Lesson video

Hi, and thank you for joining me.

My name is Dr.

Rawlinson and I'll be guiding you through this lesson.

So let's get started.

Welcome to today's lesson from the unit of 2D and 3D Shape with Surface Area and Volume.

This lesson is called 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 will be useful during today's lesson.

So you may 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 learning cycles.

In the first learning cycle, we're going to look at Contextual Problems. And then the second part of the lesson, we'll look at Abstract Problems, but let's start off with Contextual Problems. The first context is gonna be focusing on apple juice or orange juice sold in cartons and cans.

Here we have a carton of orange juice.

One litre of orange juice is sold in a carton.

Now Lucas says, "I know that one millilitre of liquid fits into one cubic centimetre," or centimetre cubed.

He then says, "There are 1000 millilitres in one litre, so the carton must have a volume of 1000 centimetres cubed." We're going to check that, and here are some dimensions for this particular carton.

Will this carton hold one litre of orange juice? And that's what we're going to work out.

Lucas says, "To work this out, I need to calculate the volume of each carton.

He then says, "If the carton has a volume of at least 1000 centimetres cubed, then it will hold one litre of juice." So you are going to help me with this.

Let's start by breaking this problem down and making sure we understand what the dimensions of this carton tell us.

The carton is a prism.

The volume of a prism is found by multiplying the area of the cross section by the length of the prism.

Looking at this prism, what is the length of it? Pause the video while you choose from those four options and then press play when you are ready for an answer.

The answer is B: seven centimetres.

What shape is the cross section of this prism? Pause the video while you choose from one of those four options and then press play when you're ready for an answer.

The shape is a pentagon, which is D from our options.

So now we've got a sense of what this carton looks like, let's find the volume of it.

And we first need to find the area of the cross section.

Now the area of the cross section can be found in different ways.

It's a pentagon, but we could break this pentagon up into different shapes.

One method could be to split it horizontally, so we have a rectangle and a triangle above it.

Let's find the area for each part.

The height of that triangle must be four centimetres.

We can get that by doing 20 centimetres, the total height, subtract 16 centimetres, which is the height of the rectangle, and that'll leave four centimetres for the triangle.

So when we find the area of the triangle, we can do 1/2 times base times height, so it'll be 1/2 times 8 times 4, which gives 16 centimetres squared.

And then the area of the rectangle can be done by doing 8 times 16, which is 128 centimetres squared.

So we know the area of each part of the shape.

To find a total area of the cross section, we need to add those together, which gives us 144 centimetres squared.

That was one method, but a different method could be to split it in a different way.

We could split it vertically.

And what we have now is two trapezium.

Let's find the area of each trapezium.

Well, the distance between the parallel lines on a trapezium is four centimetres, and we got that by dividing the eight centimetres by two.

So the area of a trapezium is half multiplied by the sum of the parallel sides, so that's 16 plus 20, multiplied by the distance between them, which is four.

When we do that calculation, we get 72 centimetres squared.

Now there are two trapezium here and they're both congruent.

So they both have the same area, which means the total area of the cross section would be 2 times 72, which would be 144 centimetres squared.

Again, same answer, we got the other way.

So back to our problem.

We have a carton of orange juice and we want to know will this carton hold one litre of orange juice? And Lucas says, "To work this out, I need to calculate the volume of each carton.

Well, the length of the prism is seven centimetres and the area of the cross section we've worked out as 144 centimetres squared.

So we now have all the pieces we need to find the volume.

The volume would be the area of the cross section multiplied by the length.

So we'd do 144 multiplied by seven to get 1008 centimetres cubed.

So yes, the of the carton is 1008 centimetres cubed, which is more than 1000 centimetres cubed.

So that means it will hold one litre of orange juice.

Here we have a new carton design now.

It's also a prism.

A pentagon is the cross section, but it has slightly different dimensions to the one on the left.

Now we are told that it will hold one litre of orange juice.

So what we're gonna focus on this time is, which carton uses the least amount of packaging? Lucas says, "To work this out, I need to calculate the surface area of each carton." To find the surface area of a prism, it helps to consider what the net of the prism looks like.

So you are going to help me with this.

Here we have four options for a net.

Which of the following nets are of this prism? And there may be more than one answer.

Pause the video while you choose and press play when you are ready for some answers.

Well, let's break this down.

The cross section of the prism is a pentagon, so it'll have two pentagons on the net and it'll have five rectangles, one rectangle attached to each side of the pentagon.

So we look at A, yes, that one works.

We've got two pentagons and we have five rectangles.

In B, we have six rectangles, which means those faces at the bottom would overlap when the net is folded together to make the prism.

For C, there isn't enough faces.

There are only four rectangles and we need five.

For D, yes, that one would work.

We have two pentagons and five rectangles and they are arranged in places, which means they won't overlap with each other when it's folded up.

Let's use this net here.

To find the surface area, we need to find the area of each of these faces and then find the total area of all of those.

Well, we worked out the area of the cross section earlier and that is 144 centimetres squared.

So we know the area of two of these faces, but what we need to work out next is what are the areas of the other faces? Well, before we can work out the areas of the other faces, we need to figure out what are the dimensions on these rectangles? So you can help me with this.

What are the dimensions of the rectangles? We have five lengths labelled A to E.

Could you please compare the net to the picture of the prism on the left and work out what is each of those lengths? A, B, C, D, and E.

Pause the video while you do that and press play when you're ready to see the answers.

Okay, let's take a look.

A is 5.

7 centimetres.

That side of the rectangle would be adjoined to one of the sloping sides at the top of the pentagon.

B will be 16 centimetres.

It is the height of the rectangle when we split it into a rectangle and triangle.

C is seven centimetres.

It's the length of the prism, the distance between the front pentagonal face and the back pentagonal face.

D is eight centimetres and E is also seven centimetres.

So now we have all the dimensions of the rectangles, we are in a position where we can work out the area of each rectangle and here they are.

Now we have the area of each face on this prism, we can find the total surface area by adding them all together.

And if we do that, we'd get a total surface area of 647.

8 centimetres squared.

So over to you to work out one yourself now.

Here we have the other carton that we saw, the new design and we have a net for this carton.

Use the net to find the surface area of the new carton.

Now you'll need to do several calculations in order to get this, find the area of each face and then add them together.

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

The area of the cross section is 108 centimetres squared, so we know the area of two of the faces.

And these are the areas of all the rectangles.

And then when you add them together, you should get 636 centimetres squared.

So back to our problem.

We wanted to know which carton uses the least amount of packaging.

Well we have the surface area of each now, and what we can see is that the surface area for the new carton is less than the surface area of the old carton.

So Lucas says, "The new carton has a smaller surface area, so it will use less packaging." So we've looked at drink cartons, now let's look at drink cans.

Here we have two cans.

They are both 33 centilitre drink cans.

Sofia says, "I know that 1 centilitre is the same as 10 millilitres.

So these cans hold 330 millilitres of drink." Lucas says, "This means that the volume of these drink cans is 330 centimetres cubed." Now these cans are made from aluminium.

Which cylindrical drink can is made from the least amount of aluminium.

So once again, this problem is looking at the amount of packaging.

So Lucas says, "We need to find the surface area of each cylinder." And Sofia says, "First I need to find the height of the apple juice can." Oh yeah, we don't know what the height of that can is.

But let's consider what we know about this can.

They are 33 centilitre drink cans.

So we know the volume is 330 centimetres cubed.

We also know that to find the volume of a cylinder, you do the area of the cross-section multiplied by the height.

We know the radius as well.

So Sofia says, "I know the volume of the apple juice can and I know the radius, so I can calculate the area of the cross section." The area of the cross section is equal to pi times radius squared, because it's a circle.

The radius is three, so it'll be pi times 3 squared, which is pi times 9, and that will be 28.

274, and there are more decimals in centimetres squared.

So now we know the area of the cross section, we can calculate the height of the apple juice, because the volume is equal to the area of the cross section times the height.

So if we substitute in the volume and the area of the cross section, we have an equation that looks a bit like this.

So to work out the height, we could do 330 divided by the area of the cross section, and that would give 11.

671 and there are more decimals.

Let's round it to 11.

7 to one decimal place.

So we now have enough information to work out the surface area of both drink cans, and Lucas says, "We need to find the cylinder with the smallest surface area." Well, to find the surface area of a cylinder, we need to calculate the area of its net.

The net would look something a bit like this, two circles and a rectangle.

And here are the dimensions that we know.

We've also worked out the areas of these circles here.

They're both 28.

3 when they are rounded to one decimal place.

So we just need to work out the area of the rectangle, but we only have one of the lengths in that rectangle so far, that's the 11.

7 centimetres.

What about the other length? Well that would be equal to the circumference of the circle because it wraps around the circle.

So Sofia says, "I need to calculate the circumference of the cross section." To do that we'd use the formula 2 times pi times radius.

If we do that we would get 18.

849 centimetres.

So let's now do the area of the rectangle by doing that number multiplied by 11.

7, and that would give 220.

5 centimetres squared to one decimal place.

So now we have all the information we need to find the total surface area.

We can do two times the area of the circle plus the area of the rectangle, and that would give 277.

1 centimetres squared.

So here's the other drink can.

Could you please use a net to calculate the surface area of this cylindrical drink can.

You'll need to break this down into lots of small parts.

Work out the area of the circles, work out the circumference, which is the length of the rectangle, and then you can work out the area of the rectangle and add them all together to get your total surface area.

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

Well, the radius is 2.

7 centimetres, which means we can work out the areas of the circles to be 22.

9 centimetres squared with some more decimals afterwards.

We can get the circumference to be 16.

9 something centimetres, and then that means the area of the rectangle will be 245.

9 something centimetres.

And if you add it all together, we'd get 291.

8 centimetres squared, when rounded to one decimal place.

So now we have the surface area of each drink can, we are in a position where we can answer our problem, which cylindrical can could be made using the least amount of aluminium? Well, the apple juice has the smaller surface area, so can be made with less aluminium than the lemonade can.

Let's go a bit further and think about cost then.

What is the difference in the packaging costs for each can? Well, we know the surface area of each can and we know the cost per centimetre squared.

So what we need to do is do a multiplication.

We can do 0.

13 multiplied by the surface area.

For the apple juice, that works out as 29 pence per can.

And for the lemonade, that works out as 38 pence per can.

That means that apple juice can cost approximately nine pence less to make.

Okay, it's over to you now for Task A, this task contains five questions and here is question one.

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

And here is question two.

Pause the video while you work through it and press play when you're ready for question three.

Here is question three.

Pause while you work through it and then press play for question four.

Here is question four.

Pause while you work through this and press play for question five.

And here is question five.

Pause while you work through this and press play when you're ready to go through some answers.

Okay, let's go through some answers.

For 1a, which cuboid has the greater volume? Well, the volume for cuboid A is 280 centimetres cubed.

For volume B, it would be 294 centimetres cubed.

So it'll be cuboid B has a greater volume.

Which cuboid has the greater surface area? Well, the surface area of A would be 328 centimetres squared.

The surface area of B would B 322 centimetres squared.

So that means cuboid A has the greatest surface area.

In question two, Sofia is going to paint 10 of these boxes and she needs to buy several pots of paint.

Each pot will cover 7.

2 metres squared and cost 18.

99 pound.

So how much will Sofia need to spend on paints? Well, we can see of the cuboid that the lengths are in different units, so we wanna make them into the same units.

And we know that the paint is in metres squared, so let's make them into metres.

We can then find the area of each face and then find the total surface area, which is 1.

5 metres squared.

And then we can work out the area of paint which is needed.

There will be 10 boxes, so it's 10 lots of 1.

5, and then we can figure out how many pots of paint is needed.

We can do that by dividing the total area by the amount of area that each paint pot will cover.

We need slightly more than two pots.

We're gonna need three pots.

So three lots of the cost would give us 56 pound 97 pence.

Then question three, the prisms are in ascending order of volume, but we want to put them in ascending order of length.

That means you need to work out the area of the cross section of each one if you're not told it, and then work out the lengths.

For A, the length would be 8.

5 centimetres.

For B, there's your area of the cross section, so the length would be 9.

6 centimetres, and C, you need to work out the area of the cross section and get the length of 8 centimetres.

So, our answer would be C, A, and then B.

In question four we have the old shape of a one pound coin and the new shape and we need to work out which one pound coin is made from less metal? Well, if the metal just surrounded a coin we'd be considering the surface area, but that's not the case.

The metal runs throughout the coin.

So we're thinking about the volume here.

So you need to work out the volume of each one.

The volume of each one is on the screen here.

And then you can see that the new one pound coin has a smaller volume so will need to be made from less metal.

And then question five, the Oak Chocolate Company used to package its sweets in cylindrical boxes.

It decided to change its packaging into hexagonal prisms. The packaging costs 0.

08 pence per centimetres squared.

How much money is saved per tube? Well, here's the surface area of the cylinder and then the packaging costs for that.

And here's a surface area for the hexagonal prism and the packaging costs for that.

So the difference of cost of packaging for one tube will be 0.

576 pence.

Now that doesn't sound like very much, but in a year, the Oak Chocolate Company makes 46.

5 million packets of sweets.

How much money is saved over the year? Well, you can do the cost that is saved multiplied by the 46.

5 million and that would give 26,784,000 pence, which is equal to 267,840 pounds.

That's quite a lot of money you have to save.

Great work so far.

Now let's move on to the second part of this lesson, which is looking at abstract problems. Here we have a cuboid.

And on this cuboid we are given all of its dimensions, so we can use that information to work out the volume and surface area.

This would be the volume.

We can multiply the three lengths together to get 3072 centimetres cubed.

And this would be the surface area.

Each multiplication we can see in brackets is the area of one of the rectangles that we can see.

And each rectangle that we can see in this cuboid has another one that is congruent to it, which is why we are timesing by two.

Then we add them together to get 1,408 centimetres squared.

But here we have another cuboid where we are not given all of its dimensions, which is why one of its dimensions is labelled x, that represents that it's unknown.

Now we don't have all the information we need to work out the volume, but what we could do is write an expression for the volume instead.

And we'll do that by using the same method we would do to work out the volume, or we just have an x in the calculation instead.

So we would multiply our three lengths together.

One of the lengths is x.

So which means we need to simplify our answer rather than actually work out what it is, and we'd get 384x centimetres cubed.

And we can't work out what the actual surface area is because we have our unknown.

But what we could do is write an expression for it.

We do all the same calculations we'd normally do.

So in the brackets we have the area of each rectangle, but some of those has an x in and then we're timesing by two because there are two congruent rectangles each time.

And then if we work out each of those multiplications, we'd get this, and we could simplify it to get this, 768 + 80x.

Now if I wanna put the units on the end of this centimetres squared, I can't just write it after the second term because then it looks like only that part is the area.

So to say that all of this is the area, I'm gonna put it in brackets, write it as, in brackets, 80x + 768 and then centimetres squared.

So over to you now.

Here we have another cuboid.

You've got two of its dimensions and one of them is unknown and labelled y.

Could you first please write an expression for the volume of this cuboid and then write an expression for its surface area.

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

Let's look at the answers.

Part A would be 128y centimetres cubed, and part B, in brackets would have 48y + 256 and then centimetres squared.

Here we have another cuboid where all three lengths are unknown and they're labelled x, y, and z.

The volume of this cuboid can be written as xyz centimetres cubed.

Which of these expressions could represent the surface area of the cuboid? And there may be more than one correct answer.

Pause the video while you choose and then press play when you're ready to see what the answer is.

The answers are B and C.

That's what you'd get if you multiplied each pair of lengths together to find the area of the rectangle and then consider all six rectangles and add them together.

Algebraically, you can see why B and C are equivalents because if you expand the brackets for the expression in B, you'd get the expression for C, or if you factorised the expression for C, you'd get the expression for B.

Here we have another cuboid, where we are told one of its lengths, but the other two lengths are unknown.

Now we could use two different letters for those lengths, x and y, for example, but if we know a relationship between those lengths, then we could use that relationship to express them both using the same variable.

For example, expressed them both as x.

For example, we could be told that the height of the cuboid is double the depth, so we could use that relationship to express them both using the same variable.

If we label the shorter one x, then that would mean the other one is double that, so it would be 2x, like this.

And then with that, we could express the volume in terms of x.

If we multiply those three lengths together, we'd get 48x squared.

And we could do the same with a surface area.

We could do all the calculations we normally do, but with x and 2x in and we'd get a surface area expressed as 4x squared + 144x.

Here we have another cuboid where all three lengths are unknown.

Once again, we could label them x, y, and z, but if we know a relationship between those three lengths, we could use one variable and that would make it easier to express its volume, its surface area, and other qualities about it.

For example, we might be given the relationship as a ratio, such as the ratio of the width to the height, to the depth being three, to two to one.

That means we don't have to use a different letter each time.

We can express them all with the same letter, using that relationship.

The relationship is three to two to one as a ratio.

We could label any of those parts as our variable, but it would probably easiest to use the smallest one, the one, and label that one as D.

If that length is D, then the other ones are 2D and 3D.

We can label it onto our cuboid.

And now we have that, we can express things like the volume in terms of D by multiplying them all together to get 6d cubed.

Or we can express the surface area by working out the expression for each area of each face and then considering all six faces and adding them together and we would get 22d squared.

Let's look at a different shape now.

Here we have a rectangular-based pyramid and we have two lengths of that pyramid which are labelled, but we don't know what they are.

So we could label one of them as x, and the other one as y using two different variables.

But if we know a relationship between those two lengths, then we could express them using the same variable.

Let's take a look.

The width of the base is half the length of the base.

Well, that means we could label one of them as x and the other one as 1/2x, use a different letter, but we can consider this relationship in a different way.

If the width is half the length, that means the length is double the width and that might be an easier relationship to deal with.

We could label the width as w, which means the length could be labelled as 2w, because it's double.

Here we have the net now of this pyramid and we've got a little bit more information given to us.

We've got the heights of two triangles there.

Given this net, write an expression for the surface area of the pyramid.

Well, we could start by labelling on our variables onto the net.

We know that the width is w and the length is 2w.

And then to find a surface area, we need to find the area of each of those faces.

If we label these ones A, B, and C, we've got three of those faces covered.

You may think why have we not labelled the other two faces? Well those other two triangles are congruent to triangles B and C, and that means they have the same area.

Area of the rectangle A will be 2w squared.

The area of the smaller triangle B, would be 25w.

The area of the larger triangle, labelled C would be 36w.

And then we can work out an expression for the surface area.

We could do the area of the rectangle plus two lots of the triangle B, plus two lots of triangle C and add them all together and we'd get the surface area as being 2w squared + 122w.

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

This task contains two questions and here is question one.

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

And here is question two.

Pause the video while you work through this and press play when you're ready for some answers.

Okay, let's now go through some answers.

With question one, we are given the ratio between these three lengths here.

So when we expressed them in terms of k, we'd get 2k for the height and 5k for the width.

And then we can use that to find an expression for the volume, which would be by multiplying them all together to get 10k cubed.

And then to write an expression for the surface area, well you could write an expression for the area of each rectangle.

You get 10k squared, 5k squared, and 2k squared.

And don't forget, there are six faces to a cuboid, so we would add them together and then multiply by two, or we can multiply each one by two and then add them together afterwards.

Either way is fine.

And you should get 34k squared.

Then question two, we need to use the net and any relationships we're given to write an expression for the surface area of the 3D solid, in terms of x.

In this square-based pyramid, we're told the height of each triangle is double the sides of the square base, which means the height must be 2x.

So then we can write an expression for the area of each face.

The area of the square would be x squared.

The area of each triangle will be 1/2 times base times height, so that's 1/2 times x times 2x, which means the surface area of the pyramid will be the area of the square plus four lots of the area of the triangle where that'll give 5x squared.

And then we have our cone where the diameter of the semicircle is four times the radius of the circular base of the cone.

Well the radius of the circular base is x, which means the diameter of the semicircle must be 4x, and we will need the radius for that semicircle, which will be 2x.

And then we can do the area of the semicircle by doing 1/2, 'cause it's a semicircle, times pi r squared, where r this time is 2x, and that will simplify to 2 pi x squared.

The area of the circular base will be pi times radius squared, which will be pi x squared.

Add them together and we get 3 pi x squared.

Fantastic work today.

Let's now summarise what we've learned.

To find the surface area of a 3D solid, you can calculate the area of its net.

The volume of a prism can be calculated by multiplying the area of the cross section by the length of the prism.

We need to check that the units of length are all the same before we perform any calculations.

And finally, when we have unknown lengths, writing an algebraic statement about the surface area or volume can be done from a diagram of a 3D solid.

Well done today.

Have a great day.