Lesson video

Hello and welcome.

It is a pleasure to see you here for this Maths lesson with me, Mr. Gratton.

Today, we'll be looking at, how we can apply Pythagoras's theorem? to solve some more real world problems? Pause here to have a quick look at some of the keywords and ideas that we'll be looking at today.

First of all, how can some real world contexts be modelled by right-angled triangles? Let's have a look how.

Andeep says he is confident in his skills when finding the length of a side of a right-angled triangle when he knows two other side lengths at least.

But what's the point? When can all of this be used in the real world? Well, Laura considers that there may be some real world situations where right angles are incredibly important and where triangles may also appear.

Can you think of any? There are many contexts in the real world that involve lengths or distances that are perpendicular to each other.

For example, most buildings will have a floor and a wall that are perpendicular to each other, or a third length that bounds an imaginary right-angled triangle.

For example, a roof, a floor, and a support pillar in the attic of a house.

And sometimes it is helpful to model these contexts as right-angled triangles in order to help find important lengths or distances within that context.

Such as this one.

As strange as it sounds, a strand of spaghetti and a saucepan, they create a right-angled triangle.

But how and why? Well, for spaghetti to cook evenly, the whole strand must submerge completely into the saucepan at the start of the cooking process.

If not, you may end up with some overcooked or crunchy parts at the end if some of it sticks out at the beginning.

So this isn't what we want, the spaghetti at the end sticks out.

Whilst in this diagram, here it is fully submerged.

This spaghetti strand will cook evenly.

What information about the spaghetti and the saucepan might we need in order to see if the spaghetti will fit fully? Well, if we know the height of the saucepan and its maximum length, which is the diameter of that circle on the bottom of the saucepan, and we are looking for that maximum length for the best opportunity for the spaghetti to fit, if we know both of those, then we can calculate the maximum distance from the top of one end of the saucepan to the bottom diagonally opposite other end of it.

If we measure the length of that spaghetti strand and find it to be shorter than this maximum diagonal distance, then it will successfully fit in the pan.

Knowing these lengths, we can model the saucepan as a right-angled triangle.

The height, 12 centimetres for the saucepan, and the maximum base, 22 centimetres, are represented by the two shorter sides of this right-angled triangle.

This diagonal maximum distance we want to test for the spaghetti strand is the hypotenuse of the triangle.

And since we know two lengths of a triangle, we can use Pythagoras's theorem to find that hypotenuse.

So, using Pythagoras's theorem, we have 12 squared for the height, 22 squared for the maximum base, equals x squared for that unknown hypotenuse.

And by solving that equation, we get 25.

1.

And so the maximum distance from one end of the saucepan to the other is 25.

1 centimetres.

And so we have modelled on this right-angled triangle the hypotenuse of 25.

1 centimetres.

This means that the maximum diagonal length of the saucepan is also 25.

1 centimetres.

The length of the spaghetti strand is 28 centimetres, which is longer than the saucepan's maximum length.

This means it will not fully fit into the pan.

There will be a small part of the spaghetti that will not be submerged into the water and therefore it will not cook evenly with the rest of that strand.

Also note how we approached this question.

We knew both the base and the height of the saucepan and we knew the length of the spaghetti.

However, we only modelled the saucepan in this triangle.

The spaghetti's length was not modelled at all.

Its length only came in later when we compared it to the saucepan.

This is because it is sensible to model only one thing, one object or one collection of related objects at a time.

The saucepan was this one fixed object and so we modelled that as a right-angled triangle.

Similar to the saucepan, here's a check.

Jun only wants to use long pencils that stick out of a pencil pot, so that he doesn't have to rummage around to find the shorter pencils.

He has this 14 centimetre pencil.

Whilst its vertical standing length is taller than the pencil pot, he's worried that the pencil will tip over into the pencil pot and be stuck inside.

Jun models this pencil pot, not the pencil, as a right-angled triangle.

Pause here to think, what will the lengths of the two shorter sides of this right-angled triangle be? And the answers are 11 and 6.

11 centimetres is the height of the pencil pot and 6 centimetres is the maximum base.

Now we know two lengths of this triangular model, pause here to use Pythagoras's theorem and find the length of the hypotenuse of this triangle and round it to two decimal places.

The hypotenuse is 12.

53 centimetres.

Now that we have calculated that the hypotenuse of the triangle is 12.

53 centimetres, what conclusion can we draw about the pencil and the pencil pot? Pause here to select the correct conclusion.

The pencil is long enough, because this 14 centimetre pencil is longer than the maximum diagonal distance of that pencil pot at 12.

53 centimetres.

Onto practise task A.

For question one, will this 35 centimetre ladle sink into or stick out of the saucepan? Pause here to label the saucepan and the right-angled triangular model and then use Pythagoras's theorem to form a conclusion.

And for question two, Izzy is practising backhand tennis shots, with her maximum shot having a length of 28 metres, by first modelling the locations of Izzy, Sofia, and Laura onto this triangle and writing down appropriate distances, explain whether Izzy's shot can reach both of them.

Pause here to do this.

And for question three, a 256 centimetre long access ramp is installed to help passengers board this maglev train.

The step to board the train is 43 centimetres higher than the platform.

Pause here to figure out how far onto the platform horizontally the access ramp will need to reach.

Okay, onto the answers.

For question one, on this triangular model, the hypotenuse is 36 centimetres, but the ladle is only 35 centimetres, meaning it will sink completely into the saucepan, which is very annoying when it happens whilst cooking.

And for question two, the distance from Izzy to Sofia is 24 metres, and from Izzy to Laura is 26.

4 metres.

Izzy's backhand shot can reach both of them because 28 metres is longer.

And for question three, using Pythagoras's theorem to find the length of a shorter side, the horizontal distance from the train to the base of the ramp, we get a distance of 252.

4 centimetres.

So, from tennis to spaghetti, we've looked at some contexts we may be able to model as right-angled triangles.

But even if they can at a first glance, there may be limitations to our assumptions.

Let's be critical and have a look at some more contexts.

But before that, sometimes there are descriptions of contexts without any diagrams to aid us.

We may have to model them as right-angled triangles completely from scratch.

For example, here's a context made from only words.

That's a lot of information.

How do we even know that this creates a right-angled triangle? How can we go about modelling something completely from scratch? Let's have a look.

This context mentions a small bird, so I plot a point on a piece of paper to represent the starting location of that bird.

The bird sits three metres horizontally away from that tree.

This is a lot of information very quickly.

We have a tree, a second object that is really tall, and it is three metres away from the bird horizontally.

So I draw a horizontal line segment and label it three metres.

In reality, it doesn't matter how long I draw that line segment.

Don't worry about the accuracy, scale, or proportions of your model.

For basic models that we will look at today, we don't need high levels of accuracy, rather just a good sketch, to give us an idea of what shapes, usually right-angled triangles, our context problem can be modelled as.

We know that the tree is tall, but we're not sure how tall.

So let's draw a vertical line segment and label it with a letter to represent its unknown height.

And finally, the bird flies nine metres to the top of the tree.

So we draw a line segment from the bird to the top of the tree and label it nine metres.

Different descriptions can and will give you different shapes when modelled.

But this one involving a bird, a tree, and the distances between them seemed to now be modelled by a right-angled triangle.

Why bother though? Because we want to find the height of the tree, the length of one of the shorter sides of this modelled right-angled triangle.

And so, using Pythagoras's theorem, we can set up the equation x squared, the height of the tree, plus three squared, the distance between the bird and the tree, equals nine squared, the flight path of the bird.

Solving this gives us that x equals the square root of 72, therefore the tree is approximately 8.

5 metres tall.

For this check, in four parts, we have a 10 metre flagpole, a long metal wire attached both to the top of the flagpole and to a hook that is a horizontal distance of four metres from the bottom of the flagpole.

Pause now to think, on this model, what does the vertical height even represent? That vertical height represents the flagpole.

And for this next part, pause here to answer, what is the length of this side of the triangle currently labelled "a"? And since we know this vertical height represents the flagpole, the flagpole is a total of 10 metres tall.

Pause again here to refer back to this context and figure out, which other side of this triangular model do we know? Do we know b or c? And what is the length of the side that we know? We know this horizontal length is four metres.

That is because this distance is the distance between the hook in the ground and the bottom of the flagpole.

And finally, pause here to use Pythagoras's theorem in order to calculate the length of the metal wire, the hypotenuse of this right-angled triangular model.

The length of this metal wire, the hypotenuse is 10.

77 metres.

There are many contexts that we can model, but are we sure, are we certain, that all of these contexts that we've looked at have right angles in them? The bird and tree example, we made a lot of assumptions about, well, everything actually, when modelling it as a right-angled triangle.

Were our assumptions fair or terrible? Meaning a right-angled triangle is a bad model that reflects this situation.

Let's make a judgement call.

First of all, we assumed that the bird flew in a straight line.

If however it flew in a wobbly path, maybe due to wind or the way that the bird flaps its wings, then we don't have a hypotenuse.

Actually, we don't have a triangle at all.

And furthermore, we also assumed that the tree stood perfectly vertically, so that there was a right angle between the tree and the ground.

But what if instead the tree was leaning? We cannot use Pythagoras's theorem on a triangle that is not right-angled.

So even if the bird did fly straight, When modelling, we have to consider whether assumptions we make are sensible.

If not, the context cannot be modelled, for example, as a right-angled triangle.

Some assumptions include assuming all lengths and distances are linear paths, straight lines, or line segments, and not curves.

Assuming two lengths or distances are perpendicular to each other, and not at a different angle.

And finally, we have to make assumptions that the two perpendicular lengths, if they are perpendicular at all, touch or intersect in some way.

Okay, everyone, here's another check.

For this first saucepan that we looked at, it was sensible to assume that the maximum base and its height were perpendicular.

It looked like the side of the saucepan touched the circumference of the circle at the base of the pan at 90 degrees.

But for this saucepan, pause here to consider what assumption clearly does not hold true, meaning we cannot model this as a right-angled triangle.

Quite clearly, none of the sides are perpendicular to each other.

If modelled as a triangle, the sides of length 12 centimetres and 18 centimetres would not meet at a right angle.

And so the third side would not be a hypotenuse that we can use Pythagoras's theorem on to find the length of.

Okay, everyone, it's time for the next practise task.

By modelling these three radio towers as vertices of a right-angled triangle, calculate whether towers B and C are close enough together that their signal can still be broadcast to each other if tower A breaks down.

Pause here to do this question.

And for question two, the captain's claim that the ship is 120 metres from shore might be wrong.

Pause here to consider what assumptions they might be wrong about.

For question three, a window is 412 centimetres above ground level.

For a window cleaner to safely reach the window, the base or bottom legs of their ladder should be a distance that is one quarter of the vertical height that the ladder will stand.

Pause here to model this situation as a right-angled triangle, calculate what length the ladder should be set to, and also write down the assumptions you are making about this context for it to be effectively modelled as a right-angled triangle.

And finally, pause here to answer question four.

Is a 35 metre reel of electrical wire going to be long enough to connect a battery to both lights on this phone mast system? Okay, here are the answers.

For question one, the distance between towers B and C is 3,061 metres, meaning that the radio signal would be lost because the towers are too far away from each other for the signal to reach.

However, we have assumed that each tower is the same height and the same height above sea level.

Whilst such minimal variations won't have a significant effect on the answer since 3,061 metres is only just over the signal limit, these small variations may still have some impact.

And for question two, the captain of the boat modelled the situation he's in, the location of the boat as a right-angled triangle, and 120 metres is a mathematically correctly calculated horizontal distance for the right-angled triangle that they modelled.

But we can see the shore is closer than the lighthouse, meaning that the boat may be less than 120 metres away.

Furthermore, the lighthouse is on a hill, meaning its true height above sea level will be much higher than 50 metres.

And for question three, here's the model.

The ladder's base should be 103 centimetres away from the base of the wall because 103 is a quarter of 412, and the ladder itself should be approximately 425 centimetres in length.

For this context, some examples of assumptions are that the wall is perfectly perpendicular to the ground.

Furthermore, that the ground is not sloped and that the ladder does not bend when the window cleaner is climbing on it.

And finally, question four.

35 metres is enough wire.

The distance from the hook to the battery is 18.

7 metres.

The distance from the battery to the top of the mast is 15 metres, giving a total length of wire needed at 33.

7 metres.

And pause here to look at some of the possible assumptions that we made about this context.

So far, we've only looked at contexts that explicitly made right-angled triangles.

But what about contexts that made composite shapes instead, where right-angled triangles are merely a part of these greater composite shapes? Let's have a look.

Not everything in the real world is a straightforward right-angled triangle.

But Laura is correct.

Pythagoras's theorem is still possible if there are right-angled triangles embedded into more complex shapes, such as this kite.

Joining the diagonals with wooden supports will create four right-angled triangles on that kite.

Laura places a 30 centimetre bamboo stick along one of the diagonals of this kite to support it during stronger winds, but she also wants to place another bamboo stick along its other diagonal.

How long should this second bamboo stick be? Can you spot where the right-angled triangles are in this kite? What sides of these triangles do we already know the length of, and which sides do we need to calculate the length of to figure out how long the bamboo stick needs to be? These two line segments represent the two bamboo sticks.

There are a total of four right-angled triangles in this kite.

Actually, there are two pairs of congruent right-angled triangles.

The top two are congruent with each other and the bottom two are also congruent with each other.

Let's focus on one unique triangle each.

This top right-angled triangle has a hypotenuse of 20 centimetres from the side of the original kite, whereas this triangle has a hypotenuse of 85 centimetres known from the side length of its congruent counterpart.

That horizontal bamboo stick is a total of 30 centimetres long, which is then cut into two equal halves as the other diagonal perpendicularly bisect the one with a 30 centimetre stick on it.

Therefore, this distance is now 15 centimetres, meaning one of the two shorter sides of both of these triangles, the shared side actually, is 15 centimetres in length.

We now have two right-angled triangles with two side lengths that are known.

We can use Pythagoras's theorem to find the vertical height of both of these triangles.

Starting with the top, the equation is a squared plus 15 squared is the hypotenuse squared of 20 squared, which can be solved to get 13.

2 et cetera.

We must keep as much precision as possible as we'll be using this value in further calculations later on.

Similarly, with the bottom triangle, we have b squared plus 15 squared equals a hypotenuse squared of 85 squared, which can be solved to get 83.

6 et cetera.

We can then join these two right-angled triangles back up at their shared vertex.

That total vertical distance is the sum of the two lengths that we just calculated.

This combined length is also the missing diagonal of that kite, the total vertical distance and length of the bamboo that we needed to attach onto the kites.

Therefore, we can add these two partial vertical distances together to give a total of 96.

9 centimetres, after rounding, of bamboo needed to complete the kite.

For this check, in three parts, I have modelled this snooker table as two non-congruent right-angled triangles.

Pause here to write down the letters of the sides that we know, and also write down the lengths of those sides.

We know that the vertical height of both of these triangles is three metres as the maximum length of that snooker table is also three metres.

We know that one of those two triangles has a length of one metre, and the combined length is 1.

8 metres, leaving 0.

8 metres for the other triangle.

Currently, we do not know the hypotenuse of either triangle, so we do not know c or d.

However, by using Pythagoras's theorem twice, we can find the lengths of those two missing sides, c and d.

Pause now to do so.

And so, the length of c is 3.

10 metres after rounding, whilst the length of d is longer, at 3.

16 metres.

And finally, Jacob says his previous record for the distance he can strike a ball was 6.

2 metres.

Pause here to decide if this new shot breaks his previous record or not.

In order to show whether he has or has not, write down an appropriate calculation.

And the total distance that the snooker ball travelled in this shot is a combination of the distance from the bottom of the table to the top, at 3.

10 metres, plus the distance that the ball travelled back down the table after bouncing off the edge of the table.

This extra distance is 3.

16 metres, for a total distance of 6.

26 metres.

6.

26 metres is further than Jacob's previous record of 6.

2 metres.

Onto this final check.

Pause here to see if seven metres is enough adhesive tape for Aisha's picture frame, and what is the minimum amount of ribbon, rounded to the nearest half metre, that Laura needs for the perimeter of her kite? Pause now.

By first modelling this tent as an isosceles right-angled triangular prism, with the entrance to the tent as that base face, pause here to calculate the total length of tent pole that needs to be fit into the lining of each of the nine edges of this tent.

For Aisha, the length of one diagonal is approximately 1.

47 metres.

If the perimeter and both diagonals need adhesive tape, the total length required is 1.

47, plus 1.

47 for the two diagonals, plus 4.

1 for the perimeter of the frame, for a total of 7.

04 metres.

Therefore, Aisha is four centimetres too short.

Whilst she does not have fully enough tape to cover everything that she needs, in reality, will a four centimetre loss be that impactful to this context? Probably not.

For question two, the perimeter of this kite is two lots of 0.

43139, plus two lots of 1.

33645, for a total of 3.

53568 metres.

There are two options for rounding here, either 3.

5 metres or 4 metres.

This is because we had to round to the nearest half metre.

If we chose 3.

5 metres, that would not be enough to fully cover its perimeter, and so 4 metres is required.

And for question three, the total length of tent pole required for all of the nine edges of lining of this tent is 31.

67 metres.

Great job, everyone, on dealing with all of these contexts in a lesson where we have modelled real world contexts involving perpendicular distances as right-angled triangles, which require Pythagoras's theorem to solve problems about these contexts.

We've also been critical that if assumptions about linearity or perpendicularity are not true, then our models of right-angled triangles may not be accurate.

And lastly, we looked at more complex problems that can be modelled as multiple triangles.

Thank you so much for all of your attention today and for all of your effort.

Have a great rest of your day, and until our next Maths lesson together, take care and goodbye.