Lesson video

Hello, my name is Dr.

Rowlandson, and I'm excited to be guiding you through today's lesson.

Let's get started.

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

This lesson is called Checking and Securing Understanding of Surface Area of Other Prisms. And by the end of today's lesson, we'll be able to efficiently calculate the surface area of other prisms. The lesson is broken into two learning cycles and we're going to start by looking at prisms with regular polygonal cross-section.

Let's begin with a simple shape like a cube.

Here we have a cube that has a length of 10 centimetres, and Sam says, "I know I can find the surface area of this cube by finding the areas of every square on its net." Let's take a look at the net.

Here it is.

Each square has a length of 10 centimetres and there are six squares there.

And each square has an area of 100 centimetres squared.

You get by doing 10 squared each time.

Now Sam says, "Could I do the same thing with other prisms too?" Let's take a look.

Here we have an equilateral triangular prism.

The length on the triangle is 10 centimetres.

That'll be the length of all the sides in the triangle 'cause it's equilateral.

And the depth of the prism is 36 centimetres.

Let's take a what it look like if we unravel this prism to create its net and look something a bit like this.

We have three rectangular faces and the depth of a prism from one vertex of the triangular face to its correspondent vertex on the opposite face is constant, meaning that each rectangle has the same length.

So we knew that the length of the prism earlier its depth was 36 centimetres, which means the length of each rectangle is 36 centimetres as well.

Its two triangular faces are both equilateral triangles, meaning the length of each edge is equal.

Each of them has a length of 10 centimetres and we can see that on each length of the triangle, but also in the rectangles means each rectangle is 36 centimetres by 10 centimetres.

We are told here that the area of one triangular face is 25 root 3 centimetres squared, and if we're not told that there are ways we can work that out as well.

But let's go with what we're told here.

25 root 3 centimetres squared.

If I wanna work out the toll surface area of this prism, now we have two triangles, so we could do 2 times 25 root 3, and that would give 50 root 3 for the area of the two triangles combined.

And we have three rectangles.

So each rectangle is 36 centimetres by 10 centimetres, which would multiply to get the area of each rectangle and that would give 360.

So if we multiply it by 3, we get the total area of the 3 rectangles as being 1,080.

So the total surface area of the prism would be the sum of those two previous answers, 50 root 3 plus 1,080 centimetres squared.

And we can leave it in insert form and that'll be the most accurate way of writing it, or we can write it as 1166.

6 centimetres squared if it's rounded to one decimal place.

Let's check what we've learned.

Here we have a regular pentagonal prism and it's been unfolded to show its net as well.

You need to calculate the lengths on the net that are labelled A and B, and you need to check the picture of the prism and the picture of the net in order to spot what A and B are.

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

The answer is 9 millimetres and 12 millimetres, and we can see those measurements taken from the prism when it's in its complete form there.

Calculate now the area of one rectangular face of this prism.

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

The answer is 108 millimetres squared, which we can get by doing 9 millimetres times 12 millilitres.

That's the measurements on each rectangle.

So now, could you please calculate the surface area of the prism altogether? You may need to look at us more information on the diagram on the left to help you.

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

Well, we are told that the area of the pentagonal face is 110 millimetres squared and we have two of those, and we've worked out the area of the rectangle is 108 millimetres squared and we have five of those.

So the total surface area will be 760 millimetres squared.

Here we have a hexagonal prism and the hexagon is a regular hexagon.

Sam says, "There's no need for me to look at the whole net of a prism with a regular polygonal cross-section.

That's because all the rectangular faces are congruent, so I only need to find the area of one of them." So, let's now look at that prism in its complete form rather than its nets.

Here we have a regular hexagonal prism and we're told that the area of the hexagon is 65 centimetres squared and we're given the lengths on there as well.

We are told that it's a regular hexagonal prism, therefore, it has two congruent hexagonal faces and six congruent rectangular faces.

One rectangle for each side on the hexagon.

So, if we look at the rectangles, we have 2 centimetres by 5 centimetres.

So each rectangular face has an area of 10 centimetres squared and we have six of those.

So we can do 6 times the area of the rectangle plus 2 times the area of the hexagon to get the total surface area of the prism, which would be 190 centimetres squared.

So let's check what we've learned.

Here, we have a regular polygonal prism.

Complete the sentence.

This prism has two congruent octagonal faces and blank congruent rectangular faces.

What number goes in that blank? Pause video while you write down an answer and press Play When you ready to see what the answer is.

The answer is 8.

It has eight congruent rectangular faces.

One rectangle for each side on the octagon.

So, with that in mind, could you please use the information on the diagram to fill in the blanks in this calculation here to find the surface area of the prism.

Pause video while you do that and press Play when you ready to see the answers.

The answers are 2 times 309.

each 309 is the area of an octagon plus 8 times 56, where 56 is the area of each rectangle, which we can get by doing 8 times 7.

So it's 2 times 309 plus 8 times 56, and that gives 1,066.

So the surface area of the regular octagonal prism here is 1,066 centimetres squared.

Now, so far, we have been given the area of the cross-sectional face, but sometimes, we need to calculate the area of the cross-sectional face as well as rectangular faces.

For example, here we have an equilateral triangular prism.

We have an equilateral triangle as a cross-section, but we are not told the area for this time.

So if we want to find a surface area, we could start by finding the area of each rectangle, which is 12 centimetres by 4 is 5 centimetres and that is 540 centimetres squared.

And we have three of those, but we need to work out the area of the triangle as well.

Its base is 12 centimetres, its height is 6 root 3 centimetres, so we could multiply those together and multiply by a half to find the area of each triangular face and that would be equal to 36 root 3 centimetres squared.

And then, to find the total surface area of the prism, we can do 3 times the area of the rectangle plus 2 times the area of the triangle.

And that would give this calculation here which would simplify to get 1,740 centimetres squared when rounded to three significant figures.

So let's check what we've learned.

Here, we have an equilateral triangular prism with different measurements on and what you need to do is calculate the area of one rectangular face and one triangular face on this prism and give your answers as a third where appropriate.

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

The area of each rectangle is 700 centimetres squared and the area of each triangle is 100 root 3 centimetres squared.

So you probably guessed what we're going to do next.

Let's now find the surface area of the entire prism.

Pause video while you do that and press Play when you ready for an answer, and give your answer to two decimal places, the answer is 2,446.

41 centimetres squared.

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

This task contains three questions and here is Question 1.

The diagram shows the net for a prism.

Each octagon has an area of 120.

71 centimetres squared and that's rounded 2 decimal places.

And we can see that the octagon is regular because it's got the hash markings on the sides of the octagon.

What you need to do is calculate the surface area of the prism and give your answer accurate to three significant figures using the measurements we can see on the screen.

Pause video while you do that and press Play when you're ready for more questions.

And here are Questions 2 and 3.

In Question 2, the diagram shows a prism, its cross-section is a regular hexagon with an area of 24 root 3 centimetres squared and the perimeter of the cross-section is 24 centimetres.

That'll be helpful.

Calculate its surface area and give your answer to 3 significant figures.

And with Question 3 you've got a prism.

Its cross-section is an equal triangle, the height of 4 root 3 centimetres, and you've got some other measurements on there as well.

And what you need to do is calculate its surface area, given your answer accurate to 3 significant figures again.

Pause the video while you work these out and press Play when you are ready for answers.

Let's now work through some answers.

Let's start with Question 1.

Each rectangle has an area of 6 multiplied by 5.

The 6 is shown as the length of the rectangle and the 5 is the length of each side of the octagon, which is also the width or the height of each rectangle, depending on which way you look at it.

And then, to get the total surface area, we'll do 8 lots of the area of a rectangle and plus 2 lots of the area of the octagon.

And that would give 481.

42 which would round to 481 and 3 significant figures and that's in centimetres squared.

And then, Question 2, we have a prism which has a cross-section that is a regular hexagon.

We know the area of the hex skin and the perimeter.

We going to calculate the surface area of the prism.

Well, the length of one edge on the regular hexagon would be its perimeter 24 divided by 6.

That gives 4, and that means that each rectangle will be 12 by 4 centimetres.

So the area of one rectangle will be 4 multiply by 12, which will be 48.

And we have 6 of those rectangles, and we have 2 hexagons.

So to find a surface area, we do 6 lots of the area of the rectangle, 6 times 48 plus 2 lots of the area of the hexagon.

So that's 2 times 24 root 3 and that would give 371.

138 and more decimals, which would round to 371 centimetres squared to 3 significant figures.

Then Question 3.

The diagram shows a triangular prism where the cross-section is an equilateral triangle.

We are told the height and we're told the base of the triangle and we're told the depth of the prism.

You need to work out the surface area.

Well, 'cause it's an equilateral triangle, all those limbs around the outside of the collateral triangle would be 8 centimetres, which means each rectangle would be 3 by 8 centimetres.

So the area of the rectangular face would be 8 times 3 is 24.

The area of the triangle face would be 1/2 multiplied by the base, which is 8 multiplied by the height, which is 4 root 3 and that gives 16 root 3.

So to find the total surface area, we'll do 3 lots of the area of the rectangle plus 2 lots of the area of the triangle.

And that gives 127 centimetres squared when round to 3 significant figures.

Great work so far.

Now let's move on to the second part of this lesson, which is looking at prisms with irregular cross-sections.

Here we have a prism that has a right-angle trapezium as its cross-sectional base.

That means that the cross-section is not a regular polygon and that means that its non-cross-sectional faces are not necessarily going to be congruent to each other.

In other words, those rectangles will not necessarily have the same area.

Let's unpeel this now and take a look at its nets.

Here's what it looks like.

We can see that we have 2 trapeziums and 4 rectangles, one rectangle for each side on the trapezium and those rectangles are not congruent to each other.

So to find the surface area of a prism with an irregular cross-section, we do need to find the area of each individual face.

Let's do that.

This prism has two congruent trapezium-shaped faces.

The area of one of these faces is equal to 5 plus 13.

That's the sum of the 2 parallel sides divided by 2 times 6, where 6 is the height and that gives us 54.

Now that's the area of one of those trapezium faces.

We also know the area of the other trapezium face is also 54 centimetres squared.

Now there are four rectangular faces which are not congruent, but each face shares an edge with the trapezium.

So for those four rectangles, we can see one length each time and the length is adjoined to the trapezium.

But for one of those rectangles, we can see another length which is perpendicular to the one which is attached to the trapezium.

That edge which is perpendicular to the length which is shared with the trapezium will be the depth of the prism and that will be the same for every rectangular face.

So they'll all have 7 centimetres.

Now we have enough information to find the area of each rectangular face.

This rectangle we labelled A is 6 by 7 centimetres.

The rectangle labelled B is 13 by 7 centimetres.

Rectangle labelled C is 10 by 7 centimetres, and the rectangle labelled D is 5 by 7 centimetres.

So the surface area would be 2 times 54, where 54 is the area of each trapezium plus 42 plus 91 plus 70 plus 35, which is the area of each of the four rectangles.

And all together, we get 346 centimetres squared.

So let's check what we've learned.

Here we have the net of a triangular prism whose depth is 9 centimetres and the triangle is not equilateral.

Could you please find the area of each face on this net labelled A to E? Pause video while you do that and press Play when you're ready to see some answers.

Let's take a look at some answers.

Face A, which is a triangle, would be a 1/2 times 21, which is the height.

We can see there times by 20, which is the base.

And that base is the same as the 20 centimetres we have on rectangle E and that would give 210.

Now face B is a rectangle, which means one of its measurements would be equal to the depth of the prism.

And the other measurement is equal to the height of triangle A.

So its area would be 21 multiplied by 9, which is 189.

Shape C is a triangle again and it is congruent to triangle A.

So it would be 210.

Shape D is a rectangle, which means one of its measurements would be the depth of the prism.

So one of its measurements would be 9 centimetres and we can see if one is 29 centimetres, so the area would be 29 multiplied by 9, which is 261.

And then shape E is also rectangle, which means one of its measurements would be 9 centimetres and the other one would be the 20 centimetres we can see on the screen.

So its area would be 20 multiplied by 9, which is 180.

So you guess what we're going to do next, calculate the surface area of this prism.

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

The answer is 1,050 centimetres squared, which is what you get when you add those answers together.

Here we have another prism, which has a pentagonal cross-section.

Now, so far, we've been finding the area of each rectangle separately and adding them all together and also adding on the two lots of the cross-section.

But Sam says, "Do I really need to calculate the area of every single face on this prism? It's so time-consuming." Hmm, I wonder is a more efficient way we can do this.

Well, there is a more efficient way to find the surface area of a prism.

We can do this if we know all of the edge lengths of the cross-sectional face of the prism.

Let's see how.

We start by looking at the two congruent cross-sectional faces and one rectangle that is perpendicular to them.

So, for example, in this case, here is one cross-sectional face, which is a pentagon and we can choose any of those rectangles 'cause they're all perpendicular to that pentagon.

But let's choose the one which is towards the front from the way we'll come at it, which is the one which is 5 centimetres by 16 centimetres.

And then also, we have another cross-sectional face underneath this prism as well.

We can then, stack every adjacent rectangular face next to each other.

So let's do that.

We have the rectangle with a length of 16 centimetres, we then have a rectangle with a length of 7 centimetres, one with a length of 10 centimetres, and then going around the corner at the top, we can see there we have another one which is 10 centimetres and another one which is 7 centimetres.

Now, we've pieced all those rectangles together to make one long rectangle because they all have the same heights.

Do you all fit together to make one single long rectangle? Can we think what the length of that rectangle would be? Well, it would be 50 centimetres if we add all those together.

And the sum of all these lengths is also equal to the perimeter of the cross-sectional face.

The perimeter of that pentagon is 50 centimetres and the length of this overall rectangle here is 50 centimetres as well.

The height of the rectangle is equal to the depth of the prism, so the area of this long rectangle is 50 centimetres multiplied by 5 centimetres, which would give 250 centimetres squared.

This area is a sum of the areas of all the rectangular faces individually as well.

So rather than finding the area of all five rectangles and then adding them up, we could multiply the perimeter of the cross-section, which is the total length of all those rectangles when we stack 'em together and multiply it by the depth of the prism, which is the height of each rectangle.

In other words, find the perimeter of the cross-sectional face and multiply it by the depth to get the total area of all the rectangles.

Let's check what we've learned with that.

On the left, we have a cross-sectional face of a prism of a depth of 4 centimetres.

The perimeter of the cross-sectional face is 102 centimetres and its area is 630 centimetres squared.

On the right, we have the net of the same prism.

We can see that cross-sectional face, which is A and C, and we have that rectangle as well.

And that rectangle is a combination of all five rectangles that will wrap around that prism.

What you need to do is find the areas of shapes A, B, and C.

Pause the video while you do that and press Play when you're ready for answers.

Well, the area of A is 630 centimetres squared and so is C, and we knew that 'cause we're told that on the left-hand side of the screen.

What about B though? Well, the length of that rectangle of B would be equal to the perimeter of the cross-sectional face, so its length will be 102 centimetres.

The height of that rectangle will be equal to the depth of the prism.

That'll be 4 centimetres, so its area would be 4 times 102, which is 408 centimetres squared.

So now we have that information.

Could you please find the surface area of the prism? Pause video while you do that and press Play when you're ready for an answer.

To get the surface area, we need to add together the three areas that we've worked out so far, and if we do that, we get 1,668 centimetres squared.

Okay, it's okay now for Task B, this task contains four questions, and here is Question 1.

You've got a triangular prism and a series of questions to answer about this triangular prism.

Pause video while you do that and press Play when you're ready for the next question.

Here is Question 2.

You have a pentagonal prism this time, which has been forward to create a net.

The perimeter of the cross-sectional face is 70 centimetres and you have parts A and B to answer for this question.

Pause video while you do that and press Play when you're ready for Question 3.

Here is Question 3.

Here we have a prism, which is the result of a regular hexagon prism being cut in half through the vertices of its two cross-sectional faces.

Find the surface area of this prism and give your answer accurate to 3 significant figures.

Now, as a hint, you may find it useful to consider Pythagoras' theorem here.

And think about how you might need to use that.

Pause video while you do that and press Play when you're ready for Question 4.

And here is Question 4.

You have two prisms. Prism A is a regular hexagonal prism and prism B has been made by cutting prism A in half through the vertices of its two cross-sectional faces.

By considering the different approaches for calculating the surface area, decide which approach you would take for each prism and justify your decision.

One approach is to calculate the area of each face and find the sum.

And the other approach is to use the perimeter of the cross-section and the depth to reduce your calculations.

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

Now go through some answers.

So Question 1.

We have our triangular prism.

How many faces does it have? It has 5 faces.

B, find the area of the triangle on top, which is labelled A.

That would be 30 centimetres squared.

Part C, find the area of the rectangle at the front, which is labelled B.

That would be 240 centimetres squared.

Part D, find the area of the rectangle at the side, which is labelled C.

That'd be 100 centimetres squared of Part E.

Find the area of the rectangle at the side, which is labelled D.

That's at the back, which we can't see.

That would be 260 centimetres squared.

So, Part F, find the total surface area of the prism.

While that would be the sum of 2 lobster triangles and the other rectangles as well to give 660 centimetres squared.

Question 2.

We have our prism, which is on the fold to make a net.

The perimeter of the cross-section is 70 centimetres and in Part A, to labelled all the known lengths and areas of this prism onto its net.

And that's what we can see on the screen here.

We have 250 centimetres squared for each of the pentagons.

We have 420 centimetres squared for the area of the overall rectangle.

And that has come from doing 70, which is the perimeter multiplied by 6, which is the depth of the prism.

And Part B, you have to get the surface area of a net and that was 920 centimetres squared.

Then, Question 3, we have the prism, which is a result of a regular hexagonal prism being cut in half through the vertices of its two cross-sectional faces.

That means a cross-section of this one is a trapezium.

The problem is we're only given the length of one of the sides for this trapezium.

We're given the height as well or the distance between two of its parallel sides, but there are three sides in that trapezium what we don't necessarily know the length of, but we can think about how this trapezium was made by cutting a regular hexagon in half.

A regular hexagon, all the sides will be the same length, all the sides will be 14 centimetres.

That means that three of the sides, not trapezium will be 14 centimetres and there's only one side we are left to know that is the longer side.

We could do that by splitting this into a triangle within our shape, within our trapezium and use Pythagoras' theorem to work out the value of X on that triangle.

Now we know the value of X is 7, then we could think about how the length of the longest HARE trapezium would be equal to 7 plus the 14 which is the length of the shorter side plus another 7, and that would give 28 centimetres.

So now we have all four lengths on that trapezium.

Now we've done that, we can find a surface area.

The area of the trapezium would be 14 plus 28, the sum of the parasites divided by two multiplied by the distance between the parasites and that would give approximately 254.

61.

And then, the area of the rectangle that wraps around this prism.

Or you could think of it as the combination of the four rectangles.

Well, the area that would be the perimeter of the trapezium times by the depth, the perimeter of the trapezium will be 4 lots of 3 plus 28.

And then when we multiply that by 22, we get 1,540, that means the total surface area would be 2,050 centimetres squared.

That's from doing the area of the long rectangle plus 2 lots of the area of the trapezium.

Then Question 4.

You have to consider the different approaches of finding the surface area of these prisms and decide which approach you would take for each prism and justify your answer.

Now your answers will differ for this 'cause this is a question about personal preference when it comes to method selection and your answers would be specific to you.

For example, you might say that find an area of each individual face and summing them is the easiest for you to understand and that's absolutely fine.

Or you might say that using the perimeter means using fewer calculations and that's absolutely fine as well.

Or you might say, "If I know some areas, then so and so and so on." Answers may differ.

Well done today.

Now let's summarise what we've learned in this lesson.

The surface area of a prism is a sum of the area of all of its faces.

The net of a prism can help find the surface area of a prism, but it can be quite time-consuming to draw the net and find things individually by using known area facts.

The area of all faces can be found and then summed to find the surface area of the prism.

But it's important to find the surface area systematically and efficiently, and we've looked at some different methods for trying to make our calculations more efficient during today's lesson.

Well done today.

Hope you have a great day.