Lesson video

Hello there and welcome to this lesson again with me, Dr.

Saada.

In today's lesson, we're going to look at finding right-angled triangles.

All you need for this lesson is a pen and paper.

So please grab these and make sure that you are ready to do some learning.

When you are, let's begin.

Let us start with this task.

Find out as much information as you can about this diagram.

Some examples have been given by students.

And I'll go through these with you.

The point C is 11, 2, has the coordinate of 11,2.

Second student says, the length of the line BD is three.

ABD is a right-angled triangle.

AD has a length of one.

Can you work out the length of AB? Pause the video and have a go.

If you finish and you want to challenge yourself, try and find out other information as much information as you can about this diagram.

Pause the video and have it go.

Okay, let's have a look at some solutions and answers.

We know that to BD is three units.

It starts on 2 on the Y-axis and ends up at 5, so that's three units.

AD is one unit, and I know that ABD is a right-angled triangle, so I can find the length of AB using Pythagoras.

Okay, Pythagoras says that, one squared plus three squared is equal to x squared.

That is 10 equals to x squared, and to find x, I square root it, and I can leave it in surd form because it's more accurate.

So x is equal to square root of 10.

Really good job, if you had this correct.

Did you get a chance to work anything else from this diagram? I wonder what you did.

For example, I looked at this and I said, one DC here is nine units.

It starts at two on the X-axis and finishes at 11 on the X-axis.

So that's knowing units, or if you just count the squares.

And now BDC is a right-angled triangle.

If I do nine squared plus B squared, that would give me C squared or this length of BC, but squared, and then I square root it.

And that would give me BC is equals to the square root of 90.

Really good job if you had this correct, well done.

And now we are going to look at a couple of examples.

Example one, the right-angled triangle PQR has short sides of length, nine centimetres and five centimetres calculate the perimeter of triangle PQR.

So to find out the perimeter we know that we have to add the three sides.

We have to go all the way around the triangle.

We know that the two sides, the two shorter sides, one of them is nine, and one of them is five centimetres.

We don't know what PR is, what the length of it.

So I called it m.

Its a right-angled triangle so I can use Pythagoras and say that m squared is equal to nine squared plus five squared.

What do I do next? Excellent, we can start the calculation with is nine squared and what is five squared? Really good, so m squared is equal to 81 plus 25, add them up for me, really good, m squared is equal to 106.

And how do I find m? Remember, I don't want m squared, I want to know m.

Really good, so m is equal to the square root of 106, and it is square root of 106 centimetre.

Now, I'm going to leave it as it is.

I'm not going to write it as decimal.

It's an irrational number.

It's not going to terminate as a decimal.

It's more precise to leave it in surd form because if I write it down as decimal now and round, it may affect my final answer.

So I'm rather keeping it as a square root or 106.

Now, I can say that the perimeter is equal to nine plus five, plus the square root of 106.

If you put this into your calculator, you get 24.

2956 and so on, really carries on because it's a non terminating decimals.

Now, the question didn't tell me what to do, in terms of rounding and accuracy.

I can see that I have whole numbers so I can round it to if I want a whole number or I can do it to one decimal place, whatever you decide to do, you have to let the person that is looking at your work, know what you're doing.

So I'm going to write down, this is equal to 24.

3 centimetres, and I'm going to say that I rounded it to one decimal place.

And now let's have a look at example number two, the hypotenuse of right-angled triangle ABC is 14 centimetre and another side length, 11 centimetre.

Calculate the area of triangle ABC.

And I have this triangle here sketched.

So I have ABC, the hypotenuse is the longest side is 14 centimetre, and one of the shorter sides is 11 centimetre.

I know it's a right-angled triangle because the question told me that.

Now I need to calculate the area of the triangle.

How do we calculate the area of a triangle? Do you remember? Good job, it's area equals half base times height.

Do I know that base and the height here? Well, the base and the height has to be a perpendicular height.

So I always like to think about them like this, if you can see my arms, yeah? So I always like to think about them like this.

They have to meet that right angle.

So two sides that meet that's right angle are AC and BC.

So I have one of them, I don't have the other one.

I don't have BC.

I can call this the height if I want, I can call it the base, it really doesn't matter, depending on how you're looking at your triangle, especially if you rotate it.

So let's call it height, I don't know what the height is, but I know it's right-angled triangle.

And I know that I can use Pythagoras to help me find this out.

So the height squared plus 11 squared equals 14 squared.

How do you find now height squared? Excellent, so h squared equals four squared minus 11 squared.

And now what is 14 squared? What is 11 squared? Really good, it's really important to know your square numbers.

So height squared is equal to 196 minus 121.

Go on, subtract them for me please.

Good job, h squared is equal to 75 and I want just height on, so hight is square root of 75 centimetres.

Really good job.

So I would leave it in surd form because it's an irrational number.

Let's go back to the area question.

So area is equal to half times base times height.

I know the base, I know the height.

So I can say it's half multiplied by the square root of 75, multiplied by 11.

Remember it doesn't matter which way we write them down because multiplication is what we need to do.

So it really doesn't matter if I say half times root 75 times 11 or half times 11 times root 75.

Now, put that into your calculator and you would get 47.

63139 and it keeps carrying on because it's a non terminating decimal.

So the question you didn't tell me about rounding at all, or any degree of accuracy.

So I'm going to say it's equal to 47.

6 centimetre squared.

remember for units, the unit for area.

And I'm going to say that I rounded this to one decimal place, so I'll write that down.

Okay, really good job.

And the last example for today's lesson, example three, find the area of the isosceles triangle below.

Now, what do we know about isosceles triangle? The two sides are equal and the two base angles are equal.

Okay, and we can see here that I have two sides are equal.

Each of them are 12 centimetre and the base is nine centimetres.

Now, I need to find the area.

And we just said that the area needs the base multiplied by the perpendicular height.

And I don't have the perpendicular height here.

So I need to draw that.

There we go.

Now, I'm going to give it a letter.

So I'm going to say it's r.

And it crosses that nine, or it meets the nine centimetre at right angle.

And when it does that bisects that side, and it gives us two equal parts.

Each of them being half of the total, so 4.

5.

So now you can see, I have two small right-angled triangles and I can use either of them, they have the same values to find the height.

So if I take one triangle and I say, r squared, plus 4.

5 squared is equal to 12 squared.

Now, I want to find the r squared.

I need to do something about this.

So r squared is equal to 12 squared, minus 4.

5 squared.

So I've rearranged, okay? What is 12 squared? Really good, 4.

5 squared.

I would be really surprised if you know, know the answer off by heart, but if you do, that's really good, if not, use a calculator, r squared equals 144 minus 20.

25, really good.

Now subtract r squared is equal to 223.

75.

And or therefore is the square root of 123.

75 centimetres.

And I'm leaving it in surd form because it is really precise, I can do that.

Now, let's go find the area.

Area is equal to half base times height.

So half multiplied by 4.

5 multiplied by the root of 123.

75.

And if you use your calculator for this, that gives you 25.

02969.

And it carries on 6 6 9.

So on being to do some rounding here, and I'm looking at my number and I'm thinking, if I round to one decimal places is going to be 0.

0.

What would be good here is for us to even revise significance fingers.

Now, how do I run this to significant figures to two significant figures, do you remember? Write it down if you do.

So you can check with your answer is correct or not.

Really good, it is 25 centimetres squared to two significant figures.

It really doesn't matter to what you rounded here, because the question didn't tell you, but you need to be sensible about what you do for rounding.

So don't do four decimal places, for example, it's not needed.

Okay, really good job, well done.

And now it's time for you to practise, applying what we've learned in today's lesson into the two questions shown here, independently.

I'm going to read the questions with you first, question one, calculate the perpendicular height of this equilateral triangle giving your answer correct to one decimal place.

And you've got the triangle there.

It's a good idea to sketch it down and do some annotation on it, okay? Question two, calculate the length of the side marked x in each of the following triangles, giving your answer correct to one decimal place.

So you can see where x is in the second diagram, okay? Pause the video and have it go.

Hello everyone, Ms. Jones here, and I'm going to go over the solution for this independent task.

Okay, so we are looking at this first triangle and we need to calculate the perpendicular height.

Now, I know that the perpendicular height is this length from this Vertex down to the base of the triangle.

And you know, once that's put in, I can see that we've got a right-angled triangle here, so I'm going to use Pythagoras theorem.

I need to know the length of this missing side.

And I know that's half of five centimetres because this is an isosceles triangle.

So this length will be 2.

5, so if I write my equation, I know that 2.

5 squared added to h squared is equal to five squared, with five being the hypotenuse here.

which is equal to 25.

I can simplify that further.

I know that 2.

5 squared is 6.

5.

So 6.

5 added to h squared will get me 25, which means that h squared.

If I take off that 6.

25 is 18.

75, which means eight is the square root of that, which got me 4.

33 centimetres rounded to one decimal place, okay? If you need to correct any of yours, do it now.

If not, let's go on to the next one.

Okay, this time we are looking at the side marked x.

We don't know the length of this side.

I'm going to call it a, but we do know that it's the same length as this side, okay? We've got those markings, they're indicating that this is an isosceles triangle.

Okay, again, so we can use Pythagoras here with 13 being our hypotenuse.

So I know that 12 squared added to a squared is equal to 13 squared, which means that 144 added to a squared is equal to 169.

So that must mean a squared is equal to 25, which means a is equal to five.

So what I've done now is find out that this length and this length is equal to five.

From there, I can use my Pythagoras theorem to work out what the hypotenuse is, which is x.

So five squared added to five squared will get me x squared.

And I know that five squared is 25.

So two lots of that there, will get me 50, 50 is equal to x squared.

So x must be the root of 50, which I've worked out as 7.

1 centimetres, when rounded to one decimal place.

Okay, I hand you back over to your teacher now, who's going to explain your next task.

And this brings us to our explore task.

It's such a lovely question.

I absolutely love this question, let's read it together.

This is a sequence of equilateral triangles, which having increasing side lengths, what are the heights of these triangles? How would the pattern continue? Okay, I'm going to give you a little hint just to get you started.

But if you're feeling super confident, please pause the video and have it go.

If not, hold on.

So we have equilateral triangles, equilateral triangles have equal side lengths.

So if you look at the first one here, this side and this side should also be two centimetres.

Now, we're looking at the heights.

So we want to look at the pattern in the heights of those triangles.

If I want to find the height of this first triangle, I can completely ignore the other ones and draw the height here, which bisects the base of this triangle into one on each side.

Now, I can use Pythagoras to find the height of this triangle.

What do you think I'm going to do with the next one? Yeah, you're right.

I'm going to do exactly the same thing.

Now, the base of the second triangle is four centimetres.

What would be the other sides? Good job, so this would be four centimetre and this would be four centimetres.

I would do exactly the same thing through the height bisect it.

So I have two on each, two centimetres on each side of the height.

And now I can use Pythagoras to find that the height of the second triangle.

I can do the same with the third one and with the fourth one, and the pattern continues.

And that's why I've hidden part of that pattern, the rest of the pattern you cannot see it.

So I want you to pause the video, sketch this diagrams in your books and have a go at this.

Okay, let's go through the answers to the explore task.

So if you've done the first try and get correctly, if you found the height using Pythagoras, you will have a height of square root of three.

I'm leaving everything in surd form.

For the second one, what did you get? Good job, square root of 12.

And for the next one.

Excellent job, square root of 27.

And next one, really good, square root of 48.

Now, just by looking at these, I can not see any specific pattern.

I can see that there are increasing, but it's not as easy for me to spot what is happening, okay? So what you can do, you can rewrite the surds.

Now, if you have not done surds yet, then that's fine.

That's one thing for you to look at, or you can for now just grab a calculator and just write these on your calculator and see what it displays you, some of the calculator should be able to display something different.

So if you type into your calculator square root 12, it should give you two, lots of square root three, so two root three.

Square root of 27 should give you, three route three, and square root of 48 should give you, four route three.

So now, can you see the pattern? We started with square root of three, then two lots of route three, then three lots of root three, then four lots of root three.

What would the next one be? Good job, five root three, well done.

And this is it from me for today.

I hope you enjoyed this lesson and found it helpful.

Can you please make sure that you complete the exit quiz for this lesson.

I would love to see your work.

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

Have a lovely day and enjoy the rest of your learning, bye.