for the total surface area." That's a great way to start.

So let's work through it together.

Let's begin with the faces where we can actually work out the area of them.

And there's quite a few of these.

There's one at the very front, which is 3 times 4 to get 12.

There's another one facing in the front, which is 5 times 4 to get 20.

There's one at the back, which we can't see where we have two known lengths, and that is 8 times 4 to get 32.

And there's one on top where we know both those lengths, that's 3 times 4 to get 12.

Those are all the faces where we know the areas, so we can find the total area of those as being 76 centimetres squared.

Let's now look at the ones where there's an unknown length.

We can't actually work out what the area is, but we can write an expression for each.

This one at the front here, it can be written as 3x + 24.

I can work that out by splitting into two rectangles, find the area of each and adding them together.

The one on the back is the same as the one at the front, that'd be 3x + 24.

This one on top here will be 4x, that's 4 times x.

And the one underneath will be 4x + 12, which we can do by doing 4 times x plus 3.

And then we can add those together to get 14x + 60.

So we have an expression for the total of the unknown areas, and we have the area of the ones that we do know.

We could combine these to get the total surface area written as 14x + 60 + 76, which is adding those together and then simplify to get 14x + 136.

So now we have an expression for the surface area of this composite solid, and we know what the surface area is.

So we can create an equation.

The total surface area is 14x + 136, and we can equate it to 213.

And then we can rearrange this to solve and find their value of x.

If we subtract 136 on both sides, we get this, and divide both sides by 14, we get x = 5.

5 again.

Let's check what we've learned.

Here, we have a composite solid constructed from cuboids, and all lengths are given in centimetres.

Calculate the total area for all the faces that have two known lengths.

Pause the video while you do this, and press play when you're ready for answers.

So it's 80.

5 centimetres squared.

There are four faces which have two known lengths, and those are the areas we can see on the screen there.

And if you add them together, that's what you get.

So now, write an expression for the total area for all the faces that have an unknown length or unknown lengths.

Pause the video while you do this, and press play when you are ready for an answer.

The answer is 15x + 70.

5 centimetres squared.

And there are four faces that have an unknown length, and that's how you find the area each of them as an expression.

So, if we have the sum of the known areas as 80.

5 centimetre squared, and the sum of the unknown areas expressed as 15x + 70.

5 centimetre squared, I'm gonna tell you now that the actual total surface area is 212 centimetres squared.

So could you use that information, please, to find the value of x? Pause the video while you do that, and press play when you're ready for an answer.

The answer is 4.

07, rounded to three significant figures.

We get it by adding together the sum of the known areas and the sum of the unknown areas, and equating it to 212, and simplifying and rearranging to solve.

Here we have a cube and a cylinder, which each have the same volume.

Now, for the cube, we know its length is 5 centimetres.

But for this cylinder, we know the height is 6 centimetres but we don't know the radius.

The radius is expressed as x centimetres.

And what we need to do is find the value of x.

Andeep will help us with this.

He says.

"We could break this complex problem down into smaller steps." Great idea.

Let's do that together.

Step one, could be to calculate the volume of the cube.

We have all the measurements we need on the cube to do that, so that'd be a good place to start.

Step two, could be to write an expression for the volume of the cylinder.

We can't work out the volume of the cylinder 'cause we don't know what the radius is, but we could express the volume in terms of x.

And then we could form an equation and solve it because we know that the volume of one is equal to the volume of the other, and that creates our equation.

So, let's work through it together.

Begin with the volume of the cube, that'll be 5 cubed, which is 125.

The volume of the cylinder, well, we could do pi times radius squared, which in this case is x squared, and then multiply it by 6.

Now simplify to get 6 pi x squared.

So we have the actual volume and an expression for the volume, and we know that those are equal to each other, so we can create an equation.

Now I got 6 pi x squared equals 125, and then we can rearrange and solve.

We can divide both sides by 6 pi, or you can divide by 6 and then pi separately if you want to, but more efficient divide by 6 pi, and then we can square root, and we get an answer of 2.

58 rounded to three significant figures.

So let's check what we've learned.

Here, we have a cylinder where you know the radius is 2 centimetres and the height is 5 centimetres.

Could you please calculate the volume of the cylinder, giving your answer in terms of pi? Pause the video while you do this, and press play when you're ready for an answer.

Well, if we do 2 times pi times 5, you get 20 pi, which means the volume of the cylinder is 20 pi centimetres cubed.

Here, we have a cone.

We know the radius of the cone is 6 centimetres, but we don't know the height, that's expressed as x.

Could you please write an algebraic expression for the volume of the cone? There's a formula at the bottom of the screen, that'll help you with them.

Pause the video while you do that, and press play when you're ready for an answer.

The answer is 12 pi x, which we get by substituting 6 and x into our formula.

So, how about I switch it around now? Here, I've told you what the height is 6 and the radius is x.

Let's do the same thing again.

Let's write an algebraic expression for the volume of the cone this time.

Pause the video while you do that, and press play when you're ready for an answer.

Well, we are substituting the same things into the formula x and 6 but not in the same places.

That means we get a third times pi times x squared times 6, and this would give us 2 pi x squared.

And now we have a cone and a cylinder side by side.

We are told that the cone and cylinder had the same volume.

The volume of the cylinder is 20 pi centimetres cubed, we worked that out previously, and the volume of the cone is expressed as 2 pi x squared centimetres cubed, which we worked out previously as well.

Use these volumes to write down an equation for x.

Pause the video while you write down the equation, and then press play when you're ready to see what the equation is.

Our equation is 2 pi x squared equals 20 pi.

So, could you please now solve the equation and give your answer rounded to three significant figures? Pause the video while you do that, and press play when you're ready for an answer.

Answer is 3.

16 when rounded to three significant figures, and that's how we solve the equation on the screen.

Okay, it's over to you now for Task A.

This task contains three questions and here is question one.

You have a cube, a sphere, and a cone, which all have the same volume and have unknown lengths on them as well.

You also have some formers that will help you along the way as well.

In each parts, A, B, and C, you are told the value of one unknown, and you've gotta use that to find the values of the other unknowns based on that information.

Pause the video while you do that, and press play when you're ready for question two.

Here is question two.

You have a composite solid which is constructed with cuboids where all lengths are given in centimetres.

You need to work out the value of x given the volume in part A, and work out the value of x given the surface area in part B.

And the values of x will not necessarily be the same in parts A and B in this situation.

Pause the video while you works out, and press play when you're ready for question three.

And here is question three.

You've got a solid constructed by joining two cones together, which each have a radius of 6 centimetres.

You don't know the heights, but you're told that the height of the top cone is three times the height of the bottom cone, and you're told that the volume of the solid is 240 pi centimetres cubed.

Now, in part A, you need to find the height of each cone, and part B, you're told about the height of the top cone is decreased by 2 centimetres, while the height of the bottom cone is increased by 2 centimetres, and you gotta use those changes to calculate the volume of the new solid.

And if you want to, there is a link to your GeoGebra file that is interactive and may help you visualise this question more easily.

Pause the video while you work through this, and press play when you are ready to go through some answers.

Let's now go through some answers.

In question one, you've got three solids which all have the same volume and have unknowns.

And when you are given the value of one unknown, you have to work out the values of the others.

One thing that might help you with this, is to write an expression for each volume and equate them like this.

That way we can take the value that we know and substitute it in to work out the volume and equate it to each of the other two expressions to work out the other unknowns.

So for part A, when we know A is equal to 4, we can get B is equal to 2.

48 to three significant figures, and C as 2.

26 to three significant figures.

For part B, when we're told that B is equal to 4, we can get A is equal to 6.

45 to three significant figures, and C is equal to 4.

62 to three significant figures.

And for part C, I'm told that C is equal to 4, and we substitute that in, we get A equals 5.

86 to three significant figures, and B equals 3.

63 to three significant figures.

Then question two.

We have our composite solid where we have unknowns and we have the value of x.

Well, we could start by writing expression for the surface area and an expression for the volume.

Our expression for the surface area will be 330 + 44x.

Our expression for the volume would be 108 + 120x.

And now we're ready to work out the answers to part A and B.

So in part A, what would the value of x be if the volume was 492? Well, we could equate our expression for volume to 482 and rearrange and solve to get x centimetres is equal to 3.

2 centimetres.

And then for part B, what the value of x be if the surface area was 506 centimetres squared? We can take our expression for surface area, equate it to 506, and then solve that to get x = 4, or x centimetres = 4 centimetres.

Then question three.

To find the height of each cone, well, we can start by writing an expression for the height of each cone separately.

First as h and 3h, as the top cone is three times the height of the bottom cone.

And because we're told the volume of the overall solid, we could write an expression for the volume in terms of h, that would be as what we can see on the screen here.

We can substitute h and 3h into our formulas along with the 6 as well, and add them together and we get 48 pi h.

And then we can equate it to 240 pi and rearrange to get h = 5.

If we know h = 5, that means the height of the top cone must be 15 centimetres, that's 3h, and the height of the bottom cone must be 5 centimetres, that's just 1h.

Then part B, if we are told that the height of the top cone is decreased by 2 centimetres, while the height of the bottom cone is increased by 2 centimetres and we need to calculate the volume of the new solid.

Well, the new heights are 7 centimetres for the bottom cone and 13 centimetres for the top cone.

If we substitute those numbers along with a 6 into our formula for finding the volume, we would get this calculation we can see on the screen here, and that would simplify to 240 pi centimetre squared again.

Great work so far.

Now let's move on to the second part of this lesson, which is forming and using algebraic expressions.

Here we have Aisha and Lucas who each have a cylindrical container filled to the top with water.

The two containers are congruent.

The height of each container is equal to its diameter.

So we can write an expression something a bit like this.

Aisha has a cone that has the same height and diameter as the container.

And she places the cone into the container causing some water to spill out of the top, and then she takes it out again.

So we can see the amount of water left in the container has decreased 'cause some of it spilled out.

Only a little bit has decreased though.

Lucas has a sphere that has the same diameter as the container.

He places the sphere into container causing some water to spill out the top as well.

Like this.

And then he takes it out again, and we can see that some of the water has spilled out so he has less water in his container.

Aisha says, "Why don't you pour the leftover water from your container into my container? Do you think it will all fit?" Hmm.

Lucas says, "I'm not sure.

Do you think it would depend on the size of the objects?" What do you think will happen, and how could you find out whether or not it would happen the way you think it would? Pause the video while you think about this, and press play when you're ready to continue.

Well, if we knew what the measurements were, we could just work out the volumes of each and work out exactly how much water is left over and that would tell us whether or not the liquid would fit in that particular situation.

But we don't actually know what the measurements are here.

So what we're trying to work out is, would the water from one container fit in the other for all situations where the height is the same as the diameter for these particular objects? And a way to work out whether something will always work is to use algebra.

Aisha writes an algebraic expression for the volume of the cylinder.

Its height is equal to its diameter.

So, its height is twice the length of its radius, which means the height can be expressed as 2r, and that means we can express the volume in terms of r.

We could do pi r squared times 2r, where pi r squared is the area of the circle and 2r is the height of the cylinder, and that would give us 2 pi r cubed, which is the volume of the cylinder.

Lucas writes an algebraic expression for the volume of the sphere.

He says, "If I express it in terms of r, then the expression for the volume would be the same as in the formula." So it'll be 4 over 3 pi r cubed.

Let's check what we've learned by doing the rest of this problem together.

And let's start by looking at the cone.

The height of the cone is twice the length of its radius.

Write an algebraic expression for the volume of the cone.

Pause the video while you do that, and press play when you're ready for an answer.

If we substitute r and 2r into our formula for the volume of a cone, we get 2/3 pi r cubed.

So, here we have three solids, a cylinder, cone, and sphere.

They all have the same height and radius.

The height of the cylinder and the cone is equal to the diameters.

And you've got some expressions on the right-hand side for the volumes.

Match each shape with the expression for its volume.

Pause the video while you do that, and press play when you're ready for an answer.

Well, the volume of the cylinder can be expressed as 2 pi r cubed, the volume of the cone as 2/3 pi r cubed, and the volume of the sphere as 4/3 pi r cubed.

So now we have our expression for the volume of each of those three solids.

Fill in the blank.

The volume of the cylinder is blank the sum of the volumes of the cone and the sphere.

Is that blank equal to, greater than, or less than? Pause the video while you choose, and press play when you're ready for an answer.

The volume of the cylinder is equal to the sum of the volumes for the cone and the sphere.

We can see that all three expressions have pi r cubed in, and the 2 for the cylinder is equal to the sum of 2/3 from the cone and 4/3 from the sphere.

2/3 plus 4/3 is equal to 2.

So yes, the volume of the cone plus the volume of the sphere would be equal to the volume of the cylinder in this situation.

So back to our context with Aisha and Lucas.

The cylinders, cone and sphere, all have the same height and radius.

The height of each cylinder and cone is equal to the diameters.

And to begin with, the cylinders are full with water.

The cone and sphere is placed into containers so that the water spills out, and then they're taken out again so there's less water in each container.

The leftover water will, is it always, sometimes, or never, fit into a single cylinder.

Pause the video while you choose, and press play When you're ready for an answer.

The answer is, always.

The leftover water will always fit exactly into a single cylinder regardless of what the diameters are, so long as they all have the same diameters and the height of them is equal to the diameters.

Okay, it's all due now for Task B.

This task contains one question and here it is.

We have our solid once again, which is constructed by joining together two cones.

The radii of the cones are equal, and they are both expressed as r.

The height of the whole solid is expressed as h, and the height of the top cone is expressed as y.

Use this information to answer parts A, B, C, and D.

And once again, there's a link to an interactive GeoGebra version of this solid on the slide, which we can see here.

Pause the video while you do this, and press play when you're ready to go through some answers.

Let's now work through this together.

In part A, we need to write an expression for the height of the bottom cone in terms of h and y.

That would be h - y.

In part B, we need to write an expression for the volume of the top cone.

Well, we have the height is y and the radius is r, so that would be pi over 3 r squared y.

And then part C, to write an expression for the volume of the bottom cone, well, we have that the height of the bottom cone is h subtract y, and it's radius is r, so that would be pi over 3 times r squared times h subtract y.

And then part D, it said "Hence or otherwise, prove that the volume of a solid constructed with two cones as in the diagram here, is always equal to the volume of a single cone that has the same radius and height." So we can do that by taking the expressions we had for the volume of each individual cone and adding them together and then simplifying it.

Now, the volume of each individual cone has pi over 3 r squared in, so we can take pi over 3 r squared as a factor, and what we're adding together are the two heights, y and h subtract y.

And when we simplify that, the y and the subtract y cancel out to get just h.

So we have pi over 3 r squared multiply by h, and that is the volume of a cone that has a radius of r and a height of h.

So that means this shows that the volume of two cones with congruent bases, is the same as a single cone with the same base and height of the two combined.

Fantastic work today.

Now, let's summarise what we've learned.

There are a variety of approaches to calculations involving shapes.

And when shapes have unknown lengths, measures can be expressed algebraically.

And from that, equations can be formed and solved to find unknown lengths, and algebraic equations can be manipulated to support geometric reasoning.

Thank you very much today.

I hope you have a great day.