Lesson video

Hello, I'm Mrs. Lashley and I'm gonna be working with you as we go through this maths lesson today.

I really hope you're willing to try your best, and even if it gets challenging, I'll be there to support you.

So let's make a start.

So our learning outcome today is to be able to apply Phythagoras' theorem to problems in 3D.

So on the screen, there are some keywords that I'll be using during the lesson.

You've met them before, but let's have a look at them together.

So the hypotenuse is the side of a right-angled triangle, which is opposite the right angle.

Phythagoras' theorem states that the sum of the squares of the two shorter sides of a right-angled triangle is equal to the square of the hypotonus.

And lastly, a cross-section is a 2D face made from cutting straight through any plane of a 3D object.

So you can see that the hypotenuse is linked to Phythagoras' theorem and will be making use of cross section in these 3D problems. You may wish to still pause the video and read them again just to make sure you're feeling confident before we make a start.

So the lesson on applying Phythagoras' theorem in 3D is gonna be split into the two learning cycles.

In the first learning cycle, we're gonna focus purely on cubes and cuboids as our 3D shape.

When we get to the second learning cycle, we're then gonna extend it to other 3D shapes as well.

So let's make a start looking at Phythagoras' theorem in 3D with cubes and cuboids.

So the Pythagorean theorem is useful in two dimensions, but it can be used on a flat surface or plane in three dimensional space as well.

So we've been able to use it on a two dimensional right-angled triangle for quite a while now.

From our keyword slide, we know that it states that it's the sum of the square of the two shorter sides that is equal to the square of the hypotenuse.

And that hypotenuse is always opposite the right-angle and is the longest edge.

So how do we make use of that in a cuboid? How can Phythagoras' theorem for right-angled triangles take place when we are looking at a cube or a cuboid.

So Jun says, "Where there are right-angles, there are right-angled triangles!" So have a look at the cuboid, where are there right-angles? Think about that for a moment.

Pause the video, sketch yourself a diagram of a cuboid and draw on any right-angles that you know are there.

Press play when you're ready to carry on.

So a cuboid is made of all rectangular faces and every rectangle has got right-angles as its interior angles.

So there are many right-angles in this cuboid.

And so as Jun says, where there is a right angle, then you can also construct a right-angled triangle.

So Juns gonna talk us through a few now.

So this is one of the right-angled triangles that we can find on the face of a cuboid and it's on the top face.

Again, it's making use of the rectangular features.

So the right angle, there is an interior angle of our rectangular face, and then we're using the perpendicular edges, the length and the width, joining up the opposite vertices as a diagonal.

So the diagonal of that face is the hypotenuse of this right-angled triangle.

There's also this one Jun says on this side of the cuboid, and it's the same premise, it's just a different rectangular face.

So our rectangle gives us the right angle, our length and our width of our rectangle.

The dimensions are perpendicular dimensions and the diagonal across that face is the hypotenuse.

And "There's lots more," Jun says, "Here's two more on the opposite faces of the cuboid." Well, we know that the opposite faces are congruent to each other on a cuboid.

So these two right-angled triangles are also congruent.

Did you spot any that look like this? So Jun says, "There are even right-angled triangles that aren't on the faces of the shape." And so this one here is a right-angled triangle.

It makes use of a rectangular cross section A, B, C, D if we label the vertices of the cuboid.

And that's gonna be really helpful when we find ourselves with some 3D problems. So here's a check for you.

Which of these is a right-angled triangle? So which of the purple marked triangles as part of a cuboid is actually a right-angled triangle? Remember what Jun said where there's a right angle, there's a right-angled triangle.

Pause the video, and then when you're ready to check press play.

B and C were both right-angled triangles.

B was the right-angled triangle that is on the cross-section that is rectangular, whereas the right-angled triangle on C is the part of the face.

Jun's considering how he's going to calculate the volume of this cuboid.

So first of all, pause the video and remind yourself about how you calculate the volume of a cuboid.

What do you need to ensure that you can do that? And then press place so we can continue with looking at how we're gonna make use of Phythagoras' theorem in this 3D problem.

So Jun says, "We need a measure for the depth." And the depth he is labelled is EH.

What other edges are also equal to EH? Pause the video and then when you're ready to move on with working out the volume, press play.

So FG is equal to EH, is the opposite edge on that rectangular face.

AB and CD are also equal to EH, because they are opposite edges of rectangular faces.

So we need to have the length, the width, and the height in order to calculate the volume of any cuboid.

AF is our height, nine centimetres, FE is our length, 25 centimetres.

So it's the width, or in this case we're gonna call it the depth that we are missing and we can't work out the volume until we know that.

So how are we going to work it out? So have a look and think about right-angled triangles and Phythagoras' theorem.

How can we find the length, EH, from the given information? Pause the video and see if you can calculate EH before I reveal the answer.

Press play when you're ready to look at it.

So EH is a short side of a right-angled triangle with FE and FH.

So did you spot the right-angled triangle FEH? We've got the hypotonus and we have a short side.

So we've got two of the three edges, therefore Phythagoras' theorem can be used.

So we're gonna use Phythagoras' theorem to calculate EH, and that is 24.

5 to one decimal place.

So then we can work out the volume of this cuboid because we've got the three perpendicular measures, the nine, the 25, and the 24.

49 that we just calculated.

It's important though that we try to use the most exact value.

So on our calculator, making use of the answer key, so it's using the full value rather than any rounded value.

The volume for this cuboid is 5511.

4 cubic centimetres to one decimal place.

We only really want to round the final answer.

So here's a check for you.

If EH equals 40 centimetres and CH equals three centimetres, do you have enough information to find the volume of this cuboid? Pause the video and then when you're ready to move on, press play.

So we do not have enough information.

CH and EH are perpendicular measures, but we need three perpendicular measures to calculate the volume.

So we are missing FE or GH or BC or AD.

They're all equivalent lengths to each other and we need that dimension in order to calculate the volume and there is no way to calculate that length.

We don't have sufficient information.

So we're onto the first task of the lesson.

And so question one, you need to just determine whether you could find the volume if you were given these specific measurements.

So regardless of what the value is, if you know those measurements, could you calculate the volume? Pause the video, and then when you're ready for the next couple of questions of task A press play.

So in question two, you've got to find the volume of these shapes made up from cubes.

So for the first one, it is a single cube.

What is its volume? Knowing that the diagonal distance across a face is eight centimetres.

For the second shape it's made of cubes.

So it's a cuboid.

And this diagonal distance across the face that we can see at the front is the square root of five.

So once again, work out the volume on question two.

On question three, it's a cuboid and you've been given three lengths, but are they the appropriate lengths? Are they the ones you need? So you need to find the volume of that cuboid as well.

Pause the video and when you're ready to go through the answers to task A, press play.

So on question one, you had to determine whether you had enough information, you had sufficient information to calculate the volume.

So BG, GH and DE.

Well BG and GH are perpendicular distances, so that's good, but DE is actually the same as BG.

So we do not have three perpendicular measures, nor do we have enough information to make use of Phythagoras' theorem to calculate the third dimensional length.

So you can't do it for A.

Whereas B, AB, FE and AE.

AB and FE are perpendicular to each other.

If you think about the fact that AB is actually equal to FG.

AE and FE are in the same right-angled triangle that we could draw on that face.

So there is enough information to be able to calculate the three dimensions.

And similarly for C, you've got BD, DG, and FG.

This one's a little bit more challenging if you actually try to work out the volume, you will need to use Phythagoras' theorem a couple of times.

So every time you can try and find a right-angled triangle, remember if there's a right angle then there will be a right-angled triangle.

And sometimes with problem solving style questions you're better off starting and doing something with the information you have and then seeing if that gets you any closer to what you're trying to actually work out.

On question two and question three both of them was to work out the volume.

So on question two, the first part of the question was to work out the volume of a single cube, given that the diagonal across the face was eight centimetres.

Well by using the definition of a cube and the properties of a cube, we know that that face would be square.

So the right-angled triangle squares have right-angles.

So we can make a right-angled triangle, which would be isosceles because of the edge lengths being equal in a square.

So we can use Phythagoras' theorem to calculate what those edge lengths would be, which would be the square root of 32 or you may have that written down as four root two in its simplest form.

And then we want to calculate the volume of a cube.

And so we are just going to cube the side length.

So we need to do four root two or cubed and that's 128 root two cubic centimetres.

When we get to the second one, it was a cuboid made of two congruent cubes.

And so we know that the length of the face which had the diagonal marked would be two times as long as the height of that face or the width of that rectangle.

So we had the diagonal, which would be our hypotenuse, and we can then set up an equation using ratio and the proportionality between the two lengths and widths.

So if you let the width be equal to X, then you know that the length would be two X.

So our equation would read two X all squared plus one X squared equals the square root of five squared.

That's Phythagoras' theorem.

And then we could solve that equation.

We'd know that five X squared is equal to five and then we'd say X squared is equal to one and therefore X equals one.

So our height of that right-angled triangle or the width of the rectangular face is one and the length would be two because it is made of two cubes.

So now we can work out the volume of that cuboid that would have a length of two, a height of one and a width of one.

Two times one times one is two.

And then on question three you were given three measures for the cuboid but not the three that you needed in order to calculate the volume, but you did have sufficient information to calculate what you needed.

So 17 was given as the diagonal across one of the faces and you were given the 15 centimetres, which is one of the short edges of that right-angled triangle that we can create on that face.

Using Phythagoras' theorem, you can calculate that length, CD, is eight centimetres and that was the missing perpendicular dimension.

So once we now know that our volume would be four times 15 times eight, which is 480 cubic centimetres.

Really well done on that task.

So we're up to the second learning cycle where we're gonna continue looking at Phythagoras' theorem in 3D, but this time we're gonna look at other three dimensional shapes.

So we can apply the Pythagorean theorem to other 3D shapes such as cylinders and pyramids too.

And Jun's just reminding us, "Remember, where there are right-angles, there are right-angled triangles!" And we need a right-angled triangle for Phythagoras' theorem to hold.

So here we have a right cylinder.

We don't tend to call them right cylinders, we just tend to call them a cylinder, but there are cylinders that are known as oblique cylinders, but we are gonna be working with right cylinders here.

So Jun says, "Where might there be right-angled triangles hiding here?" So just pause the video and have a look.

Knowing that this is a right cylinder, which means that the circular face on the top is directly above the circular face on the bottom.

That cross section is uniform and they are parallel to each other but lined up exactly.

Where, are there some right-angled triangles? Press play when you're ready to look at what Jun's found.

So Jun found this one.

Did you find this one? Or perhaps you found this one, but in a different orientation.

Jun says, "It's often useful to notice this triangle because of the diameter of the circle." And if we were trying to work out the volume of a cylinder, then we need the radius and if we can calculate the diameter then we can calculate the radius.

Did you find any others? Juns found this one.

Again, you may have found a rotation of this, a different orientation, but basically the same right-angled triangle.

So he says, "This one's good to notice because it uses the radius and centre point of the circular faces." So we'd get the perpendicular height of the cylinder, which would be our length of the cylinder, necessary to work out the area of the rectangular face or the curved surface area.

Also necessary if we wanted to work out the volume of this cylinder.

If we can calculate the radius as I've already said, then that's gonna be helpful for the volume but also for the total surface area where we'd include those circular faces.

But what about not a cylinder but instead a pyramid? So this is a different shape, it's a different three dimensional shape.

It's got the apex where those triangular faces or meet and it's got, in this case, what looks to be a square or rectangular base.

So we are looking for pairs of perpendicular lines in order to be able to find a right angle and therefore a right-angled triangle.

So can you see any right-angled triangles for the pyramid? Pause the video and think about that and then when you're ready to check what Jun found, then press play.

So Jun found this one and he says, "It doesn't look like a right angle, but that's perspective playing tricks on us." So the base is going to be perpendicular to the perpendicular height of the pyramid.

That perpendicular height is from the apex directly straight down to the base.

Those two lengths are perpendicular and hence a right-angled triangle.

Did you spot that one? Jun found this one.

So here's something to think about.

Does anything that has a side on the base of the pyramid and a side from the centre point to the top work? So just pause and think about that.

Press play if you are ready for Jim's conclusion.

Jun says, "It does! And here is a less obvious one." So from the apex, which is the top of the pyramid to the base, that would be the perpendicular height.

And so that is perpendicular to the base and therefore they are two perpendicular lengths.

It doesn't need the hypotonus to be one of the slants, one of the edges of the triangles.

It could be a line within the triangle.

So let's look back at cylinder and a check for you.

What is the correct calculation to find the length of this cylinder? Pause the video and then when you're ready to check, press play.

So it's B we're gonna make use of Phythagoras' theorem because this is a right cylinder.

So the face at the top is perpendicular to that vertical height.

So seven is the hypotonus, it's the longest edge and opposite the right angle.

So we are going to find the difference between the square of the hypotenuse and the square of the shorter side and then square root it to get our length.

Why might we need the length? Well, if we were looking to work out the volume, then we'd do pie times radius squared, and in this case the radius is two, and then we would multiply it by that length that we've just calculated.

And that would give us our volume.

So up to the last task of the lesson, and both of the questions are on the screen here.

So in question one, you need to find the volume of these cylinders to one decimal place.

So on A you've got the radius and a diagonal length.

On B, you've got the diameter and a diagonal length.

And on C you have the area of the circular face and a diagonal length.

In question two it's now about a pyramid.

So a pyramid is made up of a square and four equilateral triangles.

The apex is directly above the middle of the square.

If the area of the square is 16 square centimetres, then what is the height of the pyramid? So pause the video whilst you're working through those questions in task B.

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

So on the screen we've got the answers to one and two.

So 1A, you needed to work out the volume of the cylinder and you were given the radius and the diagonal.

So the radius is really helpful to be able to work out the area of the circle.

But we also needed the length of the cylinder and that's where the diagonal was going to be helpful and using Phythagoras' theorem.

So the part of the calculation that says the square root of nine squared minus three squared is the height or the length of the cylinder.

And then we can see that we're doing pi r squared H, which is the volume of any cylinder.

So to one decimal place, part A was 239.

9 cubic centimetres.

On B, you were given slightly different information.

You were given the diameter of the circular face.

So to work out the area of that circular face, you needed to find the radius and we'd do that by halving it.

So two squared pi is the area of the circular face.

We needed to use the diameter with the diagonal that was given.

So 11 squared minus four squared and then square rooted was calculating the length or the height of the cylinder.

We multiply that by the area of the circle and that gives us our volume.

So the volume for that cylinder was 128.

8 cubic centimetres.

On part C, you were given the area of the circular face and it was nine pi.

So we can make use of our knowledge of square numbers to know that the radius is there for three.

So if the radius is three then we can use that and the six which was given to calculate the height of the cylinder and then calculate the volume.

So the volume for cylinder C was 146.

9 cubic centimetres to one decimal place.

On question two, we were now working with a right pyramid, but you know it's a right pyramid because the apex was directly above the centre of the base.

The base was square, the triangular faces were equilateral.

So we were given that the area of the base was 16 and because we knew it was a square, then we could square root that to find the length of the square.

So the edge length of the base was four.

The triangular faces were also equilateral triangles and adjoined to the base.

So the equilateral triangles also have an edge length of four.

We're gonna make use of Phythagoras' theorem in order to work out the diagonal distance from a vertex on the base to the centre.

And that is two root two.

There's a couple of ways that you could do that.

So you could use a right-angled triangle, an isosceles right-angled triangle because of it being a square and doing four squared plus four squared and then square rooting, that would give you the diagonal distance across the base and then halving it because you only want to get to the midpoint, which is where the centre point was.

Alternatively, you could have used a right-angled isosceles triangle where the hypotonus was four.

And so squaring that gives you 16, halving it and square root in it would also give you the two root two.

The two root two is helpful because then we can have our right-angled triangle that goes from the apex to the centre, from the centre across the half diagonal to the vertex of the square.

And we would know that that edge length is four because of the equilateral triangle and we know that edge is two root two.

we can then use Phythagoras' theorem once again to work out the shorter side or spot and notice that it's exactly the same as on the base.

So the height, the perpendicular height of the pyramid was two root two.

So to summarise today's lesson, right-angled triangles can be formed inside 3D shapes.

And remember what Jun said, if there is a right angle, then there will be a right-angled triangle.

Phythagoras' theorem still holds in these situations.

So Phythagoras' theorem works on a two dimensional right-angled triangle.

So if you can find the right-angled triangle inside of a 3D shape, then you can make use of Phythagoras' theorem.

And this means we can calculate unknown lengths in 3D shapes and is a very helpful skill, especially if you need to then go on and work out the volume or the surface area.

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