Lesson video

Hello, everyone! I'm Mr. Gratton, and welcome to today's maths lesson.

It is a pleasure to see you joining me in a lesson where we will be combining our knowledge of Pythagoras theorem, similarity and ratios to solve a range of geometric problems. Pythagoras theorem is a relationship between the sides of a right angled triangle that states that the sum of the squares of the two shorter sides of that right angled triangle is equal to the square of its hypotenuse.

Whilst we might be familiar with what Pythagoras theorem is, under what circumstances do we actually need to use it? Let's have a look.

Laura seems to think that we use Pythagoras's theorem whenever we see a right angles triangle, Andeep thinks differently however, and I hope that you agree with him.

Can you think of any maths topics that you've done in the past that involved right angled triangles but didn't require Pythagoras theorem? And as Laura asks, when should we use Pythagoras theorem and when should we not? Here are some different representations of Pythagoras theorem that you might have seen before.

Andeep observes that every single time he sees Pythagoras theorem, it's talking about the sides of a right angled triangle, and the only time angles are ever mentioned are when we need to check where the 90 degree right angle even is.

And as Laura says, we use Pythagoras's theorem to find the length of one of the sides of a right angled triangle, when you know or can figure out the other two sides.

Laura has correctly written an equation from Pythagoras's theorem.

Let's see what this equation actually represents from this triangle.

These two numbers represent the two sides that you know the length of.

The 44.

5 represents the length of the hypotonus in centimetres.

Whilst the 36 represents the length of a shorter side also in centimetres.

We use the square of these two lengths to find the length of the third side of a right angled triangle, this time labelled as A centimetres.

But, how do we find the value of A using our equation from Pythagoras theorem? Well, we solve the equation.

This time given a value of 26.

16 centimetres.

This is the length of that missing third side of our triangle, the side currently labelled A.

In comparison, if you are being asked to find the size of an interior angle of a triangle, then you don't use pyrosis theorem.

You use the angle fact, the interior angles in a triangle, sum to 180 degrees, like so.

This missing angle X, plus the known angles of 36 degrees and the right angle of 90 degrees has a total or sum of 180 degrees.

Again, we can solve this equation, just not one derived from Pythagoras's theorem, giving us a missing angle of 54 degrees.

For this check, which of these triangles requires Pythagoras theorem to calculate the value of X or Y? Pause now to consider the information given and what you are actually trying to figure out.

Triangle A is the triangle that you need Pythagoras theorem to calculate the value of X.

This is because you know two sides of a right angled triangle, the 18 centimetres and the 56 centimetres, and you are trying to figure out a third side, the side labelled X centimetres.

And so we know that the leftmost triangle requires Pythagoras theorem in order to find the value of X.

Pause here to use appropriate methods for both triangles in order to find the length of X centimetres, and the size of angle Y degrees.

The hypotenuse X centimetres has a length of 58.

8 centimetres.

Whilst the angle Y degrees is actually 34 degrees.

The perimeter and area of a triangle may be things that you are familiar with, but sometimes you actually need Pythagoras theorem to find out the area, or perimeter of a right angled triangle, but which ones require the theorem? And how do you know? To find the area of a right angle triangle, you must first check if you know the base and the perpendicular height of that triangle.

The base and the perpendicular heights are always the two shorter sides of a right angled triangle.

So if you know the two shorter sides, you do not need to use Pythagoras theorem to calculate the area of that triangle.

So in this example, we have a perpendicular height of 18 centimetres, a base length of 29 centimetres, and don't forget the divide by two at the end to get an area of 261 centimetre squared for its area.

If you've checked the base and the perpendicular height, and you realise that you do not know one of them, but you do know the hypotenuse, you must use Pythagoras' theorem first before you can calculate the area of that triangle.

So, we can set up the following equation from Pythagoras' theorem.

P squared, or the unknown perpendicular height squared plus 45 squared, which is the base squared equals 72 squared, which is the hypotenuse squared.

And then we can solve the equation and we see that the perpendicular height is approximately 56.

2 centimetres.

The area of this triangle is therefore 1264.

61 centimetres squared after using that perpendicular height of approximately 56.

2, and multiplying it by the base length of 45 centimetres and then dividing that product by two.

To find the perimeter of any right angled triangle, you will need to know the length of all three sides.

Ah, you always need to know all three side lengths.

So if you only know two lengths, no matter which two they are, you will need to use Pythagoras' theorem to find the length of that third side.

So, for this left triangle, I didn't need to use Pythagoras theorem for its area, but I do to find its perimeter.

So setting up the equation gives 18 squared plus 29 squared equals H squared, where H squared is the square of the hypotenuse of that triangle.

Solving this equation gives a hypotenuse of 34.

13209633 centimetres.

The perimeter is therefore that number, plus 18, plus 29, which gives you a perimeter of 81.

13 centimetres after rounding.

For the right hand triangle, however, we have already used Pythagoras theorem to calculate the length of P, which was about 56.

2 centimetres.

So, since we already know all three side lengths, we find the sum of them.

Therefore the perimeter of this triangle is 173.

2 centimetres.

And for this check, pause now to consider in which of these triangles must Pythagoras' theorem be used in order to calculate its area.

In triangle B, we know two lengths that are perpendicular to each other, the two shortest sides of a right angled triangle.

Therefore we do not need to use Pythagoras theorem there as this triangle has a base and perpendicular height.

However, we do need to use Pythagoras theorem for triangle A, since we know one of the two shorter sides, the 21 centimetres, but not the length of the other shorter side length with unknown length P centimetres.

However, we do know the hypotenuse are 48 centimetres.

And following on from this, pause here to match the numbers to complete each of these four sentences.

You should have used Pythagoras theorem to find the length of P, 43.

16 centimetres, and the length of R, 52.

39 centimetres.

By adding all three side lengths of triangle A together, you would've got its perimeter at 112.

16 centimetres.

However, you didn't need to use any extra information at all to find the area of triangle B.

It's base, times by its height, divided by two is 504 centimetre squared.

So we've seen Pythagoras theorem is sometimes useful when finding the area or perimeter of a triangle, but what about when we are dealing with similar right angled triangles? Well, let's have a look.

If two triangles are similar to each other, then the side lengths of one triangle can be multiplied by a constant scale factor to find each length of the corresponding sides on the second triangle.

This remains true even if Pythagoras's theorem has to be used first to find the length of one of its sides.

We use Pythagoras's theorem to find the length of one side of a triangle, and then multiply that side by the same scale factor used for the other already known sides of those triangles.

For example, let's find the length of the hypotonus A using Pythagoras theorem.

My equation is A squared equals 15 squared plus six squared, which when solved becomes 16.

15549 centimetres.

Since we are not at the end of a question yet, let's keep it to as many decimal places as possible to improve our precision.

You can also leave it as the square root of 261 if you feel comfortable with said form.

Right, now we know all three sides of triangle X.

Let's figure out the scale factor from triangle X to triangle Y.

Choose a pair of corresponding sides that you know both lengths of.

In this case, I have chosen the corresponding heights of 15 centimetres and 30 centimetres.

It's clear therefore that the scale factor is a times by two scale factor from X to Y.

So I take the hypotenuse from triangle X and multiplied by two to get the hypotenuse on triangle Y.

So this hypotenuse is 32.

3 centimetres after some rounding.

Here's a quick check that will test your understanding of similar triangles.

These two triangles are similar.

Pause here to figure out the scale factor from triangle P to triangle Q.

The corresponding side of both triangles that I know both lengths of are these two, the four centimetres and the 24 centimetres.

This is because the sides are adjacent to both angles, 37 degrees and the 90 degree right angle.

I then do a division, 24 divided by four is six.

And so the scale factor from P to Q is a multiplied by six scale factor.

Using Pythagoras' theorem, pause here to calculate the length of the hypotenuse of triangle P.

The hypotenuse squared is three squared plus four squared, three squared plus four squared is 25, and the square root of 25 is five centimetres.

And so the hypotonus of triangle P is five centimetres long.

And so, with all of the information we've looked at so far, pause here to find the length of the side labelled K.

K is the hypotenuse of triangle Q.

And since the scale factor from P to Q is times by six, all I need to do is take the hypotenuse from P, which is five centimetres, and then times it by six to get the hypotenuse of Q, which is 30 centimetres.

Here's practise task A.

For question one, find all missing side lengths and angles and thus find the area and perimeter of these two triangles.

For each thing that you find, consider whether it required the use of Pythagoras theorem or not for calculations such as the area and perimeter that required other values that you've already figured out.

Did those values also require the use of Pythagoras theorem? Pause here to fill out this table.

These three triangles are similar.

Pause here to calculate the lengths of T and V.

And question three, are triangles L and M similar to each other? Pause here to analyse these two triangles and write a statement explaining your answer.

And here are the answers.

Pause here to check your answers for question one.

For question two, to find the length of V, you didn't even need Pythagoras theorem, only the knowledge that the scale factor from the middle triangle to the largest triangle was times by four.

To find the length T, you first needed to find the length of the hypotenuse of the middle triangle at 23.

259 centimetres.

Furthermore, the scale factor from the middle triangle to the smallest triangle was a divide by three or times by a third.

And so for T, one third of 23.

259 is approximately 7.

8 centimetres after rounding.

And for question three, these two triangles are not similar.

I can use Pythagoras' theorem to find the hypotenuse of triangle L at 70 centimetres.

The hypotenuse of L is twice the length of the hypotenuse of M at 70 centimetres compared to 35 centimetres.

So the scale factor from L to M should be times by a half for all three sides if both triangles are similar.

However, no other side on M is half the length of any side on L, meaning the scale factor is inconsistent or more accurately, there is no scale factor between these two triangles because they are not similar with each other.

Okay, let's put right angle triangles into a different context.

One where you may have seen right angle triangles before, but for a different purpose.

Let's use Pythagoras theorem for points on a coordinate grid.

Pythagoras theorem is incredibly useful in calculating the shortest distance between two points on coordinate grid.

But how when there is no triangle, and we only have one line segment? One we don't even know the length of.

Pause here to think about or discuss how a right angled triangle can be made from this diagram.

A horizontal ray can be drawn from one of those points, and a vertical ray can be drawn from the other point.

Can you see the triangle yet? That point of intersection between the two rays is that third vertex of the right angled triangle, here.

These rays are actually very useful.

The point where the two rays intersect is always the vertex of the right angle of the right angle triangle itself, this is always true.

And Andeep is spot on.

The two sides made from the horizontal and vertical rays always produce the two shorter sides of the right angled triangle.

And so the distance between those two points, the last remaining side of the triangle is always the hypotenuse.

The distance between the original two points is always the hypotenuse, because it is always opposite the right angle that we know is always made by the intersecting two rays.

But how can we use Pythagoras' theorem? There are still no sides on this triangle.

Ah, but yes, there is! The lengths of these sides are determined by the number of units up and across using the grid squares themselves.

So the height of this triangle, that's two squares up, so two units.

Similar for the base, it is five squares across.

And so that length is five units.

And now we're back to a familiar site.

Using Pythagoras theorem gives us the hypotenuse squared, equals two squared, plus five squared, and solving it gives us a hypotenuse of 5.

385 centimetres.

This is then also the shortest distance between those two original points.

And for this check, the distance from point A to point B can be found by creating a right angled triangle with length AB as its hypotenuse.

Pause here to figure out the horizontal base length and the vertical height of this right angles triangle.

The base is five units long, and the height is three units high to create this triangle.

And so using Pythagoras theorem, pause here to calculate the shortest distance from A to B.

And so, the shortest distance is 5.

83 units, the length of the hypotenuse of that triangle.

Okay, let's explore a question that requires us to look at multiple right angle triangles in order to find many distances or in this case, many sides to another shape.

The question states, calculate the perimeter of this shape, but in order to calculate the perimeter of any triangle, you need to know the lengths of all three sides of that triangle.

And currently we only know one, that base horizontal length that is 10 centimetres long.

So for each unknown side length, we need to construct a right angled triangle.

So, in this question, we need two right angled triangles, one with X as its hypotenuse and one with Y as its hypotenuse.

It is good practise to construct and use Pythagoras' theorem for one triangle at a time.

So to find the length of X, we use Pythagoras' theorem on triangle A to give X squared, the hypotonus squared equals five squared plus seven squared, which when solved gives us X equals 8.

60 units.

And similarly, to find the length of Y, we use Pythagoras theorem on this right angled triangle, giving Y squared equals five squared plus three squared, where Y squared is the hypotenuse of this second right angled triangle.

When we solve this, we get Y equals 5.

83 units.

And so, the challenge in this question was to spot that we needed to construct two right angle triangles in order to find two lengths.

Once we know all three side lengths, we sum them together to get a perimeter of 24.

43 units.

Okay, for this check, pause here to consider which of these is a suitable equation to find the hypotenuse of triangle A.

The horizontal length of triangle A is six units, and the height of the triangle is seven units.

And so option C is the correct answer.

And for this same triangle, pause here to calculate the other length, the other side currently labelled with an E.

The right angled triangle from this side length has a base of one unit and a height of seven units, giving a hypotenuse of 7.

07 units.

For question one of this practise task, use Pythagoras' theorem to find the shortest distance between the following pairs of points.

Pause here to do so, and some aligned segments have been drawn to help with parts A and B.

For question two, by using Pythagoras' theorem possibly multiple times for each shape, calculate each shape's perimeter.

For question three, it's time to get creative.

Write down three coordinates.

These coordinates are the vertices that make a triangle with a perimeter between 20 and 22 units in length.

Show that this is correct using Pythagoras theorem.

Pause to do both these questions.

And for the answers, question one, we have A is 6.

7 units, B is 4.

5, C is 8.

1, D is eight, and E is 12.

2 units.

For Part D, you didn't even need to use Pythagoras theorem.

The distance from C to D is a horizontal distance of eight units found simply by counting the number of squares along.

And for question two, pause here to compare your calculations to the answers that you see on screen.

And for question three, a very well done if you were able to show that the triangle from your three coordinates had a perimeter between 20 and 22 units.

For the one example that I have given, its perimeter is approximately 20.

47 units.

And for this last cycle, let's see how Pythagoras theorem is used frequently with ratios.

What could I mean by that? Well, I am certain that you see several similar shapes involving right angled triangles on a daily basis without even realising it.

In fact, you are probably looking at one right now.

What do I mean? Well, most TV screens, computer monitors and mobile phone screens are similar in their shape.

This is so a movie can be played and it looks perfectly normal on one screen without it looking distorted or stretched on a different screen that is much wider when compared in proportion to its heights.

For most common widescreen TVs, monitors and phone screens, they are in the 16 to nine aspect ratio, meaning for every 16 units across, there are nine units up.

The length of one unit is what changes between each screen.

So for a small phone screen, each unit will be tiny, but there will still be 16 equal units across and nine of those same equal units up.

The same applies for a massive TV screen.

Just those nine and 16 units will be larger.

And whilst that might be interesting, what does it have to do with Pythagoras theorem? Well, most of the time, when you want to buy a TV, monitor or phone, the size of the screen is described with only one distance given.

For example, this phone screen is seven inches.

But what does that distance represent? Not its horizontal or vertical distances.

The length describes the diagonal of the screen, and so we can find the length and the width of a screen by using Pythagoras theorem and the 16 to nine aspect ratio.

Our method goes like this, work with only these unit ratio measurements to build a full picture of the lengths involved in one type of measurement.

Once we've established the width, height and diagonal lengths measured in units from a ratio, then we can build a ratio table to compare it to actual measures such as centimetres or inches.

So for a 16 to nine units screen, its diagonal length can be calculated by X squared, the diagonal squared equals 16 squared plus nine squared, which solves to get 18.

36 units across its diagonal.

Now we know the width, the height, and the diagonal in this units measurement.

We can represent all of this information in a ratio table, with all of the unit measurements in one row.

So the 16 is the width of the screen in units.

The nine is the height of the screen in units, and the 18.

36 is the diagonal of the screen also in units.

If we are told that this screen is seven inches along its diagonal, then we can put this information into the appropriate part of the ratio table, but on a different row, as this measurement is in inches this time and not units.

So the seven goes into the diagonal length column, but in the inches row.

I then use proportional reasoning to figure out that my multiplier to get from units to inches is seven over 18.

36.

I can then multiply the length of the screen by seven over 18.

36 to convert it from units to inches, giving 6.

1 inches for the width of the screen.

The same can be done for the height, nine times seven over 18.

36, giving a height of 3.

4 inches.

This phone with an aspect ratio of 16 to nine is 6.

1 inches wide, and 3.

4 inches tall, and is exactly seven inches along its diagonal.

For this check, we're going into the past.

Before widescreen monitors, retro monitors had a width and height in the ratio four to three.

By using Pythagoras theorem, pause here to find out how many units long the diagonal of this screen actually is.

Some of you might have noticed that this is a 345 Pythagorean triple.

The diagonal of this computer screen is five units long.

For this same screen I have a ratio table.

The first row shows the width, height, and diagonal measured in ratio units.

The second row will eventually show the measurements, but this time in inches.

The box the monitor came in said it had an 18 inch screen.

Pause here to figure out where would 18 go in this ratio table? If a screen is said to be 18 inches, that describes the diagonal, and so the diagonal of the screen is 18 inches long.

Pause here to find the multiplier K that converts from the ratio unit measurement to the inch measurement.

To find my multiplier, I do 18 divided by five, which is 3.

6.

And so my multiplier to go from the units measurement to the inches measurement will be a multiply by 3.

6.

So, consequently, pause here to calculate the width and the height of this screen.

And the answers are four times 3.

6 is 14.

4 inches, whilst three times 3.

6 is 10.

8 inches.

Okay, onto the final practise task.

For question one, pause here to complete each sentence by choosing the appropriate word below for each of these three 16 to nine aspect ratio widescreen TVs.

And for parts B and C of question one, complete this ratio table, and using rules of similar shapes, calculate the width of screens A and C.

Pause to do this.

And for question two, and ultra widescreen TV has an aspect ratio of 22 to nine, the TV is advertised as 35 inches with a two inch tall stand.

Pause here to figure out whether this TV with stand on will fit inside Izzy's TV cabinet.

And finally, question three, which of these three TV's, each advertised as having a screen which is 28 inches, has the largest screen area? Pause now to find the areas of all three TV screens.

And here are the answers.

Each of the TV's is similar to the others.

This is because their width and height are in proportion to each other.

The scale factor from screen A to B is a multiply by 1.

5.

Using the ratio table, the width of screen B is 10.

459.

To find the width of screen A, I would multiply 10.

459 by the multiplier two thirds or divide it by the multiplier 1.

5, to get 6.

97 inches.

Similarly, to find the width of screen C, I would multiply 10.

459 by the multiplier, five over two, or 2.

5 to get 26.

15 inches.

For question two, Izzy's cabinet is wide enough, but with the stand on, it is not tall enough to fit the TV.

And for question three, the retro TV has the largest area, even though they all have the same diagonal length.

Thank you all for your hard work in a really challenging problem solving lesson where we have decided whether using Pythagoras theorem is necessary or not across a range of mathematical questions on topics such as perimeter, area and similarity.

We've also used Pythagoras theorem to find the shortest distance between two points on a coordinate grid and to find the perimeter of a polygon on a coordinate grid.

And finally, we've used Pythagoras Theorem with the ratios to consider the dimensions of TV and phone screens given a diagonal length.

You deserve a break after all of that fun maths challenge.

And so, until our next maths encounter together, stay safe and goodbye.