Lesson video

Hello, I'm Mrs. Lashley, and I'm really looking forward to working with you as we go through this lesson today.

So in today's lesson, we're gonna be working with perimeter, area, and volume to solve a variety of different questions and problems. There are gonna be words that we're gonna use during the lesson that you will have met before and hopefully feel familiar with already.

But if you wish to pause the video now and reread the definitions just to make sure that you're ready as we start the lesson.

Our lesson has got three learning cycles.

The first part is surface area and volume.

So working between the two of them and seeing if they can be connected in any way.

The second part is composite solids, and the last part is problems in context.

So we're gonna make a start on learning cycle one, which is all to do with surface area and volume.

So the volume of both the cube, which is being represented here in net form, and cylinder are equal.

And we want to find the radius of the cylinder.

So we've got a cube, which we know is a 3D solid, is a prism, and we've got a cylinder, which is like a prism, but isn't a prism.

And the volume of both of them, so the amount of space the 3D shapes take up is the same.

So when we are faced with a problem, and we're gonna be faced with many problems throughout today's lesson, we should always start by considering the information that we have, what we've been given, and what other information that can then give.

So not to worry too much about what you need to do, but think about what you can do with the given information.

So Jun is doing just that with this same problem.

He knows that all of the faces of a cube are square, and if they are square, that means all the edges are equal.

The question told us it was a cube.

And so, "What do I know about a cube," is what Jun has first thought about.

What does he know? Well, he knows the faces are square, and then he's followed that up with, "Well okay, I know that squares have equal edges." Jacob has followed that up to say, "Well, the net shows that four of those edges are 20 centimetres." Because they are connected as a net and they are all square, so the edges, the four edges there have a total of 20 centimetres.

And then that then leads on to say, okay, so if I now know that four edges as 20 centimetres, and because they are square, they are equal, that would mean that one edge is five centimetres.

So now I've got enough information to get the volume of the cube.

Jun has correctly calculated the volume as 125 cubic centimetres.

So we've started with a net, where we've just got this distance, this length from the top to the bottom, as 20 centimetres.

But because it's a cube, because the faces are square, because squares have equal edges, we've then got down to the length of one edge, and from that we can get the volume of the cube.

And Jacob has then linked that back to our problem, that that means that the cylinder has a volume of 125, because we were told that they had equal volumes.

So by getting the volume of the cube, that's also given us the information of the volume of the cylinder.

The height of the cylinder is four, and therefore the area of the circle is 31.

25.

Because the volume of a cylinder is calculated by doing the area of a circle multiplied by the height.

So he has done division to undo the multiplication of the height in the volume, and then we're left with the area of the circle.

Jacob has then gone, "Well, okay, so if the area of the circle is pi R squared, and that equals 31.

25, then I can use all my knowledge of inverse and solving equations to get the radius as 3.

2 centimetres to one decimal place.

And we've then answered the question, find the radius of the cylinder.

So we started from a net of a cube and all the information we knew about cubes to get the volume of a cube.

We knew the volume was the same as the cylinder in this particular question.

We've used the height of the cylinder to get the area of the circle.

And from the area of the circle, we've got all the way to the radius.

So here is a check.

Using that thought process of, what do I know and what does that then tell me? So a cube has a volume of 27 cubic metres.

"What information can you deduce from this?" Pause the video, and then when you're ready to check, press play.

Firstly, from the volume of a cube, you might cube root it to get that the edge length is three metres.

Because three multiplied by three multiplied by three is 27, three cubed is 27.

And that's using the fact that a cube has square faces or the edges are equal, so the volume of a cube is just the edge length cubed.

But if you know the edge length, then you also know.

So if you've gone from the volume to the edge length, from the edge length, you would also then be able to deduce that the area of one face is nine square metres.

Because it's a cube, the face is a square, and three squared is nine.

That also means that if you know the area of one face, then you know the surface area.

Because it's a cube and all of the faces are square, if you know that each face has an area of nine, and there are six faces on a cube, nine multiplied by six is 54.

So you now have gone from the volume to actually knowing the edge length, the area of a face, the surface area from just that one piece of information.

And this is because it was a cube.

A cube has got these equal edges, and that's partly why we can get so much.

This one you may have missed, but you from this, if you know the edge length, then you can also get the perimeter of the net.

A cube has got 11 unique nets, however, each one of them has got the same amount of edges on the exterior, on the perimeter.

So it's 42 metres when the edge length is three.

So from that, can you work out how many edges are on the boundary of a net of a cube? Okay, so let's continue with problem solving with volume and surface area.

So here we've got a cuboid.

So it's no longer a cube, so it's not about equal edges.

A cuboid has a volume of 585 cubic centimetres, a height of five centimetres, and a length of 13 centimetres.

What is the surface area of the cuboid? On the screen right here, we've just got a diagram of a generic cuboid with the width, the length, and the height labelled.

The volume of a cuboid is the area of the cross section, which is a rectangle, so length times width, and then multiply it by the height, so that's length times width, times height.

The surface area of a cuboid, surface area is the total of all the faces areas.

And a cuboid has got pairs of congruent rectangular faces.

So width times height would give you one of the rectangles, and then we're gonna double it because there are two of them.

Width times length will give you a second rectangular face, and we are gonna double it 'cause there are two of them.

And then length times height is the third rectangular face, and then double it because there are two of them.

We're gonna sum them together to get our surface area.

So there are the formulas that we're gonna be working with in this problem.

We cannot calculate the surface area until we have all three dimensions of the cuboid.

And in the question, we were only given two of the three, but what we was also given was the volume.

So from the volume, we can get that third length, which in this case is the width.

So the volume was 585, the height was five, the length was 13.

So if we substitute that in and solve, we get the width to be nine.

Now we have all of the information we need to work out the surface area of the cuboid.

So these problems, think about what you have and what that means you could work out.

And from that point, you're probably closer to having enough information to answer the problem you're trying to.

So now we know all three dimensions of the cuboid, we can substitute that into that formula of the surface area, and get the surface area of 454 square centimetres.

Here is another problem to do with surface area and volume.

So, "The surface area of this cuboid is 488 square centimetres.

All edges are an integer number of centimetres.

Areas of two of the faces are given.

What is the volume of this cuboid?" So the problem we're trying to answer is, what is the volume of the cuboid? What do we know? Well, we know the surface area, and we know two of the faces of the cuboid.

Sam is reminding, or has recognised, and hopefully you recognise this as well, that they have three pairs of congruent rectangular faces.

So every cuboid, the sort of top and the bottom face are identical.

The front and the back are identical.

The left hand side and the right hand side are identical, the sort of opposite ones.

So if we label that third face, which is gonna be the third pair of congruent rectangular faces, as X, 'cause we don't know it yet.

So if we give it an algebraic term.

Then we can write out the surface area as 84 plus 84 because there are a pair of 84s, the top and the bottom, plus 112 plus 112 'cause there are a pair of those, plus x plus x because there are a pair of those, and that equals 488.

We were given that in the question.

Here's now just a linear equation which we will be able to solve.

Simplifying and evaluating the numbers, the sum, we get to 392.

And X at X is 2X simplified, and that still equals 488.

So we can now solve this linear equation.

Subtract 392 from both sides, divide it by two, the X is 48.

So we now know that the third rectangular face has an area of 48.

So we've moved forward a little bit.

We've still not got the volume of the cuboid, but we've worked out something.

And working out something will probably lead you closer to the right answer.

So between each of the faces, there is a shared edge.

And if you think about when we unfold a 3D shape into its net, the shared edge is a dimension that is part of both of the faces.

And so this is gonna be a common factor.

This is because the area of a rectangle, this is a cuboid, so the faces are rectangular.

The area of a rectangle is two numbers multiplied together, is a product of two numbers.

And a product of two numbers gives you a multiple, and they can be known as the factors.

So the shared edge will be a factor of both 84 and 48, for example.

That shared edge that has just appeared on the diagram we're gonna call X.

If you went through and thought of all the factors of 84 and all the factors of 48, the ones that they have in common, the ones that overlap would appear in both of your lists.

And so that shared edge is going to be one of those.

The length of our cuboid I've called Y, and that is the shared edge between 84 and 112.

So that Y value is one of the common factors of 84 and 112.

Finally, the third dimension of our cuboid is the shared edge between the face 48 and the face with the area 112.

The X multiplied by Y is 84 'cause that is the shared edge, which is the width of our cuboid, and Y, which is the shared edge, which is our length.

So X multiplied by Y has to equal 84 to give us the area of the rectangular face, that's the sort of top of our cuboid.

So, which numbers from the list that we already identified make this true? So, which of them have a product of 84? There's two options.

Six multiplied by 14 gives you 84, and seven multiplied by 12 gives you 84.

So right now, we know that X is either six centimetres or 12 centimetres, and Y is either seven centimetres or 14 centimetres.

We also know that Y multiplied by Z is 112.

We need to to go through the same process of from those two lists of values for Y and for Z, which numbers from the lists make a product of 112? But remembering that Y is seven or 14, we have already narrowed it down to seven or 14.

Which pairs would give you a product of 112? Well, seven, if Y is seven, Z could be 16, that is a product of 112.

Or if Y is 14 Z, could be eight, that would give you a product 112.

So Z is either eight or 16.

So now, we know that XZ is 48.

And X is either six or 12.

We calculated that at the beginning with the 84.

Or Z is eight or 16, and we just figured that out using the 112.

So X is either six or 12, Y is either seven or 14, and Z is either eight or 16.

Which X and Z pair gives us 48? Hopefully you can see that, the six multiplied by eight gives you 48.

Which means that X is six, Y is 14 because six times 14 gives you the 84, and Z equals eight.

And if you check, 14 times eight was 112.

Now we know the dimensions of our cuboid.

So we've gone from our surface area, to our face areas, to then our dimensions, our lengths, widths, and heights, which means we've got all of the edges, so we can calculate the volume.

The volume of this cuboid would be 672 cubic centimetres because six times eight is the area of the cross section, and then multiplied by 14 is the length of the cuboid.

You're now gonna do some practise to do with surface area and volume.

So question one, the surface area of this triangular prism is 240 square centimetres.

What is the length of the prism is part A, and hence calculate the volume of the prism is part B.

Pause the video whilst you work through that question.

When you're ready for the next questions, press play.

Question two and three are both on this slide.

So question two, you've been given the volume of a cylinder and its height.

You need to work out the cylinder's surface area.

So again, go through that thought process of, "Okay, from that information, what can I work out? And then from my new information, what else can I work out?" And that potentially will be what you need to work out.

Question three, surface area of this prism is a right trapezoidal prism.

Work out the volume of the prism.

Pause the video whilst you're doing that.

And then when you're ready to go onto the last question for this task, press play.

Question four is similar to that one where you needed to get from the surface area to the volume of a cuboid.

So the areas of three faces have been given, the edges are integer amounts of centimetres, what is the volume of this cuboid? Pause the video whilst you're working that one out.

And when you come back, we'll go through the answers to all the questions.

Question one, you needed to get the length of the prism in part A.

And that was necessary for you to be able to work out the volume of the prism.

So this was a little bit more structured in how you were gonna get through the problem.

So you can see on the screen we've got, the surface area is the total of all of the faces areas.

So two times the area of a triangle, because you've got two congruent faces on the prism.

And then three different rectangle areas because it was a scalene triangle.

Add them all up and solve the equation to work out that the prism had to have a length of six centimetres.

Once you knew the length of the prism, then you could get the volume by doing half times based on perpendicular height because that will give you the area of the cross section, which in this case was a triangle, and multiply it by the length, to give you 180 cubic centimetres.

Question two, the volume of a cylinder was given and its height was given, you needed the surface area of the cylinder.

So from the volume and the height, you could work out the radius.

And then you needed the surface area of a cylinder.

So recalling that a surface area of a cylinder is the two circular faces areas.

And then the curved surface becomes a rectangle when laid out.

And importantly, the rectangle's dimensions are the circumference of the circle, because it wraps around the circumference of the base, times by the height of the cylinder.

So substituting the radius, you knew the height of the cylinder from the given information, the surface area is 180 pi square centimetres.

Question three, the surface area of this prism was given to you.

There was a missing dimension, which was the length of the prism.

It was a right trapezoidal prism, so you did need to recall the area of a trapezium formula.

There were four different rectangles that were part of the surface area, because the trapezium didn't have any equal edges on it.

So again, you need to just set up an equation with the unknown in there, which was L, which was the length of the prism, and solve it to get that the length was eight.

Once you knew the length of the prism was eight, then you could calculate the volume area of the cross section, which was the trapezium, and then multiply it by eight.

The answer was 592 cubic centimetres.

Finally, question four.

You needed to figure out what the common factors were between each of the faces so that you can come up with the dimensions of the cuboid, so then you could work out the volume.

12, five, and nine, hopefully you got as your dimensions.

So that volume was 540 cubic centimetres.

Now we're onto the second part, which is to do with composite solids.

We've got some composite shapes, two dimensional shapes.

And they're all known as composite shapes because they are created or formed from other 2D shapes.

The first one could be thought of as two rectangles.

We might call this a rectilinear composite shape.

The middle one you could think of as a rectangle and a triangle.

And the last one you could think of a sector of a circle and a square.

So they are compound shapes or composite shapes because they are made or formed using other shapes.

In a similar way, there are composite solids.

So what we just looked at were two dimensional shapes, but now we're thinking about the three dimensional shapes, so solids.

Some are prisms and some are not.

So remembering that a prism needs to have a polygon as its cross-section.

So the first two are.

The last one is not a prism, as it doesn't have a polygon as its cross-section, but does have the same style of structure that that cross-section will run throughout the whole length.

It's not a cylinder either, because it isn't just a circle, so we're gonna call that a composite solid.

So a quick check on that.

There are six 3D solids on the screen, and which of them classify as composite prisms? Pause the video whilst you're figuring it out.

When you're ready to press play and check the answer, come back.

So the first one is a composite prism because it's got a uniform cross section running throughout, and that cross section is a polygon, which makes it a prism.

It's composite because you could probably describe it as rectangles and triangles to create it.

The next one isn't, we could call it a composite solid, but we wouldn't call it a prism because there is not a uniform cross section, which is a polygon, running throughout it.

That's the same on the third one.

This one is also not a prism.

This one isn't a prism, but this one would be.

So this cross-section is made as a polygon, it's a composite polygon, and it's a prism because of the uniform cross-section running through the length or the height of the whole solid.

Here we've got Jacob and Lucas, and they're having a discussion about composite prisms and composite solids.

So Jacob said, "Composite prisms are prisms. So we must calculate the volume and surface area the same way." Lucas wonders that, "Can we calculate the volume and surface area as separate prisms and then total them?" So because they are composite, they are built from other shapes.

If you think just about the two dimensional concept, the rectangles, or the rectangles and the triangles.

So Lucas is asking, "Can we calculate their volume and the surface areas as separate prisms, and total them to get the composite prisms volume or surface area?" Jacob thinks, "Yeah, that seems to be an idea that would work." Here is an L-shaped prism or a composite rectilinear prism, and all the lengths are measured in centimetres.

So the L shape is the cross-section, which is a polygon, that is the reason it's a prism.

And we can think of it as a composite rectilinear prism, so that L shape is created from rectangles.

The volume of a prism, which hopefully you already remember, is the area of the cross-section multiplied by the length.

So here is just a sketch of the cross-section, not worrying about the depth or the length of the prism, but just the cross section.

The area of the cross section is 17.

And therefore, the volume of the composite prism would be 17 multiplied by four, because we can go back to the original diagram, we can see it's got a height or a length of four.

So it's got a volume of 68 cubic centimetres.

If we focus it on two separate prisms, like Lucas suggested, we can split our composite shape here.

And think of it as two, A and B, two prisms that have been joined together to create the composite prism.

The area of the cross section of A is a rectangle, so five.

The volume of A is therefore 20, five by four, because we know that the height is four, or the length is four.

Doing exactly the same thing, B is another rectangle, four times three gives us 12, the volume would be 48.

And the volume of the composite prism, so thinking about what Lucas said, so can we work them out as separate prisms, which is what we've done here.

It's rectilinear, so two rectangles.

And then summing them gives us 68, which is the same volume that was calculated previously.

The volume was the same.

And the volume was the same because, actually the calculations were the same.

So in the first one, where we just kept it as one composite prism, so we did the area of the cross section and multiplied it by four, that's the second line.

And the first line of calculation is where we split it into two cuboids.

So five times one was the area, the cross section multiplied by four, to get the volume of that first cuboid, and four times three.

And hopefully you can see there that they are the same, one is just factorised out, the multiplication of four.

There is a common factor of four in both parts of the calculation.

"This makes sense.

A cake cut in slices doesn't change the total amount of cake." When you start with a cake, and then you slice it up, the amount of cake is the same.

And that's all that we've sort of done.

We've taken our prism and we've sliced it up into two cuboids in this case.

So here is a check.

The volume of this prism is 240 cubic centimetres.

It can be split into two identical trapezium based prisms. What is the volume of one? Pause the video whilst you work that out.

And then when you're ready to move on, press play.

I would hope you've got 120, and that's because they were two identical, so we can just divide it by two.

So now we're gonna start thinking about the surface area.

So we've seen with the volume was the same regardless if we did the area of the cross-section multiplied by the length, or if we split it into other prisms that can create that composite shape.

"So is the surface area of the composite solid the same as the total of the surface area of its parts?" What he's saying is, so does it still work for surface area? We're back on cakes again.

And this time, Lucas is saying he doesn't think so.

And that's because if you were to ice the whole cake, the outside of the whole cake, or ice individual slices, you would use different amounts of icing.

So we've got this same L-shaped or composite rectilinear prism back.

The surface area of a prism is going to be two times the area of the cross-section, because there are those base, that's the base on its parallel face.

Plus the perimeter of the cross-section multiplied by the length.

And the perimeter is because the rectangular faces, being that this is a right prism.

The rectangular faces that join up the cross-sectional faces run round the perimeter of the cross-section.

So an efficient way to get the surface area is the perimeter of the cross-section multiplied by the length.

There's just one large rectangle.

So keeping it as a full composite solid, in this case a prism, the area of the cross section is 17.

We had to work out for the volume previously, so we've got 17 back.

The perimeter of the cross section, we need to work out some missing edges from the diagram there, but we get 24.

So the surface area of the prism is two lots of the cross-sectional face, 'cause that's the top and the bottom in the way that this is orientated.

And then it's the rectangle that wraps around the outside, which has a length of the perimeter, 24, and a width of four, which is the height of the prism.

That comes out as 130.

So when we keep it as a composite solid, as a composite prism, the surface area is 130.

So going back to Lucas, that's your whole cake.

If we have iced the whole cake, including the bottom, you've used 130 square centimetres of icing.

So now if we go back to splitting it into two cuboids because it is a composite solid, and working out the surface area of both of those, do we still get 130 when we total it? Lucas doesn't think we will, because his analogy about cake, if we were to ice those two separately, we would use a different amount of icing.

So the area of A, cross-sectional area of A is five.

Again, we did that on the volume.

The perimeter of A would be 12.

The surface area is therefore 58.

So the surface area for cuboid A is 58.

Repeating this for B, we get a surface area of 80.

So we've done the cross-sectional phase, we've doubled it, and then we've got the perimeter of the cross-sectional phase times by the height of the prism.

When you total those, the combined surface area is 138.

So we have got a different surface area in those two methods.

So Jacob has said, "You were correct, and the problem occurs where the two prisms join." And so where the volume is just about how much space there is, where the surface area is about the outside.

And so if we have covered both of them separately, when they join up there is the issue.

And that is identified there in grey.

And that area that we have included in our surface areas for our individual prisms is no longer on the surface of the composite prism.

That face where they are joining is internal, and it has been accounted for twice because you've you've included that is a face on the prism A, and it's also part of a face on prism B.

So it's been included in our calculations twice.

So when we did the surface area of the full prism without cutting it into separate prisms, we got 130.

That particular surface that connects the two has an area of eight.

And that is the difference between our combined surface area, so our two prisms added together, and our surface area of the prism.

A quick check on that.

"The surface area of a composite solid is equal to the combined surface area of the solids that created it." True or false, and then justify your answer.

Pause whilst you're working that out.

And then when you're ready to move on, press play.

I'm hoping you went for false.

We just saw that it didn't work.

And that's because the solids created will join and hide some of their surface area.

That what was on the surface of the prisms then becomes internal on the composite prism.

So a task on this, firstly you need to match up the composite solid with its parts.

What has created it? If you were splitting it up into different solids, what would you be splitting it into? Pause the video whilst you're doing the matching.

When you're ready for question two, press play.

Question two, you've got a composite prism here, you need to work out its volume.

So pause the video whilst you're working on that.

And then when you're ready for the answers to both question one and question two, press play.

Question one was the matching up with composite solids and its parts.

Use the lines to check that you've got those correct.

Question two was volume of a composite prism.

We saw with volume, it didn't matter.

You could have split it as two separate prisms, or three separate prisms potentially on here, and then added them together.

Or you could have just treated it as one, worked out the area of the cross section, and then multiply it by the length, which is what I've done here.

Your answer should be 13,112 millimetres cubed, or cubic millimetres.

So we're now up to the last part of this lesson, which is to do with problems in context.

So working with perimeter, area, and volume, but in context.

So the first problem we're faced with is about painting the outside of a garage.

So we've got masonry paint, which is the type of paint you would use on brickwork, et cetera.

And it comes in a certain size of tin.

There are some dimensions.

And then we've got three Oak pupils who have got their own opinions on how many tins are necessary to do this work.

Who do you think is correct? You might wanna pause whilst you think about that, read it through carefully, and then we'll start going through the answers.

So, Aisha has said that, "The volume is 70 cubic metres, so five tins are needed." Well, she cannot be correct, purely on the fact that she's worked out volume.

Even if the volume is correct, that's not what's necessary here.

Because the context is about painting the outside, so it's not about how much space is within the garage.

So this question is not about volume, and therefore Aisha is incorrect.

Here we've got Izzy, and Izzy has calculated the surface area of a cuboid with these dimensions correctly.

But in the context of the question, you are painting the outside of the garage.

So you are not going to be painting the bottom of the garage as a cuboid, because that would be within the ground.

So you do need to consider, in the context of every question, although it might be similar to surface area in terms of what you are calculating, are all of the surfaces involved? Which leaves us with Laura, and Laura has done it correctly.

So Laura has only worked out the area that she needs to paint.

Laura has done it correctly because she has worked out the area of the surfaces that she will be painting.

And then, because if you go into a hardware shop, you cannot buy parts of tins.

Even if you only need half a tin, let's say, you can't go in and only buy half.

So rounding it up to get how many you'd actually need to go and buy to do the job that you're trying to do.

This check, you need to sort the problems into whether it is a length problem, an area problem, or a volume problem.

So pause the video whilst you're working through that check.

And then when you're ready, we'll see the answers, and hopefully you'd have got them all in the right places.

So the problems have now been sorted into the relevant places.

So whether it is a length problem; so replacing the fence, putting Christmas lights up, framing a picture.

All of those are to do with the length; how much fence, how much boundary you would need to be covering, the length of that.

Lights going around the sort of perimeter of your roof line is how many lights you need in a length.

And then when we get to the area, covering a wall in tiles, the wall has an area.

And the amount you would need is dependent on how much area you are covering.

Similarly, for carpeting a floor, the floor is an area, think of it as a 2D shape.

And so how much a carpet you need depends on how much area there is.

That leaves us with the amount of water being collected in a water butt, in a garden running off the roof.

So that's a sort of capacity and volume, how much water can fill up the water butt.

And lastly, how much air in a balloon.

So the balloon starts with no air, and then you pump it up, and the volume is increasing as the air goes inside.

So here is another problem that's in context, and it says that the juice from a full carton is transferred into a cylindrical glass.

How far is the juice from the top of the glass to the nearest millimetre? And Alex has identified that this is a question about volume, and it's because the juice fills up the space inside the carton.

So because volume is about how much space a 3D shape takes up, the juice is taking up that space.

So that's why we know it's a volume question.

So we can start by calculating the volume of the carton.

We were told it was a full carton, so we need all of the volume.

It's a cuboid, we're gonna do length times width time height, and that comes out as 455.

When you pour into the cylindrical glass, the amount of juice is not changing, it's unchanged.

The volume is going to stay unchanged in terms of the space occupied.

This question is actually about it filling up a glass, but it doesn't get full.

It wants to know how far from the top of the glass we get.

So we have been told that the cylindrical glass has a diameter of seven, and a total height of 15.

The 15 is not necessary for the next part of the question though, because it's the height that we want to work out, the height of juice within the glass.

So volume is 455 because it's just the amount of juice being transferred.

Assuming that no drips, no leaking happens.

The volume of a cylinder is pi R squared H.

So we know that the radius, which is what R is, is 3.

5.

And so we can set that up as an equation to work out the height is 11.

82 centimetres.

And this is where the 15 comes in, because we want to know how far from the top of the glass.

Well, the glass is 15 centimetres tall, and the juice has started from the bottom and gone up 11.

82.

So what is the space remaining is how far from the top.

And so a subtraction, 3.

2 centimetres to one decimal place or to the nearest millimetre.

A check, "If both of these vessels are full, which would have more juice?" Pause the video whilst you work that one out.

Okay, so vessel A is a carton, is a cuboid, the volume is 432.

B is a cylindrical glass again.

It's got diameter of six, so we need to change that to a radius of three.

Pi R squared H gives you 144 pi, which is approximately 452.

4.

So the glass would hold more juice.

So now you're gonna do some practise for problems in context.

So for every question, I want you firstly to think, what question is this? Is this a volume question? Is it a surface area question? Is it a perimeter question? Is it like a surface area but not all of the faces? So each question, make sure you're deciding on the context and what that actually means for the maths you're gonna calculate.

Pause the video whilst you have a go at question one and two.

And then when you're finished with both of those, press play, and you'll go onto question three.

Okay, so question three.

One factor that affects the rate of dissolving is the contact between the solute and the solvent.

So here are some sugar lumps, which is the solute, of the same mass.

When put in the water, which is gonna be your solvent, A, which would have the most contact, and B, which would have the least contact? On the right you've got a table that shows you the length, width, and height.

All of those are in millimetres, they're all cuboids.

Pause the video whilst you're working that out.

And when we come back, we'll go through our answers.

Question one.

Firstly, I'm hoping you worked out it was surface area.

So these mini Swiss rolls are being covered in chocolate.

So the outside, the surfaces are going to be in contact with the chocolate.

So this is why it's a surface area question.

They are cylindrical, or we're gonna base them on a cylinder for this problem.

And so you need to get the surface area of one, and then multiply it by 24 to get the amount of chocolate used.

Question two, "A metal pole is melted down and remoulded into cubes with edges of five centimetres.

So how many cubes can be made from this pole?" So this is a volume question, because it's the metal that is just gonna be reformed into a different shape, the amount of metal isn't changing.

So we need to get the volume of the cylinder, and then the volume of the cube, or what the cubes would be.

And then you can do a division to work out how many you can make.

And the answer is 55.

Three, the most contact and the least contact.

So again, this is about water touching the outside of the sugar cube, so surface area.

So D would have the most contact, it would have a surface area of 2008 square millimetres.

And C would have the least amount of contact, and that would be 600 square millimetres.

So to summarise this lesson, this lesson is about problem solving with perimeter, area, and volume.

And there's often a link between volume and surface area, as they are both calculated using the lengths of the prism or the cylinder.

The volume of a composite 3D shape can be calculated by totaling the volume of the solids that created it.

However, that is not true for a surface area.

So the surface area is not the total because of that joined face that becomes internal.

And most importantly, in my opinion, is when you're faced with a problem identifying whether the question is length, area, or volume, is the first stage, especially when it's in context.

Well done in this lesson, and I look forward to working with you again in the future.