Lesson video

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

It's a pleasure to see you joining me in a lesson where we will be combining our knowledge of Pythagoras's theorem similarity and ratios to solve a range of geometric problems. Pythagoras's 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.

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

Pythagoras's theorem is incredibly useful in calculating the shortest distance between two points on a 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-angled 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's 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 are back to a familiar site.

Using Pythagoras's 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-angled triangle.

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

And so using Pythagoras's 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-angled 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 hypotheses and one with Y as its hypothesis.

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

So to find the length of X, we use Pythagoras's theorem on triangle A to give X squared, the hypothenuse 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's theorem on this right-angled triangle, giving Y squared equals five squared plus three squared, where Y squared is the hypothenuse 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-angled 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 hypothenuse 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's 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's theorem possibly multiple times for each shape, calculate each shapes 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's 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 as 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's 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's 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's 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 length 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 screen 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's 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 3-4-5 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 this screen is 18 inches long.

Pause here, define 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, an ultra widescreen TV has an aspect ratio of 22 to nine.

The TV is advertised as 35 inches with a two-inch tall stand.

Pauses 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 TVs 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 TVs 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 2/3 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's theorem to find the shortest distance between two points on a coordinate grid and to find the perimeter of a polygon on a coordinate grids.

And finally, we've used Pythagoras's 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.