Lesson video

Hello, everyone.

I'm Mr. Gratton.

Thank you so much for joining me for another maths lesson.

In this lesson, we will look at the concept of surface area and how to find the surface area of a cuboid.

Pause here to familiarise yourself with the definitions of net and surface area.

And pause here to familiarise yourself with some of the other keywords that we'll be looking at today.

But before we look at the methods to find the surface area of the cuboid, let's look at what is a net.

And what does a net of a cuboid even look like? This is a three dimensional object.

The net of any 3D object is a two dimensional representation of all of the faces of that object.

And so, this is a net of this three dimensional cuboid.

All of the faces of any three dimensional object when represented as a net must have edges that are touching the adjacent edge of another face.

This edge is touching both the rectangle to the left and right of it.

They must be touching and not separated.

Therefore, this is not a net of the cuboid to the left.

A cuboid is a three-dimensional shape that has exactly six rectangular faces.

Each face has a matching congruent face on the opposite side of the cuboid.

So for example, the front and the back are congruent to each other like so, the left and the right are congruent to each other and the top and the bottom are both congruent to each other.

If a three-dimensional representation of a cuboid has these properties, then the net must also have those exact same properties, six faces in total, and three pairs of congruent faces.

So the back and the front must both be congruent.

The right and the left must both be congruent and the bottom and the top must also both be congruent to each other.

Okay, quick check.

Which of these five possible nets are actually a net of a cuboid? Pause now to have a look through all five possible nets.

B and e are the only two correct answers.

A only has five faces and c has too many with seven.

Remember, you must have exactly six rectangular faces.

In d, two of those rectangles are congruent, a second two are also congruent, but the final two are not congruent to each other and therefore it cannot be a cuboid.

A further property of a net is that if two rectangular faces are touching each other, then the edge connecting these two faces must be the same length like in this diagram here.

However, if the edge was a different length for the bottom and the right faces like this, then the 3D shape would have a gap like a letter box and therefore would not be considered a cuboid.

Which of these possible nets are actually nets of the cuboid and which ones are not? Pause now to have a go.

The only correct answers are a and d.

B and c have rectangles whose edges do not properly and completely line up with exactly one other rectangle.

We've dealt with cuboids, but what about a cube? Well, a cube is a very specific type of cuboid where all six of the faces are exactly the same shape, always a square where each edge is exactly the same length, same as every other edge in the cube.

And so in this cube, if one edge was 16 centimetres, then every other edge would also be 16 centimetres in length.

And because every square has a length and a height that is exactly the same, the areas of every single square that composes the net of the shape will also be the same.

For example, 16 times 16 is 256, and so that one square has an area of 256 centimetre squared.

But because of that, every single other square also has an area of 256 centimetre squared.

Only some arrangements of six squares when placed into a possible net actually do fold up to make a three dimensional cube.

One of the best ways to see which combinations of six squares actually do fold up to make a three-dimensional cube and which ones do not, is to physically play around with them, to physically fold each potential net up to see if they become cubes or not.

Such as this one, if I were to fold up this possible net, would it make a cube or would it make something different? Let's have a look.

This does fold up to become a cube, therefore the original net is a proper net of a cube.

One quick check question, how many squares are there in the net of a cube? Pause now to have a think.

There are six squares as a cube is made out of exactly six square faces.

Onto some practise.

Each group of cubes, so group A, group B, group C, D and E, each group contains only nets that actually fold up into cubes or non nets that will not fold up into cubes.

By spotting any patterns or similarities within each group can you identify which groups do and do not fold up into cubes? Pause now to answer all four parts to this question.

For question 1A, there are many possible observations.

One of these is that in group A, each of them has a horizontal line of four squares with exactly one square above and one square below it, whereas group C have five or six squares in a row, which is more than any other group in this diagram.

For part b, groups A, B and D all fold up into cubes, whereas C and E do not.

And for part c, some potential rules you could come up with include no arrangement has five or more squares in a row or a column.

Furthermore, it is impossible for squares that are distributed into only two rows or two columns to ever make a cube.

However, it is possible for some arrangements of six squares distributed into at least three rows and three columns to make a cube.

And for part d, the one net not covered in this diagram is as follows.

Notice how this one does fold up into a cube.

And now that we know what a net is, let's see if we can calculate the surface area of the cuboid using its net.

As a quick recap, the area of one rectangle is the length multiplied by its width.

And so the area of this rectangle is 20 times 16 equals 320 centimetres squared.

Here are five other rectangles.

Let's find the area of each one.

This top rectangle has an area of 8 times 16 equals 128 centimetres squared.

Whilst this central rectangle does not have a length or a width labelled, we can find that its length, looking at the far left is 20 centimetres, and its width, looking all the way down is 8 centimetres.

Therefore the area of this central rectangle is 160 centimetres squared.

We know that a property of a cuboid is that it has three pairs of rectangles where a rectangle in a pair is congruent to the other rectangle in that same pair.

Therefore, the missing three areas will be the same as the three that we have already calculated.

Pause here to see if you can figure out which missing area matches with which rectangle and here are the areas.

Next up, if we then connect each of these six rectangles together so that each edge is touching the edge of a another rectangle, well, it'll become the net of the cuboid like so.

If this is the cuboid after it's been folded up, which rectangle in the net matches with which face on the cuboid? Let's have a look.

The 160 centimetre squared would go on the top, with the 128 going at the front and the 320 going from the right.

The surface area of the cuboid is the sum or addition of the areas of all six rectangular faces.

And so I'd have to add together all six of the areas of all six rectangles that I calculated earlier.

Therefore, the total surface area of this cuboid is 1,216 centimetres squared.

Note that because we are dealing with a surface area that the units should be written as centimetres squared, not centimetres cubed, even though we're dealing with a three-dimensional object.

It's also possible to use the grid squares inside a rectangle to find the total surface area of the cuboid after it has been folded up.

For example, this rectangle is a 5 by 4 rectangle and so its area is going to be 20 square units.

This rectangle is a 3 by 4 rectangle, and so its area is gonna be 12 square units.

Similarly, this one is a 3 by 5 rectangle, and so this area is 15 square units.

Notice how each multiplication has at least one number in common with a different calculation.

For example, the 5 times 4 and the 3 times 4 both have 4 as part of its calculation.

This will always be true as adjacent rectangles will share one common edge and with it the length of that edge.

And so, the length of that edge will be shared between two different rectangles and two different calculations.

As with before, because the net of the cuboid is composed of three pairs of congruent rectangles, we can match the missing three areas with the three that we've already calculated.

Pause now to figure out which areas go with which rectangles.

And here they are.

The total surface area of this cuboid will be 12 plus 12 plus 20 plus 20 plus 15 plus 15.

What do you notice about these six numbers? Well, there are three pairs of numbers.

This is because there are three pairs of congruent rectangles.

For example, the rectangular face at the front and the back are both congruence to each other.

Another way of representing this is through a factorised form to represent the surface area of the cuboid.

Rather than saying 12 plus 12, 20 plus 20, 15 plus 15, we can say two lots of, open brackets, 12 plus 20 plus 15.

Both methods will give you the same answer of 94 centimetres squared.

Okay, onto a check.

These six rectangles can be connected to create a net of a cuboid.

What is the area of the rectangle marked a.

Pause now to use the other measurements on screen to inform your decision.

And so, the length and the width of this rectangle are 9 and 6 giving an area of 54 square units.

If I attach all six of these rectangles together, I get the net of a cuboid.

Find the area of the faces marked with b and c.

Pause now to think of your answers.

And the area of b is 27 units squared.

For c, notice how c is congruent to the rectangle in the top left hand corner of this net.

I know that it is six units wide and therefore the area of this top left hand rectangle is 18 units squared.

And therefore c at the bottom of the net is also 18 units squared.

If a face of the net of a cuboid isn't clearly labelled, then try and find the congruent opposite face and then apply that same area to the rectangle that you were originally looking at.

Okay, onto some practise questions.

For question numbers 1 and 2, let's look at this diagram.

For question 1, which of the two cuboids shown as nets has a great surface area? And for question number 2, which one is the cube? And how do you know? Pause now to have a look at both of these questions.

And for question number 3, by first of all finding the area of each individual rectangle, calculate the total surface area if this was turned into a net and folded up into a cuboid.

Pause now to give this a go.

And for question number 4, here's that different representation of the surface area of a cuboid.

By finding at least three areas of three different rectangles, calculate the total surface area of this cuboid after the net is followed up into 3D.

Pause now to do this.

And for question number 5, the face labelled with this area is a square.

Knowing this, calculate the total surface area of the entire cuboid.

Pause now to give this question a go.

Okay, onto the answers.

For question number 1, b has the greater surface area by two square units.

And for question number 2, the answer is also b.

B is the net of a cube because all of the faces are squares each with an equal area of four square units.

And for question number 3, three of the faces of this cuboid have areas of 84, 60 and 140 units squared.

I know that the remaining three faces also have these exact same areas, and so the total surface area of the cuboid is 568 units squared.

For question number 4, three of the faces on the cuboid have an area of 375, 120 and 200 centimetres squared.

The total surface area is gonna be double the sum of those three numbers at 1,390 centimetres squared.

And for question number 5, if the area of a square is 49 centimetres squared, then both the length and the width are both 7 centimetres.

Therefore the areas of every rectangular face is gonna be 77 centimetre squared with the one congruent square area being 49 centimetres squared.

Therefore its total surface area is going to be 406 centimetres squared.

And now that we're familiar with what a surface area is and how it relates to the net of a cuboid, how can we find the surface area of a three-dimensional representation of the cuboid without the net as a representation to guide us? So let's have a look at some of the areas of the rectangles on this cuboid represented in net form.

We have 5 times 12, which is 60 square units.

Directly underneath it, we have 36 square units.

And to the right of that we have 15 square units.

And as always, these three areas are the same for they're three corresponding rectangles.

If I were to fold this net up into a cuboid, it would look like this.

Let's match the face on the cuboid with the rectangle on the net, starting with the 36 square units.

That would go on the bottom of the cuboid.

And one of the dimensions of this cuboid is going to be 3 units with one of the other dimensions, 12 units.

This is because 3 and 12 are the width and the length of the rectangle that got us to the 36 unit square area of that face.

And next up, let's have a look at this 60 square units rectangle.

That's going to be on the right hand side of this cuboid.

We already know the 12 units, that is the width of this rectangle and 12 times 5.

So 5 units is the height of this cuboid.

Whilst we already know the base, the height and the depth of this cuboid, let's use one of the other rectangles just to check that we've got everything correct.

So let's use this 15 square unit rectangle.

We know that 15 is 3 times 5.

Well clearly on this front rectangle, the width of this rectangle is 3 units and the height is the same 5 units as at the back of the cuboid.

The height is the same at the front and at the back of any cuboid.

And whether we have the net given or not, these properties of any cuboid are always true.

The area of each face is the multiplication of the lengths of its two edges and the face on the opposite side of the shape is congruent and has exactly the same area because it is composed of edges of exactly the same length.

Okay, how do I use the information given to calculate the area of the back face of this cuboid? Well, the edge lengths used to calculate the area of the front face are the same lengths used to calculate the area of the back face.

And therefore the back face of the cuboid has an area of 6 times 15 equals 90 units squared.

Okay, onto a quick check.

Which two edge lengths define the hidden face marked with the letter a.

Pause now to have a look at all of the dimensions given on this cuboid.

And the answers are 12 and 7.

Therefore the face labelled a has a total area of 84 square units.

And onto the next check, which two edge lengths define the hidden face marked b.

Pause Now to give this question a go.

And the answers are 30 and 7, giving a total area of b at 210 square units, The surface area is the sum of all six faces.

In this example, we will add together each face one by one starting with the front, which is 14 times two, which is 28, and the back is exactly the same at 14 times 2 is 28.

Next up, let's have a look at the right which is 5 times 2 which is 10, and the far left is also gonna be 5 times 2 equals 10.

The bottom is going to be 14 times 5 which is 70.

And so the top is also going to be 14 times 5, which is 70.

The total surface area is going to be 28 plus 28 plus 10 plus 10 plus 70 plus 70 for a grand total surface area of 216 centimetres squared.

That is the correct answer, but that was a lot of steps to get to this answer.

However, there are other approaches to finding the surface area of any cuboid, each equivalent to adding together each of the six faces one at a time, and therefore giving the same answer.

But these other methods require fewer steps and therefore can be a lot quicker and efficient.

Rather than adding the same size face twice, we can double the area of one face to account for its opposite face.

And so, rather than doing the calculations for front and back separately, let's do front and back is 14 times 2, times 2 again because there are two of them.

This multiplied by 2.

Make sure you are accounting for both the front and the back faces.

And similarly, for right and left, we can do five times two for one of the faces times by 2 again because there are two congruent faces for a total area of 20 units squared for the left and the right combined.

Last up, we've got the top and the bottom, which is 14 times 5 times by 2 again to account for both the top and the bottom.

If we were to add together 56, 20 and 140, we would get to the exact same 216 centimetre squared answer that we had using the original method.

There is an even more efficient method to the previous one because we save the doubling for doubling only once right at the very end of the calculation after we've added together each of the individual areas.

So for this method, let's have a look at front and back and only look at the front.

The front is 14 times 2, which equals 28.

We're only gonna look at the right this time, 5 times 2 equals 10, and we're only gonna look at the bottom 14 times 5 equals 70.

We are not gonna look at the back, the left and the top explicitly.

We are just gonna add together the front, right, and bottom to get 108 units squared.

But now right at the end we are going to double to get the same total surface area of 216 centimetres squared.

We double the end to account for the back, the left and the top that we did not explicitly calculate earlier in our method.

We could also represent this exact same method using brackets, 28 plus 10 plus 70 in brackets with a multiplied by 2 outside the brackets to multiply the total of everything inside it.

The final correct answer is exactly the same as before at 216 centimetre squared.

This method is both the most efficient and it accounts for all the hidden faces in a representation of a cuboid that does not explicitly show all six faces.

Just because I can't see the back, the left and the bottom face of the cuboid, it does not mean it's not there.

We still have to account for them.

And so we first of all add together everything that we can see, 90 plus 300 plus 120, and then we double that amount to account for the three faces that we can not see.

And so the total surface area is going to be 1020 centimetre squared, which is double the area of the three faces that we can see.

And here it is again represented in factorised bracketed form, the two outside the bracket that doubles the 90, 300 and 120 accounts for the back left and bottom that we cannot see.

For this check, what are the values of a, b, c, and d.

And pause here to consider both dimensions of this cuboid as well as areas of some of its faces.

And the answers are as follows.

For this next check, similar again, what are the missing lengths and areas and also a different but important missing value that you need to account for.

Pause now to have a think about what these missing values could be.

And the answers are as follows.

Notice how the answer to part a is not a length or an area.

It is a multiply by 2 that accounts for the fact you are finding the areas of both the right and the left faces of this cuboid.

E is exactly the same, it is a times by 2 to account for the fact that you're looking at the bottom and the top faces of the cuboid.

C and d are length that can be written in either order.

What are the values of a, b and c for this question? Pause now to give this a go.

And the answers are as follows, 35, 84 and a total surface area of 358 centimetres squared.

Rather than grid squares that you may be more familiar with cuboids can also be drawn on isometric paper made of tessellating triangles rather than the grid squares.

This helps to draw shapes more convincingly in 3D like this cuboid.

Lines going in this direction with a negative gradient show the width of a cuboid, lines going vertically upwards show the height of a cuboid and lines going in this direction with a positive gradient show the depth of a cuboid.

We define one full edge in any direction as one unit of distance.

For example, the height of this cuboid is one, two, three, 3 units high.

And by that same, logic we have a width of one, two, three, 4 units and a depth of 5 units.

As with before, we can find the areas of the front, the right and the top faces, the three faces that we can see on this diagram by multiplying two of the height, the width, and the depths together.

And then we can add all three of these areas together and then double it to account for the back, the left and the bottom faces that we cannot see.

Therefore, the total surface area of this isometrically drawn keyboard is 94 square units.

Okay, some quick chats for understanding about isometrically drawn cuboids.

How many units high is this? cuboid? Pause to have a look.

And the height is 3 units.

What about the depth of this cuboid? Pause now to consider how many units deep this cuboid is.

And the answer is 6 units deep.

And finally, how wide is this cuboid? Pause now to give this question a go.

And the answer is 2 units wide.

Okay, here's your final set of practise questions.

For question 1a, what are the missing lengths labelled a, b, and c? And for part b, what two lengths should be multiplied together to find the following areas for d, e, and f? Pause now to give both these parts of question one a go.

Onto question number 2, what are the missing lengths and missing areas that you need to calculate to find the total surface area of this cuboid? Pause now to find all the missing values.

And onto question number 3, here is a different representation of the different calculations that you need to do to calculate the total surface area of this cuboid.

Pause here to figure out which values go where.

Here's a different representation again.

This is the most efficient method to find the total surface area of a cuboid filling all missing values.

Pause now to do so.

Onto question number 5.

This cuboid is actually a cube.

By calculating the areas of the front, right and top faces find the total surface area.

And for part b, Jun says, there is a quick way of calculating the total surface area of a cube.

Is Jun correct? If so, show your more efficient calculation to find the total surface area of specifically a cube.

Pause now to do this.

And for question number 6, these two cuboids represent two cargo crates that need to be covered in waterproof paint.

Which cargo crate requires more waterproof paint in total? Pause now to consider this question.

And for question number 7, here are two cuboids drawn on isometric paper.

Each unit on this paper is equal to 1 metre.

By how many square metres does cuboid b have a greater surface area than cuboid a? Pause now to give this last question a go.

And onto the answers, a has a length of 15 centimetres, b of 5 centimetres, and c of 7 centimetres.

For the areas the calculations are 7 times 5 for d, 15 times 5 for e, and 7 times 15 for F.

And for question number 2, pause here to have a look at all of the answers.

And for question number 3, the total surface area is 1,056 square metres.

Whereas for question number 4, the total surface area is 592 inches.

And for question number 5a, the surface area of this cube is 150 metres squared.

For part b, Jun is correct.

This is because each face of a cube is the same.

Therefore the area of one face, 5 times 5, or 5 squared can be multiplied by six for all six faces.

And for question number 6, b requires more paint because it has a total surface area that is 2 metres squared greater than a.

And for question number 7, b has a greater surface area by 12 square metres.

Thank you so much for all the effort you've put in to today's lesson where we have seen that a net of a cuboid is composed of exactly six rectangles with each adjacent pair of rectangles having a common edge of equal length.

We've also seen that the surface area of a cuboid can be calculated from its net representation with the surface area as the sum of all six rectangles.

We've also seen three representations of calculating the surface area of the cuboid and which one is the most efficient.

And we've also seen how to calculate the surface area of the cuboid from isometric drawings of cuboids.

That is all for today's lesson on surface area.

Thank you so much for joining me, Mr. Gratzna.

And until next time, have a great day.