Lesson video

Hi, there.

Welcome to today's lesson, with me, Dr.

Saada.

Where, we are going to look at the Pythagoras theorem on the Cartesian plane.

For today's lesson, you need a pen and paper.

Please make sure that you are set somewhere quiet so you can concentrate in today's lesson.

When you're ready.

Let's begin.

I would like you to try this question.

Use Pythagoras's theorem to find the lengths of the of the lines and sort them from the shortest to the longest.

Now, my little hint for you that you've done this in lesson three, when you looked at tilted squares.

So, if you still have your notes and you're not too sure you can refer back to them.

Pause the video and have a go.

Okay.

If you need a bit of help, I'm going to give you a hint here.

Let's look at one of these lines.

I'm going to select pink one.

Let's look at the pink one.

I can make a right angle triangle with side lengths two, and two for the shorter sides.

I'm going to take that, I will just draw it here in the white space so can see it clearly.

So, I have this line and I have two by two.

Now, remember what Pythagoras theorem says, I can, look for the square of the two shorter sides and the square of two is four, the square of two is four.

If sum them up, they should give me the square of the longest side.

So, four plus four is eight, therefore, the length of the pink line here is the square root of eight.

That should give you a hint.

Now, pause the video and have a go.

Okay.

Let's mark your work.

Okay.

And now, let's mark your answers.

Okay.

The first one, you should have had square root of two for the green one.

What did you get for, the next one? Square root of five.

Excellent.

The pink one with already done together.

So that's this collaborative eight.

We worked it out.

What about the next one? That purple line.

What did you get for it? Good job.

Square root of 13.

What about the orange one? Excellent, Square root 17.

And blue one? Good job.

Square root of 10.

Really, good.

Well done.

We are going to look at example number one for today's lesson.

Find the distance between the points, two, one and four, three.

We can see here, that I already have my X and Y axes and I have the line drawn there.

So, I have the point two, one and I have the point four, three.

And I connected them using a line just to show you, how we're going to use this to find the distance between the two points.

So first, I'm going to look at here and see, what do I have? I have two, look at the other shorter side, and I have two places across the Y axes.

So I went to through the X axis two places and up the Y axes two places.

And I can see that I have, what looks like a right angled triangle.

So I can use Pythagoras.

What does Pythagoras tell me? Good job.

So two squared plus two squared equals a squared.

And I'm goin to call this side like of length, the letter a, just like I can work it out.

I could have called it any letter.

Two squared plus two squared is? Excellent.

Four plus four is equal to a squared.

And therefore, I know that a squared is equal to eight, But I just want the length.

I just want a, so what do I need to do? Good job.

Square root of eight.

That gives me a, So, I know now that the distance between the two points, two points here, is a is equal to square root of eight.

Let's look at example number two.

Find the distance between the points, negative three, one and four, negative three.

So we have to be, we have to be really careful because we've got the negative signs this time.

Okay.

So, I have here labelled the two points.

This one, the first one at negative three, one and the second one at four negative three.

Okay.

Now I want to find the distance between them.

I can just to make things a little easier for me.

I can connect them, and say, this is the distance between the two points.

This is what I really want to work out.

Okay.

You don't have to, but you can.

Now, in order to work this here, I, am going to count across the X axis and see what is the distance from one point to the other along the X axis? And that is seven units.

Because, I can count the number of squares or because I can see that from four to go to zero, I need four places.

And then too negative three, that's another three places.

So seven places all together.

What about across the Y axis? Okay.

So I went up one, two, three, four places.

And I can also think about it, that I'm going from negative three to zero that's three places.

And then, one up that's four places altogether.

So, I didn't have to really draw trying at this time, now I know that there is what looks like a right right angle triangle there.

With an eight and seven.

Because if you look here, this is their X coordinate of the first point negative three.

To get from negative three to four, without having to draw the X and Y axes without having to draw the line.

To get from negative three to four, negative three to zero, that's three places, Zero to four, that's four places.

Think about the number line.

If you are at negative three and you want to get two, four, always use zero, as your reference point.

From negative three to zero, you have three places.

And from zero to four, you have four places.

So, altogether the difference in the X coordinate is seven.

And now, let's look at the Y coordinate.

The Y coordinate we are moving from one to negative three.

Okay.

So again, on your number line, you sketch it.

Imagine you have negative three, and you have one, think about, your reference point as zero.

Negative three to get to zero that's three places.

From zero to one that's one.

So, overall we have done four.

So the difference in the Y coordinate is four.

Which is what we have seen in the diagram.

When, when we counted the squares or the gaps.

Okay, so now, Pythagoras theorem tells us that, four squared plus the seven squared, the two shorter sides is equal to, I called it L, so L squared.

What is four squared? What is four squared? 16 plus 49 is equal to L squared.

Now, at the math, 65 equal L squared.

And therefore, L is square root of 65.

So, the distance between the two points, is the square root of 65.

I'm leaving it in the surds form just because it's more precise.

The question says, give you answer in a specific format, or if they ask you to simplify or to round it, you will have to do that.

Okay, I would like you now to mark and correct your work.

Okay, for the first one, the difference in X is three, the difference in Y is four.

So I have three squared, squared plus four squared is equals to d squared.

I called it d.

Now, I know that nine plus 16 is equal to d squared.

If you added them up correctly, you should have 25 is equal to d squared.

And therefore, the distance d is equal to the square root of 25, which is five units.

Good job.

Next one.

The difference in the X, X coordinate, one to three, the difference is two.

And the difference in the Y coordinate is four.

So we have two squared, plus four squared is equal to d squared.

What did you get next? Great, well done.

You get four plus 16 equals to d squared.

And therefore, d squared is equal to? Really good.

20.

And then, four d is equal to square root of 20.

So d is equal to the square root of 20 units and I'm leaving it in surd form.

Next one.

The difference in the X coordinate was one, and the difference in the Y coordinate is four.

So, I wrote down one squared plus four squared equals to d squared.

This gives me, one plus 16 is equal to d squared.

And therefore, d squared is equal to 17.

So, d is square root of 17 units.

Good job.

If you have question one correct.

Now, what can you see? What is similar between the three questions or the three parts of question one.

In each of them, I use the three, seven.

The first one was zero, three, then one, three, then two, three.

What happened to the distance? The first one, the distance was square root of 25.

It was quite big.

The first point was at zero, three.

Was at the origin of the graph.

It was quite far, from three, seven.

Then, I moved on to one, three.

So I moved closer to three seven, and then two, three, even closer.

You can see that the distance becomes a smaller.

Started with the square root of 25 and square root of 20, then the square root of 17.

Good job.

Okay.

And, question number two.

The distance between the points seven, zero and 15, C.

So, this time for the second point, I didn't tell you the coordinates.

I told you the X coordinate.

I did not tell you the Y coordinate.

Instead, I wrote down C is 10 units.

So, the distance that we have been calculating in the previous questions is given this time, as 10 units.

Find the possible values of C? Let's write down what we know.

Eight squared plus something squared is equal to 10 squared.

I know that the difference in the X coordinate, from seven to 15, it's eight.

So eight squared, plus something squared.

I don't know what that something is because I don't know what the difference is in Y, and I cannot even calculate it.

Because I know, one point is at zero and the other, is at something else.

I don't know what it is.

So, I'm going to just leave it as a box there, is equal to 10 squared.

Because, I know that the distance, this time, that slanting side, this time, is given to me, as 10 units.

So equal to 10 squared.

Now let's do some rearranging here.

I can say now that somethings squared is equal to a 100 minus 64.

100 from the 10 squared and the 64 from that eight squared.

Let's work this out, now.

Something squared is equal to 36.

We're getting somewhere now, right? It looks like something that we can work out.

What do you think that something is? Really good.

We square root it, that something is equal to six units.

So I now know what do I know? I know that, the difference between the Y coordinates of the two points is six units.

One of them is zero.

Now, if one of the points is zero, and the difference is six units from it.

Where would the second coordinate be? So where would it be? Let me just show you.

So, if I have one point here and that is at seven, zero.

And I tell you that my next point is six units away from the Y axis.

Okay.

So if the Y coordinate here is at zero, then, the next point surely can't be at 15.

Good job.

Six.

Okay.

So it can be C is six.

The questions said find the possible values.

So, it tells me that there is more than one value.

What else will I have it? Like, I can not have anywhere there, because if I have any 15, one, 15, two, it's not going to work.

It's not going to give me a difference in Y of six.

What would work.

If I have something, an X coordinate and that I have? Excellent,.

If I have negative six.

Because the difference from zero to negative six will still be six.

So, the answer is C is equal to six or C is equal to negative six.

A huge well done, If you had this answer correct.

If not, please make sure that you're doing your corrections, so you can learn, how to do this properly and independently for next time.

And there we are with a really, really interesting explore task.

Let's read it together.

Use the X and Y axes to estimate the radius of these, of these circles.

Can you accurately find the radius of these circles from the marked coordinates? The marked coordinates are, the ones marked with little circles.

You've got blue, green, purple, and, and blue one.

Okay.

Now I would like you, if you're really confident to pause the video and have a little think about this.

If not, don't worry.

I'm going to give you some hints.

And there we go.

This is the first hint.

Okay.

So, I drew a line here from the centre of the circle to, to the green point, the green coordinate.

Just to show you that this is the radius.

Can you work out the length of this green side? Of the, the same with the blue, the yellow and the purple.

So, can you, knowing the coordinates because you can read them of, can you find the lengths of each of those lines here? Okay.

And these lines represent the radius of each circle that is connected to.

Have a little think, if you can.

If you think that this is helpful, please make a start by pausing the video now.

If not, I'm going to give you a second hint.

And there we go with your next hint.

Okay.

So, I have the lines connected from the centre, of the grid.

So from the origin to, to the point, or to the coordinates that we had labelled.

And now, I drew some right angled triangles that, now, using those right angled triangles, and what you know about by Pythagoras theorem, can you find the lengths of the yellow, the purple, the green, and the blue, lines here? Using the coordinates and using Pythagoras can you find the lengths of each? From the lengths, you will you be able to estimate the radius of each circle, Pause the video and have it go.

Okay Let's go through the solutions to this really, really interesting question together.

If I look at, in the first one here, it is the square root of 13.

I know this because if I look at the two shorter sides, one of them is three squared, that's nine, plus the other one is two squared, so four.

Nine plus four is 13.

Therefore, the length is the squared of 13.

Okay.

Now the square root of 13 is between the square root of nine, and the square root of 16.

The value of the square root of nine is three and square root of four, 16 is four.

So, square root of 13 is really, between three and four.

So if this side here, is the blue line, if it's a square root of 13, this tells me that, the radius of this circle here, so the second circle that we have, that going round.

The radius of that is somewhere between three and four.

Is this correct? Well, if you look here, from the origin using the X axis, we can see that the radius is actually somewhere between three and four.

So this tells me, that using Pythagoras and using the length of the, of the line, I managed to, tell what? The radius of that circle once.

Let's have a look at the next one.

So we're going to look at this yellow one.

The length for this one is the square root of 25, which is five.

The reason for this, if, again three squared is nine, and four squared, four squared 16, so it add up to 25.

So the side length is the square root of 25.

Now, again, if you check using, from the origin, you will see that, that circle has a radius of five.

Goes all the way up to exactly five.

Can you see that? Okay, let's look at the next one.

So we'll look at that green triangle now.

The green side has a square root or a length of square root two.

Because I have one squared plus one squared which is two.

Square root, squared of two.

And, I know, that the square root of two is roughly 1.

4 something.

And if you look at that line there, it gives you, it shows you, that the radius actually goes to between for the smaller one, let me use this, so here it goes up to somewhere in between one and negative two.

So it's just about negative 1.

4.

But the length is obviously not negative positive.

Okay.

The purple one, has, has already appeared.

So, if you use Pythagoras, this gives me a square root of 37 for the length here.

And the square root of 36, we know is six.

So the square root of 37 is going to be really close approximately 6.

1.

If you look at the radius, there, it goes just over six.

So, roughly 6.

1.

Really, good job if you had this as correct.

Well done.

It was not an easy question.

It was quite challenging, but some really, good mathematical thinking happening here.

And this brings us to the end of today's lesson.

Huge well done, for all the efforts and the mathematical thinking and learning that you have done during this lesson.

You should be so, proud of yourself.

I would like you now to do the exit quiz.

I would love to see your work.

So, if you would like to share your work with us at Oak National, please ask your parents or carer to share your work on Twitter, tagging @OakNational and #LearnwithOak.

And that's it, from me today.

Enjoy the rest of your day.

Bye.