Lesson video

Hello, I'm Mrs. Lashley, and I'm really looking forward to working with you throughout this lesson.

So during this lesson, we're gonna be working out the surface area of prisms, and trying to do it in the most efficient way.

On the screen, there are some keywords that you will have met before in your learning.

I would suggest you pause the video now, read through them and re-familiarize yourself because we will be using them throughout the lesson and it's important that you know what they mean.

Today's lesson has got three learning cycles.

The first part, we're gonna be looking at nets of right prisms. Then the second learning cycle, we're going to be working out how to get the surface area of those prisms. And finally, we're gonna think about how to do it in the most efficient way.

So we're gonna make a start with the nets of right prisms. So previously, you would've learned about nets.

You've potentially made one yourself.

So a net can be folded up into a three dimensional shape, and there won't be any overlap.

So an example on the screen is for a cuboid.

The net is the two dimensional figure.

And when you would fold it up, then we would get the solid, which in this case is a cuboid.

Here, you can see the stages of the folding.

So from that two dimensional flat surface, lifting or folding one of the faces, and then continually folding it so that we see that cuboid.

Here is some colour to indicate the faces, just to maybe make that a little bit clearer for you.

So we've got a blue rectangular face, the red rectangular face and the green rectangular face.

And hopefully, you can follow that through the diagram so that you can see how those faces are folding into the solid.

So when we look at a 3D shape, we can describe it by using its faces, thinking about how many of each face there are, what shapes they are, and that will help us when we try to compose the net.

So here's a cuboid, and on the cuboid, we've labelled three of the faces A, B and C.

The reason we've given them different labels is because they are different.

So we can say that there are six faces in total.

So we can see three from this view of the 3D solid, but there are three hidden faces as well.

So that makes a total of six.

All of the faces are rectangular.

This solid is an isometric view, but actually the faces are rectangular.

And being a cuboid, there are three pairs of rectangles.

So that's why we've got A, B and C, because there is another face which we cannot see from this view that has exactly the same dimensions as rectangle A, and that's the same for B and C.

So now we've considered all of the faces of the solid, we can then compose the net.

This is a net that would be for this cuboid.

And we can see the labelling of A, B and C.

And hopefully, you can notice that they're sort of one away from each other because they are opposite each other on a cuboid.

So here's a check for for you.

Izzy has described the net of a solid and said that there are five faces.

So which of the three solids that you can see on the screen would she be describing? Pause the video whilst you think about that.

And then when you're ready to check your answer, just press play.

Hopefully you went for C.

So A and B have both got six faces or would need six faces on their net, whereas C only has the five.

Here's another check for you.

So this is the solid from part B of that check.

And I want you now to describe its faces.

Pause the video whilst you're doing that description.

And when you're ready to check it, just press play.

In total, there are six faces.

So you may have said there were six faces in total.

And then if we consider the faces individually, there are two identical trapeziums. You can see one of them there.

And then there would be four rectangles, and none of them are going to be identical.

And that's because none of the edges of the trapezium have been marked as equal lengths.

So we have to assume that every edge of that trapezium has a different measurement, and therefore the rectangles will be of different dimensions.

So on the screen, we've got a net that's gonna fold up to be an L-shaped prism.

And from the net, you can see that it's made of two of the L shapes and six rectangles.

In a moment, it's gonna animate to show you it folding up.

So hopefully, it will help you to imagine how the net folds into the L-shaped prism.

So you could see there that those six rectangles have sort of wrapped around the L, and the L shape is actually an irregular hexagon.

Hexagons have got six edges, and that's why we've got six rectangles to wrap around the outside.

The reason we're thinking about these nets is because we're trying to think about the surface area of the prism.

And so when we need a net, we don't necessarily need it accurately drawn.

Instead, a sketch will be enough to help us with our next stages of the calculation.

So it doesn't need to be to scale.

So I want you to look at these next three sketches and decide which one is the best sketch for allowing us to do any further calculations.

So do you think A, B or C? Which of those three sketches for the net of this prism conveys the important information? Which one is best? Well, B.

So A had none of the dimensions, so we wouldn't be able to do any calculations.

And if it isn't drawn to scale, this is all in metres, a big task and a big net to draw it to scale.

So without any dimensions labelled onto the net, it's not very useful at all.

C has got all of the dimensions, but there's just too many of them.

They're not necessary.

B is has just got enough information that we would then be able to go and use it for further calculations.

So when it comes to drawing your own nets, you need to put dimensions on it, but you don't need to put dimensions in absolutely every position.

You just need to make sure there's enough information that you know the dimensions for all of the faces.

So a check for you is which of the dimensions go where on this net? It's a right triangular prism.

So all of those faces that are not triangular are rectangles.

I would like you to pause the video and work out where the dimensions are for the net.

Press play when you're ready to check.

So this is where the dimensions should be.

So from the solid image, we can see that the prism has a length of four.

So that's the rectangle's length, and that's why that's gonna go on the four.

All three of those rectangles have got the length of four centimetres.

Then we've got the triangle.

It's actually an isosceles triangle.

We know that because two of the edges are 3.

5 centimetres.

So we've got one labelled on the triangular face, and then we've also got the one which when it folds, will touch that edge.

And then lastly, the base of the isosceles is 1.

5 centimetres.

So that's the smaller rectangle.

On this net, we can see, or hopefully you've recognised, that because it's an isosceles triangle, there are two identical rectangular faces.

So for this part of the lesson, you're gonna now do your task.

And the task is to match the net to its prism and sketch in the missing net and prism.

So one of the nets hasn't got a prism to match to, you need to draw it.

And one of the prisms hasn't got its net, again, you need to draw it.

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

And then when you're ready to check your answers, press play and we'll go through it.

Okay, so here are the answers.

We've got a cuboid net, which hopefully you've matched to the cuboid.

We've got a triangular prism, which hopefully you've matched to the triangular prism solid.

The last net that was given to you was a pentagonal prism.

And so you needed to draw a pentagonal prism, and there's an example of how you may have drawn it.

And then lastly, the prism without its net was a trapezoidal prism.

So you needed to draw a net for a trapezoidal prism, which needed to have two trapezium faces, which would be identical, and four rectangular faces if it was a right trapezium prism.

So we're now gonna move on to the second learning cycle of the lesson, which is all about calculating the surface area of a prism.

We've worked up to this point with drawing and recognising nets of prisms, and they're a really useful way to help us to calculate the surface area.

So on the screen is a right-angled triangular prism.

The surface area is the total of all of the faces' areas.

And we've seen that net is all of the faces drawn, and actually into a compound shape.

If you think about it as a two dimensional figure that's just made of other shapes, which is the definition of being a compound shape.

And the reason we went through nets and the reasons that we are gonna use them is because they can be a really helpful way to work out the surface area for any prism, but also for other 3D shapes that you'll do in the future.

So this right triangular prism has got five faces.

There's two congruent right-angled scalene triangles, and that's the sort of two ends, if you like, the base face.

And then there are three different rectangles.

The reason there are three different rectangles is because it's a scalene triangle.

So each of those edges is a different length.

So therefore, the rectangles would be of different dimensions.

They all have something in common, which is the length of the prism.

But their other dimension, their width if you like, will be different because it's scalene.

So if we drew a net, a sketch of the net, this could be one of them.

So we sort of unfolded the solid and thinking about the orientation that they would be.

So we sort of folded down the triangular faces.

And then those rectangles that wrap around have laid down in the middle.

So this is our compound shape.

If you look at it, it's a compound shape that's made of three rectangles and two triangles.

You could split it in different ways, but that's the way we're gonna split it because of the faces that have created it.

So to work out the surface area of that triangular prism, we're gonna work out the area of each face first.

So we've labelled the net A, B, C and D.

The reason we haven't gone to E, there are five faces, but the reason we haven't gone to E is because the face D, the triangles that we've labelled D are identical.

So we don't want to do too much work, we don't wanna do extra.

So A is a rectangle.

So the area of a rectangle is length times width, as we know.

So we've just got to identify the dimensions.

B is another rectangle.

So again, length times width, but with the correct dimensions, making sure you're matching them on the net.

C is our last rectangle on this net.

And then D is a triangle.

So for a triangle, you need 1/2 times base times perpendicular height, and multiplied by a 1/2 is the same as divided by two.

There are two of those.

So I've just repeated my calculation there.

Maybe you can figure out a more efficient way of doing that than writing both of them down.

So to find the surface area, because the surface area is the total of all the faces, we're now gonna sum up those faces.

We've got five areas to total because there are five faces making our net.

And so in this case, the surface area for this triangular prism is 66 square centimetres.

It's still got square centimetres as its unit because we've just totaled areas, we've not multiplied by another length.

So the dimensions of the unit hasn't changed.

We've just summed up areas.

So you've now got a check to do.

So missing parts.

Some answers, some calculations on the total surface area.

It's for a right trapezoidal prism.

You've got the net on the left.

So pause the video, take your time, you can use a calculator for this.

And then when you're ready to check the answers, press play and we'll go through them.

So A was a rectangular face, and the area was 900 square millimetres, B was another rectangular face.

Its dimensions were 32 and 50, and that comes out with an area of 1,600 square millimetres.

So now we've got, you've completed the answer for the area of A, you've calculated the area for B, and the other ones were already done for you.

So we've got all of the areas as the individual faces.

Now we need to do the sum to get our surface area.

The surface area is 7,670 square millimetres.

So nets are made from the faces from the solid.

And the surface area is not reliant on the arrangement of the faces, only the total areas.

So if you think about when we draw a net, sometimes there is more than one way of drawing the net, but the surface area would still be the same because it has been created using the same faces.

And those faces therefore have a fixed area, and the total therefore doesn't change.

So the arrangement of the faces is not what's important for the surface area calculation.

Laura's identified that means that we don't actually need to draw a net every time we want to work out a surface area.

And this is quite important because as your prisms become a little bit more complicated or even a different 3D shape because working out the surface area from a net works for other 3D shapes, not just prisms, as they get more complicated, drawing the net can be more difficult.

So Alex has said, "Yeah, it's just the polygons that are involved "that we need to be able to work out the surface area." And so remembering that polygons is about the shape, the 2D shape with straight edges.

Lucas is right though, we need to be careful that we don't miss any in this way because with a net, you're gonna have thought about the shapes that have created the net, how many there are, if there's any pairs or any duplications of the same face.

But if you are drawing them individually, there is a potential that you're gonna miss some out.

So you do need to listen to Lucas's advice and make sure you remember them all.

With this example, we're gonna go through the surface area.

But instead of drawing the net, we are gonna just consider the polygons that have created or would create the net.

So here, we've got a right prism.

So the right prism means that all of the faces that are joining the two parallel base faces are rectangles.

So there are two trapeziums and there are four rectangles.

So the two trapeziums, we can identify here as is the base face.

Okay, that's the uniform cross section that runs through this prism.

And here, we've just sketched it as its shape with the dimensions added on.

So we've taken that from the solid and we've drawn that there.

The four rectangles, well, we've got four different rectangles here because if you look at that trapezium, none of the edges have the same length.

So there's no repeats.

So we have got four different, unique rectangles.

What they have in common is the length of the prism.

So they're all gonna have that six metre side, but it's gonna be their width that is changing.

So the six metre by 5.

4 metre face is the slanted one.

Then we've got the one that's to the back left, six metres by five metres.

And then we've got the face that we can sort of see here highlighted in pink, which is six metres by 4.

5 metres.

And then lastly, we've got that sort of smaller one, six metres by two metres.

So once we've pulled the information off and we've sketched out individual faces, and if there are repeats, we only have to draw them once, but we do need to make sure we are noting somewhere that there is duplication, otherwise you might forget them.

We now need to work their areas out to add them back together to get our surface area.

So in the same way we did for the net.

So for our trapeziums, so recalling our trapezium formula, that's 1/2 times the sum of the parallel sides and then multiplied by the perpendicular height between the parallel sides.

Looking at the screen, you can see that we've got two times, and that's because there are two of these trapeziums. So we need both of their areas included in the surface area.

Then we've got our four different rectangles.

So we can work those out by doing length times width.

So the first one, 32.

4.

The second one, 30.

Then 27, and then 12.

So we've now calculated the areas of the individual faces.

And any that were duplications, we've included and combined them together.

So the area of the rectangles, if we add them up as like a sort of subtotal of our surface area, 101.

4.

And then to get our surface area, we need to add together all the faces.

So we've got the two trapeziums plus the rectangles, and so the surface area is 132.

9.

So just to check, thinking about what polygons make up the solid, which of these could be the faces of a triangular prism? Pause the video whilst you're really thinking about that and being careful about what's on the screen.

And then when you're ready to check that or you're not sure and you wanna come and have a check, then press play.

I would hope you've gone for B.

And I've also written the reasons that A and C do not create a triangular prism.

We're now at the point where you're gonna do a practise all about surface areas of prisms. So on the screen is question one, there's two parts to it.

On the first part, I want you to just consider that prism that you've got an isometric drawing for, and how many of each and describing, and then you're gonna draw them, the faces, with the dimensions.

Pause the video whilst you're completing question one, and then when you're ready for question two, press play and we'll move on.

So here, we've got question two, and you need to work out the surface area for this right trapezium prism.

It's completely up to you how you do this.

If you want to draw a net and do it that way, then of course you can.

Remember, it's just a sketch of a net.

You need to put specific dimensions on.

If you want to do it by just considering the different shapes that make up the faces, that's absolutely fine as well.

So pause the video whilst you're working out the surface area, press play, and then we'll go through the answers to question one and question two.

So here, we've got question one back again.

So you needed to work out how many irregular octagons were faces on the prism, and the answer is two.

And there would also then be eight rectangles.

We know it's eight because it's an octagon, and the octagon needs to have a rectangle on every edge of it.

There are eight edges on an octagon, so therefore there are eight rectangles.

Part B was then to draw all the unique faces and add on the dimensions.

So you've got the two irregular octagons, and then you would have four rectangles that were actually identical.

So that would be a one by four.

Then there would be three that are two by four.

And there is one that is four by four.

It's sort of like that back face.

So that's actually the one that is a square.

Question two, as I said, there were a couple of ways you could have done this.

You would've got all of the same areas as these if you did it on a net.

I've just done them as individual faces.

So we've got two of each of the trapeziums, remembering the formula, 1/2 times the sum of the parallel sides, and then multiplying by the perpendicular height.

And then we've got four different rectangles because, once again, the four edges of the trapezium are different.

And so the surface area, the important bit to check first of all is that the surface area is 48.

8 square centimetres.

Okay, so we're on to the last learning cycle of the lesson, and this is now about being efficient when finding the surface area.

So we now know how to work out the surface area, but is there a way just to make it a little bit more speedy in circumstances and just be a little bit more efficient? We always wanna try and do it in the most efficient way.

So if a right prism had this space, what can you tell about the faces? Pause the video whilst you have a think about that, and then we'll go through the answers when you come back.

I just want you to consider that.

Like we was doing in learning circle two, if we think about the faces, but we haven't got a 3D solid image here, we've just got the shape of the base.

What does that tell you about the faces? Firstly, we know that that base would have two of those, and in this case, it's a scalene triangle.

So we'd have two identical scalene triangular faces.

It's then gonna have three further faces that are rectangles because it is a right prism.

And each rectangle is gonna have the same perpendicular height because that is the length of the prism, but it will have different widths to their rectangles because of it being scalene.

So thinking the same question, but this time if the base was a rectangle, again, if you wish to pause and have a think about that by yourself before I go through it, absolutely.

If not, I'm gonna start going through it now.

So again, we would have two identical rectangular faces.

So every base in a prism, there are two identical ones.

This is a rectangle.

It's gonna have four additional rectangular faces because it is a right prism, they are rectangular.

Each of those four additional ones will have the same perpendicular height.

Again, that is the length of the prism.

But we are gonna have two pairs of identical rectangles because a rectangle, the properties of a rectangle is the opposite edges are equal in length.

So those opposite edges have got the same rectangles coming from them in our solid.

And then finally, what if the base was an equilateral triangle? These are just a few prisms that there are, we're just focusing on these three at the moment, but don't think that they are the only prisms. So same question, what do we know about the faces of a right prism where the base is an equilateral triangle? There's gonna be two identical equilateral triangular faces.

It has three faces that are rectangles because it's a right prism.

And all of those three are gonna be identical to each other.

And that's because the perpendicular height's the same because of the length of the prism.

But also, the equilateral triangle has got three edges of equal lengths.

So the three rectangles are identical, they are congruent to each other.

And that means that they all have the same area.

The reason we've gone through this is because considering the properties that make up the prism and the shape that is the base can really help us with being efficient when finding the surface area.

So as a check for you, and I want you to think about which of the following right prisms have repeated faces, besides the base, the cross-sectional face because we know that all prisms have got two identical cross-sectional faces.

So ignoring them, are there any other repeated faces for these four prisms? So firstly, the isosceles triangular prism.

Then the isosceles trapezium based prism.

A prism with a kite base.

And a regular pentagonal prism.

So think about, use the diagrams if that's helpful, which of these have got repeated faces, using the properties of the base? Pause the video whilst you have a go at that and then when you're ready to check it, press play.

So the first one, the isosceles trapezium prism does have repeated faces, and that's because an isosceles has got two edges of equal length.

B would also have repeated faces.

An isosceles trapezium based prism, by being an isosceles trapezium, that means they also has got two equal edges, and therefore there would be two rectangles of the same dimensions, same area.

A kite would also have repeated faces because adjacent edges on a kite are equal.

And a regular pentagonal prism by being regular, means that all of the edges are equal.

So all of the rectangles that would join those bases would be the same, would be congruent rectangles.

So we've got a right regular hexagonal prism, and there's an image there of the solid.

The area of the hexagonal face is 210 square centimetres.

The length or the height of the prism is seven centimetres.

And the edges of the regular hexagon are nine centimetres in length.

So Jun and Andeep are both going to calculate the surface area of this right hexagonal prism.

You can see there, Jun's working out, and he's gone for there are two copies of the hexagonal face and there are six rectangles that are congruent.

He's done two times 210, the given area of the hexagons, and then seven times nine is the dimensions of each of those rectangles, and multiplied it by six because there are six of them, and summed up those areas to get 798 square centimetres.

So that's Jun's working out for the surface area of this right regular hexagonal prism.

Here, we've now got Andeep's calculation, and you can see he gets the same answer.

Andeep's done it by adding together the hexagon plus a rectangle plus a rectangle plus a rectangle plus a rectangle plus a rectangle plus a rectangle plus the hexagon.

So he's sort of worked his way around the shape, thinking about the faces that have created it.

Worked out their individual areas, and then he's added it up to find the surface area.

The definition of the surface area is the total of the faces' areas, that's what they've done.

But whose method is the most efficient? So looking at those two methods, whose would we say is the most efficient to get the correct answer? They both got the correct answer, neither of them made any errors, but which one was just more efficient? I'm hoping you agree with me that Jun's is, and that's because Jun has used the regularity of that hexagon.

We were told it was a right regular hexagonal prism, which means those six rectangles were identical.

So rather than working them out individually and then adding them up, using the repeated addition that multiplication does for us and multiplying by six.

And similarly, there are two hexagonal faces that have the same area.

So just doubling the area rather than working them out.

We didn't have to work these out in this example, but not adding them up separately.

So here is a check for you.

Again, there's some blanks you need to fill out and get the final total surface area.

Whilst you're doing it, you need to think about where the efficiency is coming from, which faces are repeated, how many are there of them, and that's part of the calculation.

Pause the video whilst you're doing that.

And then when you come back, we'll go through the missing digits and then the final surface area.

The first number that you needed to fill in was 16.

Then you needed to add on 16 times, and that was 12.

So there's five missing there.

And then how many 10 by 12 rectangles are there? Well, there needs to be five rectangles in total.

We know that the bottom is the 16 times 12.

We've had two five by 12s, so that's three in total, which means there's two left to find.

So to get the surface area, if you were struggling with the missing numbers, hopefully you still worked out that surface area.

So totaling it up, the surface area of this right pentagonal prism is 808 square centimetres.

Continue with being efficient with our methods to get the surface area.

So the rectangular faces create a tube when the net is folded.

And this is the case for all right prisms. So if we start here with a net, and unfolded, and then we sort of lift it up and lift the other one up, that's the tube that I'm talking about, that those rectangular faces have sort of wrapped around.

And then we can close in our ends and we end up with here a triangular prism, but the rectangles have wrapped up to become like a tube around the outside of those bases.

So this right triangular prism can be calculated by finding the area of each face and totaling.

We've been through this during the lesson already.

So here's my net.

I've got my dimensions labelled.

So we've got two triangular faces.

That's what the uniform cross section, 1/2 times base times perpendicular height, and doubling it because there are two of them.

And then we've got three rectangular faces.

And so we've got a four by eight, another four by eight, a 5.

7 by eight.

And you can hopefully see that on the net.

That gives us a total of 109.

6.

And now we can find our surface area by adding together both the triangles and those three rectangles.

So we now have a surface area of 125.

6 square centimetres, this area.

And there's absolutely nothing wrong with what we just did there, but we're trying to be more efficient.

So what we're gonna see here is those three rectangular faces.

And some of you might have spotted that actually, there were two identical ones as well.

So we could have sped up the process by not doing four times eight add four times eight.

We could have done two lots of four times eight.

But even more efficient than that, that those three rectangles, if you think about this like a compound shape, the net, how else could you have split it up? Well, we could have split it into two triangles and one large rectangle.

And what would the dimensions of that rectangle be? Well, the perpendicular height, or in this case, it looks like a width, is gonna still be eight.

And that's the length of the prism.

But how long is that rectangle? Well, that rectangle is the sum of all of those edges.

Four add the four add the 5.

7.

The total is still the same.

The area of that large rectangle is equal to the area of the three rectangles summed together.

And so our surface area is still the same.

But this becomes slightly more efficient.

Instead of doing three rectangles, we're only working out one.

Did you notice, because Sofia's just noticed that actually that sum of the edges is the perimeter of the triangle, because that's what the tube does.

It wraps around the outside of the triangle.

Each edge of the triangle has a rectangle that meets it.

And so this is the same for all right prisms. The perimeter of the cross-sectional face, which sometimes we call the base, is the length of all of the rectangular faces combined into one because it wraps up as a tube.

So there's lots of colour here, and that's trying to identify to you where the faces meet when it folds up.

And then that hopefully shows that those rectangles down the centre, so in this case there's six because the cross-sectional face is a irregular hexagon, we sometimes call it an L shape, but there are six edges to it, so we've got six individual rectangular faces, and as it wraps around to create that tube, it wraps around the perimeter of the L or the irregular hexagon.

So the length of that central part of our net is the perimeter, and this can really help us with being efficient.

So a quick check with this one.

So this right prism, which there's a solid view there, is unfolded into its net.

And what would the rectangles' dimensions be? So don't worry about the cross-sectional faces, just what is the dimensions of that rectangle on the net.

Pause the video whilst you think about that.

And then when you're ready to check it, press play and we'll move on.

So the dimensions would be D, which is the length of the prism, and then when it's by the perimeter of that cross-sectional face.

So if you go round the perimeter and sum it up, in its simplified form, it's two A plus two B plus C.

So we're gonna go through finding the surface area for this right equilateral triangular prism.

And we're gonna do it using three different methods.

So the first method is to find the area of each face, and then sum them up to get the surface area.

The second method is to find the area of each unique face and then some multiples of those.

So using the properties of the equilateral triangle.

And then the final method is this one where we're gonna use the perimeter.

So it is a prism with a length of five centimetres, and the equilateral triangle is a perpendicular height of 3.

5 and edge length of four.

So the first method is where we're just gonna find the total of all the areas.

It's a triangular prism, so there are five faces.

We know from the the last slide, the dimensions, the perpendicular height was 3.

5, the edge length of the triangles was four and the prism had a length of five.

So for the triangle, it's 1/2 times four times 3.

5, which gives you an area of seven.

For the rectangles, it's five by four, which gives you an area of 20.

We can sum all of those.

We've got two triangles, three rectangles.

We get the surface area of 74.

If we use method two, which is to use congruent faces to be more efficient, so multiples of the ones that are the same, well, it's a prism, so we've got two identical triangular faces.

So that gives us an area of 14.

And then because it was an equilateral triangle, every rectangle was identical.

So we've got three of those 20s, which give us 60, and 14 add 60 is 74, so we get the same surface area, but it's a slightly a more efficient method by using the properties of the shape.

And then, lastly, we're gonna use the one rectangular face method, where we are thinking about it as a long tube that wraps around, and it's wrapping around the perimeter of the cross-sectional face.

So we need to work out the area of the triangles in the same way we did for the last two.

And then we need to think about it as one long rectangle.

Because the edges of the equilateral triangle were four centimetres, then the length of that one rectangle would be 12.

So the area of the one rectangle is 60, add the 14 still gives us 74.

So a check for you.

A right prism with this rectal linear composite cross-section is 10 centimetres long.

So we haven't got the solid shown here, but we tell you that the prism is 10 centimetres.

Which of the following are correct methods to find the surface area of the prism? So pause the video, go through them carefully, or it might be that you want to work it out yourself and then compare it.

Is there one that matches what you did? Press play when you're ready to check.

So B and C are both valid methods.

B is more efficient, and that's the one that's used, that there are two of these composite rectilinear faces.

And then one long rectangle, which is the perimeter of the rectilinear composite shape times by 10, which is the length of the prism, whereas part C is doing all of the individual faces, and then would find the sum of those.

So on the last bit, which is for you to do, the surface area of the following four right prisms. Pause the video whilst you're working through those four questions.

And then when you're ready, we'll go through the answers.

So on the screen, you've got the answer to question one.

So we might call this an L-shaped prism.

It's an irregular hexagon.

I'm using the method where I've used one rectangle and two of the cross-sectional face, but your answer should be 128 square metres.

Question two, we've got another L-shaped prism.

This one, you may have used the efficiency of repeated faces because there was sort of symmetry to this shape, but your surface area, 522 square millimetres.

Question three was an isosceles triangle.

The base was an isosceles triangle.

So again, you might have used repeated faces here if you were doing it by thinking about the individual faces, but the surface area should have been 264 square centimetres.

And lastly, this was an irregular pentagonal right prism.

The area of the irregular pentagon had been given to you, but you needed two of those within your surface area, the front and the back, if you like.

And then there is however you've worked out the area of the other rectangles.

The surface area in total was 115 square metres.

So to summarise today's lesson, which was about surface area of prisms, the surface area of a prism is the sum of the area of all faces.

The net is one method that can be a helpful way to find the surface area.

To be more efficient, you can try to use properties of the shape to find if there are any congruent faces, and therefore not needing to work out areas repeatedly.

But the best method is the one that you understand best and can apply reliably to different prisms because prisms come in wonderful different cross-sectional faces, and so you need to have a method that you can rely on time and time again.

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