Lesson video

Hello, my name is Dr.

Rowlandson, and I'm thrilled that you're joining me in today's lesson.

Let's get started.

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

This lesson is called Advanced Problem Solving with Further Surface Area and Volume.

And by the end of today's lesson we'll be able to use our enhanced knowledge of surface area and volume to solve problems. Here are some previous keywords that you may be familiar with and we'll use again during today's lesson.

So you might want to pause the video if you want to remind yourself what any of these words mean.

And then press Play when you're ready to continue.

This lesson contains two learning cycles.

In the first learning cycle we're going to look at contextual problems that involve volume, and the second learning cycle, we're gonna look at problems involving multiple aspects of mathematics put together.

Let's starts off with contextual problems involving volume.

Here we have a diagram that shows a cuboidal juice carton.

The carton is partially filled with juice where the height of the juice is indicated by the line segments partway up the carton.

Here we have Laura.

Laura tips and turns the carton in different orientations, and we can see those on the screen here.

In which orientations would the juice be at the same height inside the carton? And can you explain why? Pause the video while you think about this and press Play when you're ready to continue.

Let's take a look at this together then.

Laura says, the volume of the juice remains the same each time.

Two key things that change though as I move the carton are the height of the juice and which face is the base.

Here's a pair of orientations where the height of the juice inside the carton would remain the same.

Can you see why? Laura says, the volume of the juice is the same.

The base is congruent in each orientation of the carton.

Therefore, the height of the juice is the same.

So what would cause the height to change? Well, here is a pair of orientations where the height of the juice inside the carton would change.

Can we see why they would change here? Laura says, the volume of the juice is the same but the bases in each orientation has a different area.

Therefore, the height of the juice would differ.

In which orientation would the height of the juice be greater, and why does that happen? Pause video while you think about these questions and press Play when you're ready to continue.

Well, the higher the juice would be greater when the carton is in orientation A than it is in orientation E.

And let's see why.

Laura says, you can work out the height of the juice by dividing the volume by the area of the base.

She says, the area of the base in orientation A is less than the area of the base in orientation E, which we can see from the diagram.

Dividing the same volume by a smaller area would lead to a greater height.

So, the juice is higher in orientation A than it is in orientation E.

She also adds that dividing the same volume by a greater divisor would lead to a smaller height.

So let's check what we've learned.

Here we have a cuboid carton that is partially filled with juice.

It's tipped and turned to each orientation we can see labelled A, B, and C.

For which orientation, or orientations, is the height of the juice the same as at the start? Pause video while you choose and press Play when you're ready for an answer.

The answer is B.

In that orientation, the base is congruent to the base of the original orientation, so the height would be the same.

Here we have a diagram that shows a cuboid juice carton, which is partially filled with juice, and Laura changes the orientation of the carton, which is shown in the diagram.

Which statement would be true? Is it A, the height of the juice will rise, is it B, the height of the juice will lower, or is it C, the height of the juice will remain the same? Pause video while you choose and press Play when you're ready for an answer.

The answer is A, the height of the juice would rise.

The base in the new orientation would have a smaller area, by the looks of it, but in its original orientation, so the juice would rise.

So let's now explore this problem further with some measurements included.

The diagram shows a cuboid carton that is partially filled with juice.

The height of the juice is indicated by the line segments partway up the carton and all lengths are given in millimetres.

The orientation of the carton is changed as shown in the diagram now.

And what we wanna do is find the height of the juice, H, in its new orientation.

Now we're going to work through this together in a moment, but perhaps pause the video and think about what steps it might take along the way while we work it out.

And then press Play when you're ready to continue.

Let's work through this together now.

A good starting point could be to find the volume of the juice first, and that is the same in both orientations.

Now, the volume of the juice will be less than the volume of the cuboid, 'cause the juice only partially fills the carton.

But what's quite helpful is that the juice is in the shape of a cuboid while it's inside the carton in these orientations.

Now, with the orientation on the right, we don't know what the height is, that's what we're trying to work out.

But with the orientation on the left we do, so we can get all the numbers we need for the volume on the left.

We can do 60 multiplied by 90, which gives you the area of the base, and multiply it by 140, which gives the volume of 756,000 millimetres cubed.

When the carton is tipped over, the volume of the juice remains the same, but we have a new height.

We can get that height by dividing the volume by the area of the base in its new orientation, and the area of the base is 90 times 200.

When we do that division, we get 42 millimetres.

So let's check what we've learned.

Here we have a diagram that shows a cuboid carton that is partially filled with juice.

The height of the juice is indicated by the line segments partway up the carton, and all lengths are given in millimetres.

Find the volume of the juice.

Pause video while you work through this, and then press Play when you're ready for an answer.

The answer is 810,000, which we get by multiplying 60 by 90 by 150, which is the height of the juice.

So now we know the volume of the juice is 810,000 millimetres cubed.

The orientation of the carton is changed.

Let's find the height of the juice in its new orientation.

Pause video while you do this and press Play when you're ready for an answer.

The answer is 67.

5 millimetres, which we get by dividing the volume by the area of the base in its new orientation.

Let's go a bit further and work out what is the difference between the two heights of the juice? Pause video while you work that out and press Play when you're ready for an answer.

The answer is 82.

5 millimetres, which we get by doing 150, which is the original height, subtract 67.

5, which is the new height.

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

This task contains two questions and here is question one.

The diagram shows a cuboid carton that is partially filled with juice.

The height of the juice is indicated by the line segments that are partway up the carton, and all lengths you can see here are in millimetres.

Izzy rotates the carton three times, and in each orientation the base is parallel to the floor.

Based on this context, could you please answer questions A and B? Pause video while you do this, and press Play when you are ready for question two.

And here is question two.

This time we have a container which is a prism.

Its cross-section is constructed with a rectangle and an isosceles triangle.

All units you can see are given in centimetres.

Now the container is partially full with liquid.

The height of the liquid is indicated by the line segments partway up the container.

The container is rotated 180 degrees.

I want you to do is work out what is the height of the liquid H when the container is in its new orientation? Pause video while you work this out and press Play when you're ready for some answers.

Okay, let's go through some answers.

Question one.

Which of the three rotations would cause the greatest change in height for the juice? Well, it would be rotation three.

Now, you could work that out by working out the height of the juice each time and then looking for the greatest difference, or you don't necessarily need to do that.

You could just work out the area of the base each time and look for the greatest change in the area of the base, from the largest base to the smallest base.

And that would be in rotation three.

And part B says, how much does the height of the juice change during the rotation that you identified in part A? We identified rotation three.

Well, this time we will need to work out the heights.

So the volume of the liquid will be 1,728,000 millimetres cubed.

And then the height of the liquid before rotation three will be 60 millimetres.

The height after rotation three will be 180 millimetres, and the difference would be 120 millimetres.

And then question two.

We have our container which is a prism, but it is a composite solid or a compound solid.

And that's what makes this question a little bit trickier.

What is the height of the liquid after it's been rotated 180 degrees? Let's work through this.

We can start by finding the volume of the liquid in the first place, and luckily that is in the shape of a cuboid.

So we can do that by doing 20 times 18 times 20 to get 7,200 centimetres cubed.

Going forward it'll be helpful to work out the height of the triangle.

That would be six centimetres, which we get from doing 30, subtract 24.

And then when we look at the diagram on the right, if the height of the triangle is six, the distance between where the triangle ends and where the liquid ends, that would be H subtract six.

So, after the rotation, the area of the triangle at the front is 60 centimetres squared, which we get by doing 20, multiply by six and then times them by a half.

The area of the rectangle, well, we can't work out 'cause we don't know what the value of H is, but we can write an expression by doing 20 multiplied by H subtract six, which we could simplify to get 20 H subtract 120.

So that means the total area of the cross section will be the sum of those two parts, which simplifies to 20 H subtract 60, and that means the volume would be 18 times 20 H subtract 60.

And we know the volume is 7,200.

So we can create an equation now with that.

Solve the equation by adding 1,080 to both sides and then divide both sides by 360 to get H equals 23.

Great work so far.

Now let's move on to the next part of this lesson, which is problems involving multiple aspects of mathematics put together.

Here we have a triangular prism, and we're going to find the volume of it.

Oh, hello, Izzy.

Izzy's gonna help us with this problem, and she says, I need to find the area of the cross section which is a triangle, but I don't know the perpendicular height of the triangle, so I can't use the formula, area equals half times base times height.

Hmm.

If we can't use that formula, is there another formula that we know for finding the area of triangles? Well, Izzy says, I have an angle between the two known side lengths, so I could use the formula, area equals a half times A times B times sin C, where C is the angle between the lengths A and B.

So let's work through that together.

The area of the cross section is a half times five times six, which are two side lengths on either side of the angle times sin of 60 degrees, and that gives us 15 root three over two centimetres squared.

Now we could write that as a decimal, but we need to be careful not to round our answers early, 'cause it'll create an inaccurate answer further on.

It can be most helpful to leave it in surd form until we get to our final answer.

That means we can then do the volume of the prism, is 15 root three over two multiplied by 10.

And that would give 75 root three centimetres cubed, and that would be in its exact form, or we can write it as 130 centimetres cubed to three significant figures.

This time Izzy is finding the surface area of the triangle prism.

What extra information does Izzy need in order to find the surface area? And how could she obtain that information? What could she do? Pause the video while you think about this and press Play when you're ready to continue.

Well, let's take a look.

Izzy says, I need to know the lengths of all three sides of the triangle in order to calculate the area of all the rectangular faces, and one length is missing.

So that's the information she needs.

How could Izzy get it? She says, I could use a cosine rule to find the missing length.

So let's now work through this together.

To get the missing length, if we label it X to begin with, we could do X squared equals five squared plus six squared, the two lengths we know on the either side of the angle.

Subtract two times five times six times cos of six degrees, and then we can simplify that and solve it to get X equals root 31.

And once again we could write a decimal, but it's best to leave it in surd form, because that's the most accurate until we get to our final answer, then we can decide about rounding afterwards.

So now we can label that as root 31 centimetres, and we can start working out the surface area.

The area of each triangle, well, we worked that out earlier to get 15 root three over two centimetres squared.

The areas of the rectangles, well, one rectangle will be root 31 times 10, which is 10 root 31.

Another rectangle, the one underneath, will be five times 10, which is 50.

And the rectangle that's behind which we can't see will be six times 10, which is 60.

So, the total surface area would be two lots of the triangle plus each of the rectangles, which this calculation we can see here.

And this simplifies to get 192 centimetres squared when we round our answer to three significant figures.

Now, Izzy says, the total area of the rectangles could have been calculated by multiplying the perimeter of the cross section by the depth.

So we could reduce those three calculations to something slightly more efficient.

So the total area of the rectangles using this method would be to get the perimeter first, do root 31 plus 50 plus 60, and then multiply that answer by 10, and that would give 10 lots of 110 plus root 31, which we then use in our calculation, and we get the same answer as 192 centimetres squared to three significant figures.

So let's check what we've learned.

Here we have a triangle.

Find the area of this triangle and give your answer accurate to three significant figures.

Pause video while you do this and press Play when you're ready for an answer.

The answer is 7.

52 centimetres squared when round to three significant figures, and there's our calculations.

So let's now look at this triangular prism.

We know where the area of the cross section is 7.

51 and there are some more decimals there, so you can be a bit more accurate with the next calculation.

Calculate the volume of the prism and give your answer accurate to three significant figures.

Pause video while you do this and press Play when you're ready for an answer.

The answer is 60.

1 centimetres cubed, and there's our calculation.

We take our area of the cross section and we multiply it by eight.

And here we have a triangle where you know two lengths and the angle between them.

But you need to find the value of X, which is the length that is opposite the angle.

Give your answer accurate to three significant figures.

Pause video while you do this and press Play when you're ready for an answer.

We can get this by using the cosine rule.

Here's our calculations, and we get our answer of 4.

59 when round to three significant figures.

So with that in mind, could you please calculate the surface area of this triangular prism, and some of the measurements which we worked out earlier are shown on the screen.

Pause video while you do this and press Play when you are ready for an answer.

The answer is 116 centimetres squared.

And there's our calculations.

Let's now take a look at another problem which involves a cuboid and some different aspects of mathematics.

Here we have a cuboid where we're only given one of its lengths, the five centimetres, but we are also given some additional information as well.

We are told that the volume of the cuboid is 60 centimetres cubed, and we are told that the length of the base is one centimetre greater than the height.

And what we're going to do is work out the height of the cuboid.

Here comes Jun, who's going to help us with this question.

He says, I could let X centimetres equal the height of the cuboid, and that's a great idea, because when we have unknowns that we want to work with, we can use algebra to help us along the way.

So if the height is equal to X centimetres, then the length of the base would be X plus one centimetres because the length for the base is one centimetre greater than the height.

Jun then says, I could create an equation to find the value of X.

So let's do that now.

We know the volume is 60 centimetres cubed.

To get the volume I would multiply all three of those lengths together, five times X times X plus one, and that will just create this equation here.

60 equals five X multiplied by X plus one.

We could expand the brackets, and then we can see we have a quadratic here, so we can solve it by rearranging it so we have zero on one side of this equation.

And to solve this quadratic, could do it by factorising it.

If we factorise it, well, the coefficient of X squared is greater than one.

So we could use this method here, where we do five multiplied by negative 60 to get negative 300.

And then think about two values which add to get five, the coefficient of X, but multiply to get negative 300.

That would be 20 and negative 15.

And we can use those two numbers to split the middle term.

That would give us zero equals five X squared plus 20 X, subtract 15 X, subtract 60.

Or you can have those two middle terms your way around if you want to.

But now we're in a position where we can factorise the first two terms and factorise the last two terms. That would give us this here.

We can now see that X plus four is a common factor here.

So we can take that out as a factor and get five X subtract 15, multiplied by X plus four.

And with that we can create two equations, because if these two things multiply to get zero, then either five X subtract 15 is equal to zero, or X plus four is equal to zero, or both.

We can now solve both of these to get these two values here.

X equals three or X equals negative four.

Now, Jun says, a cuboid cannot have a negative length, so X can't be negative four, it must be three.

And as the height is X, it means the height must be three centimetres.

Now we have an answer, but Jun says, I could have solved the quadratic equation in a simpler way.

And can we see why? If you look back at this step here, we have zero equals five X squared plus five X subtract 60, Jun says, at this stage I could have simplified my equation by dividing everything by five.

If we do that, we get this equation here.

This is a much simpler quadratic to solve.

We can factorise it by getting X plus four multiplied by X subtract three, and then create two equations from that, and solve them to get the same answers again.

Let's check what we've learned.

Here we have a cuboid where you don't know any of the lengths, but you do know that the length for the base is five centimetres greater than the height.

Write an expression for the base and height of the cuboid.

Pause video while you do this and press Play when you're ready for an answer.

There are lots of different ways you can do this.

A straightforward way will be to label the height X centimetres, which means the base will be X plus five centimetres.

But other way could be to label the base as X centimetres and label the height as X subtract five centimetres.

The difference between those two expressions is still five, or you may use other letters or you can use other expressions that also have a difference of five as well.

Let's settle with this one.

Where the height is X, the length of the base is X plus five and the depth is four.

Write an expression for the volume of the cuboid.

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

When you multiply these lengths together, you can get either of these two expressions.

The one on the top is a factorised version and the one on the bottom is an expanded version.

I'm gonna tell you now that the volume of the cuboid is 56 centimetres cubed.

Find the height of the cuboid.

Pause video while you do that and press Play when you are ready for an answer.

The answer is two centimetres, and there's the work on the screen for how to get it.

Okay, it's over to you now for task B.

This task contains three questions, and here is question one.

You've got a triangular prism with some measurements given to you.

Find the volume of the prism and find the surface area of the prism, given your answer accurate to three significant figures.

Now, you may need to use other aspects of mathematics in order to do these things, because you may not have the information you need straight away.

Pause video while work through this and press Play when you're ready for question two.

And here is question two.

You've got a cuboid which has a volume of 108 centimetres cubed.

Now, you're only told one of the lengths, that is the two centimetres you can see there.

But you're also told that the length of the base is three centimetres greater than the height.

And what you need to do is first find the height of the cuboid and then find its surface area.

Pause video while you do this and press Play when you're ready for question three.

And here is question three.

You've got a cone, and you've got some useful formulas at the bottom of the screen there that relate to cones.

The volume of the cone is equal to six pi centimetres cubed.

The height is twice the length of its radius.

And what you need to do is calculate the total surface area of the cone, given your answer accurate to three significant figures.

Pause video while you do that and press Play when you are ready for some answers.

Let's now go through some answers.

Question one, part A.

The volume of the prism is 108 centimetres cubed, and there's the work of how to get to it.

In part B, to find the surface area of the prism, you'll first need to work out the missing length on the triangle, which you can do with the cosine rule.

That gives you 4.

1 and some more decimals.

I recommend not to round those decimals until you've definitely got a final answer, and you get the surface area, we can do that calculation there with our answer in it, to 186 point something, and that rounds to 187 centimetres squared to three significant figures.

Now question two, we need to find the height of the cuboid.

A good way to start could be to write some expressions for lengths we don't know.

There's lots of different ways you can do that, but let's go with X centimetres for the height and X plus three centimetres for the length of the base.

We can use that to create an equation which is a quadratic.

So we can rearrange, factorise, and solve to get the height of the cuboid as being six centimetres.

And then we need to find the surface area of it, which we can do by labelling on the lengths we know now that the height is six centimetres and the length of the base is nine centimetres, which is six plus three.

And use those to find the surface area of 168 centimetres squared.

And question three, we have a cone which we need to find the surface area of, but we're not given any of its lengths.

We're given that the volume is six pi centimetres cubed, and we're given that the height is twice the length of its radius.

So we need to use that information to find some missing lengths first, and then use that to find the surface area.

If the height is twice the length of its radius, then in the formula for volume of a cone where it says H, we can replace that with two R, two times a radius.

That means we've got an equation which is six pi, the volume of the cone, equals to a third times pi times R squared times by, and rather than H, two R.

That part on the right of the equation is taking it from the formula for the volume of a cone.

We could simplify and rearrange that equation to get R cubed equals nine, which means R is equal to the cube root of nine and the height H is equal to two lots of the cube root of nine.

Again, you could write out a decimal, but you've gotta be careful not to round anything too early, otherwise that answer will be inaccurate at the end.

Best leave it as the cube root of nine until we've got our final answer.

So are we yet in a position where we can find the surface area of this cone? The formula on the bottom left there has pi times R times L.

Well, we know a value for R, but we don't know a value for L yet.

We need to work that out.

We're can do that using Pythagoras' theorem, which we can see on the screen here.

We can create a right angle triangle from that cone where L is the hypotenuse, and the height and the radius are the two shorter side, which are perpendicular to each other.

So we could substitute that into Pythagoras' theorem and rearrange it to get 4.

65 and some more decimals for the length, L.

And now we have everything we need to find the surface area of the cone.

The curved area is equal to pi RL, and then also we have a circle at the bottom underneath the cone as well, which is pi R squared.

Substitute everything we know and we get 43.

9874, and so on, which we can round to get 44 centimetres squared, round to three significant figures.

Fantastic work today.

Now let's summarise what we've learned.

The volume of any solid can be calculated by a known method.

The volume of liquid inside a container remains the same when the orientation of the container is changed, but the height might change.

The surface area of any solid can be calculated by a known method, and the volume or surface area can be used to find missing lengths of a solid, and form an equation based on known information about a solid can sometimes help you to find missing information.

Well done today, hope you have a great day.