# Lesson video

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- Hi, there.

And welcome to another lesson with me, Dr.

In today's lesson, we will be looking at manipulating surds.

Don't worry if you don't know what that means, you will by the end of today's lesson.

You will need a pen and paper.

Please take a moment to grab these and clear away any distractions, and find a quiet place where you won't be disturbed during this lesson.

Okay.

The try this for today's lesson is the following.

Decide if the following equations are true or false.

You have seven equations, I want you to read each one and I want you to think about them carefully, and then decide if each one is true or false.

If not, I will give you hints.

So if you're ready to pause, you can pause in three, two, one.

Okay.

And for a little bit of a hint, look at each equation and try and work out the value for that surd.

So square root of nine is? Good job, it's three.

So you can write down three.

Plus square root of four.

What's the square root of four? Good job.

Is that equal to roughly square root of 13? Remember we looked at estimating the value of surd in lesson two.

So you may need to just look back at your previous notes from lesson two.

With this hint, you should be able to make a start.

So pause the video, have a go.

Pause in three, two, one, go.

Welcome back.

It's time now for us to go through the solutions to try this, you can mark and correct your work as we go along.

So let's have a look at the first one.

Square root of nine plus square root of four equals square root of 13.

Well, I know that the square root of nine is three, and the square root of four is two.

If I add them up, that gives me five, and five is the square root of 25, so we know it cannot be square root of 13.

Therefore the first statement is false.

The second one, square root root of nine minus the square root of four equal root five.

What did you write down? Good job.

So, square root of nine is three, we're subtracting square root of four, subtracting two, that gives us one.

And we know that one is definitely not square root of five.

'Cause square root of four is two, so square root of five is more than two.

Therefore the statement is false.

Let's look at the next one.

Square root of nine is again, three, multiplied this time by the square root of four, which is two, and three times two is? Good job, six.

And if I want to write six as a square root of something square root of 36 is six, therefore this one is true.

Really good.

Next one.

Square root of four multiplied by square root of four equals four.

This is one of my favourite questions, okay? Square root of four is? Two, good.

And square root of four is two, multiply them that gives us four.

Next one, square root of nine plus square root of nine equal.

And we read this as two lots of square root nine, or two root nine but it means two lots of square root nine.

So square root of nine is three.

Does that give us two lots of three? Two lots of square root.

Yes, it does.

Square root nine gives us two lots of three therefore this is true.

Next one, two root nine equal to root 36.

Well, what does two root nine mean? We've just done that, it means two lots of three because root nine is three and that gives us six.

Is that the same as root 36? Yes, it is.

Therefore, this is true.

And the last one, most likely the trickiest, we have square root of nine out of four, and we can see that the square root is so big that we're square rooting everything.

The fraction is underneath the square root completely.

So we are not just squaring the nine, we are not square rooting, sorry, the nine.

We're not just square rooting the four, we're square rooting the whole fraction.

So now I can rewrite this as a square root of nine out of square root of four, because I'm square rooting the numerator and I'm square rooting the denominator.

What is the square root of nine? It's three.

What's the square root of four? Is two.

So I can write this as three out of two rather than three out of four, and therefore the equation was false.

Okay.

You've done an amazing job so far.

Let's move on to the connect task.

In today's lesson, we said we will be looking at manipulating surds.

So let's have a look at this task.

If a does not equal b, decide if the following equations are always, sometimes or never true.

The first one, the square root of a plus the square root of b equal the square root of the sum of a plus b.

In order for us to decide whether this equation is true or false, we need to substitute some numbers instead of a and b.

So this is what we are going to do.

Instead of a, we're going to have the square root of four plus the square root of 16 and examine if this is equal to the square root of 20, because the 20 is the sum of four and 16.

Now what's the square root of four? Good.

And what's the square root of 16? Really good.

Is that equal to the square root of 20? Do we know what the square root of 20 is? Really good thinking.

So we know that the square root of 25 is five, so the square root of 20 is definitely less than five.

Okay? We also know that the square root of 16 is four, so square root of 20 is somewhere between four and five.

So maybe roughly 4.

5.

Square root of four is two plus four, definitely does not equal 4.

5.

Okay? It's six.

It's definitely greater than square root of 20.

So we know that this equation is not necessarily true.

Let's try it with another number.

Let's try it.

I know that the question said when a does not equal to b, but let's try it with when a and b are the same to see does it actually work? Let's look at square root of one plus the square root of one and see does that equal square root of two? Well, what's the square root of one? Really good, it's one.

So one plus one, well, one plus one equal to two, it definitely does not equal square root of two.

So we know that this first equation is false, does not work.

So the square root of a number plus the square root of a number, does not equal the square root of the sum of the two numbers.

Moving on to the second equation.

We have square root of a multiplied by square root of b equal to, the square root of the product, ab.

So I'm going to start with square root of nine, multiply that by square root of four.

According to the equation, the answer should be the square root of 36 because square nine multiplied by four is 36.

So we're going to test, is this the case? The square root of nine is three, multiplied by square root of four is two, and what does that equal to? Excellent.

It gives us six and the square root of 36 is six.

So with the root nine and four this equation has worked.

Let's try another one.

Square root of 10 multiplied by square.

Sorry, square root of 100 multiplied by the square root of 25 equal to the square root of 2,500.

Is it still working with really, really big numbers? What's the square root of 100? Really good.

What's the square root of 25? Excellent.

And 10 times five is 50, the square root of 2,500 is also 50.

So we have seen now that it works with these two numbers, let's try another one.

And I am trying this one where a and b are the same number.

I know that we said we are not going to look at it, but just to see, does it work all the time? So square root of eight multiplied by the square root of eight, is it equal to the square root of 64? Let's have a look.

Well, I can write this down as the square root of eight, all squared, because if I multiply a number by itself, that's like me saying I'm squaring it.

So eight times eight is eight squared, so square root of eight multiplied by the square root of eight is the same as square root of eight all squared.

Okay? And if I want to do that, that is equal to eight because I am square rooting and then squaring.

So I'm doing the inverse operation.

So that's eight, the square root of 64 is also eight.

So then it also work even if a and b are the same.

So the square root of a multiplied by the square root of b is equal to the square root of the product, ab.

All the time, every time.

So now I can confidently say that the square root of five multiplied by the square root of eight is equal to the square root of 40.

Okay, let's have a look at the next one.

Square root of a minus the square root of b, is that equal to the square root of the difference between a and b, so a minus b.

I'm just using some numbers.

So square root of, instead of a, I'm going to use 16.

And instead of b, I'm going to use nine.

Is the answer square root of seven? Because 16 minus nine is seven.

Well, what's the square root of 16? Really good.

Now what's the square root of nine? Excellent.

So this is four minus three.

Well, we know what four minus three is, it's one.

Is it equal to square root of seven? We definitely know that one is not the same as square root of seven, one is the same as square root of one.

We also know that the square root of four is two, we know that the square root of nine is three, so we know that the square root of seven is somewhere in between.

So two point something.

So we know that this here does not equal the square root of seven, and we know that this doesn't work.

Let's try it with a different number and see if it works.

Let's go for a slightly bigger number, shall we? So square root of 169 minus the square root of 144, is that equal to the square root of 25? Well, the square root of 169 is? Really good.

13 minus 12.

And it definitely does not equal to five.

So 13 minus 12 equals five, it does not equal five, so this equation is not true.

Okay? It's never true, it does not work.

The square root of a number minus the square root of another number does not equal the square root of that difference between those two numbers.

Let's have a look at the next one, the square root of a all squared is equal to a.

We've touched on that earlier on today in that previous example when we looked at square root of eight.

So let's use a number.

I'm going to go with 16.

So the square root of 16 all squared, is that equal to 16? Well, what's the square root of 16? Really good, four.

And four squared is equal to 16.

It does work.

It works with any other number.

We worked it out with eight earlier on, and basically what's happening here, we are square rooting something and then squaring it.

So we're doing the inverse operation which gets us back to the same number.

It's like saying I add two from.

I have a number but I add two and then I subtract two.

I end up with that original number, and this is what I've done here.

I square rooted the number and then I squared it, so it gets me back to the original number.

Now let's look at the last one where we have division.

So it's written as a fraction and the fraction line, that vinculum line represents division.

Square root of a out of square root of b, is that equal to the square root of the whole thing, a out of b? We had a look at something similar to this in the try this as well, but let's generalise it more, okay? So if I say, what's the square root of 100 divide by the square root of four? Is it the same as the square root of 100 out of four? What's the square root of 100? Good.

And the square root of four? Really good.

So that's 10 out of two.

Now I'm square rooting 100 out of four.

What is 100 divide by four? Really good.

So I can now write this as a square root of 25.

Okay.

Now what is 10 divide by two, 10 out of two? That's five, really good.

And what's the square root of 25? Excellent.

So now I can say that this is equal to this, so this does work.

And it works if you try it with any other number, it's always true.

So let's summarise from this task.

If we have the square root of a plus the square root of b, the answer is not the square root of the sum of a and b.

Similarly, when it comes to subtraction, the square root of a minus the square root of b, does not equal the square root of that difference between a and b.

So it does not equal a minus square root of a minus b.

If we are multiplying the square root of a multiplied by the square root of b is equal to the square root of the product of a and b, so product of ab, and if we are dividing the square root of a divide by the square root of b is the same as the square root of a divide by b.

So it works for multiplication and division, not for addition and subtraction.

Square rooting and squaring on the inverse operation of one another.

So if we apply one and then we apply the second one, we are doing the inverse, which gets us back to the original number that we started with.

It is time for you now to have a go at the independent task, please post the video to complete it.

When you're done, please resume the video so we can mark and correct the work together.

Off you go.

Welcome back.

How did you find it? Really good.

Well done for all the resilience that you are showing by trying to do this all on your own, you should be really, really proud of yourself.

Now let's mark and correct these together.

Decide if the following equations are true or false.

The first one, square root of three plus the square root of three equal square root of six.

What did you write down? Really good.

Okay, the square root of three plus the square root of three is two lots of square root of three, it does not equal square root of six.

Therefore, this one is false.

The second one square root of 15 plus square root of eight equal to 23.

Okay, excellent.

It is also false.

What about next one? Square root of 18 minus the square root of eight, is it square root of 10? Really good.

It is not.

Next one, square root of 15 plus square root of 15 equals two root 15.

What did you write down? Excellent.

It is true.

Next one, root eight multiplied by square root of eight equals eight? Good job, because square root of eight times square root of eight is square root of 64, and square root of 64 is equal to eight.

Really good.

Square root of five multiplied by the square root of three is equal to the square root of 15.

That's also correct, and it is correct because we know that if I multiply the square root of a by the square root of b, the answer is the square root of that product, ab.

Next one, square root of 15 multiplied by square root of 10 equals square root of 25.

Okay.

So this is just goes against what we have just said.

So it's falls the square root of 15 multiplied by the square root of 10 should be square root of 250, but it's an easy mistake to get into.

So just make sure that you're reading the questions carefully and double checking that if I multiply what happens.

Next one.

Two root eight equal to root 16.

That is not correct because two square root of eight means root eight plus root eight, and that is two square root of eight.

Whereas the square root of 16 is just four.

So it's not correct.

Just because we've got the number outside it doesn't mean we'll multiply the number and the number that is under the square root sign, so we have to be really careful with this one.

And next one, square root of 15 divide by square root of three is equal to the square root of five, that is correct because we said that square root of a divide by square root of b is equal to the square root of a out of b.

And the last one, square root of the fraction, 36 out of 64 is equal to 3/4.

Well, good job.

It is correct.

The square root of 36 out of the square root of 64 can be written as square root of 36 on its own, square root of 64 in the denominator.

We know that the square root of 36 is six, the square root of 64 is eight, and six out of eight can be simplified to 3/4 as a fraction.

You should be super proud of yourself.

Well done, for getting this done, and let's move on to the explore task.

You should be super proud of yourself.

Well done, for getting this right.

Let's move on to the explore task.

For our explore task today, I thought let's mix it up between surds and some shapes because we've been working for about four lessons now on surds, so let's just apply it to a different topic.

So let's read the question together.

An equilateral triangle of side length four centimetres and a circle has been used to create the two images below.

What are the areas of each circle? And you've been given the two diagrams with very, very little information there.

And as a challenge, can you construct similar images? So, if you are feeling confident about this question and you think you can make a start on it, please pause the video in three, two, one.

If not, I will give you some support.

Okay, so for support, I'm going to ask you to look at that first diagram for me.

I'm going to show these three lines here.

Okay, I've drawn these three lines here and created three equal triangles.

Okay? You can start by finding the area of one of these triangles.

How do you calculate the area of a triangle? Really good.

It's base multiplied by the height divide by two.

You know what the base of the triangle is, it's been given to you, because the base is four.

You cannot see the perpendicular height, so you need to do something about it.

Where is that perpendicular height? Okay, so we can draw lines now from the centre of the circle to the base, and they meet at 90 degrees.

Now you do have the base and you have the height.

You should be able to write an equation to represent that.

Now with this hint, please pause the video and have a go at this.

Pause the video in three, two, one.

Welcome back.

Let's have a look at the way I answered this question.

Now there are more than one method to tackle this question.

I started by looking at the three equal triangles that I've created, and I said to myself, well, the side here, the base is four centimetres, let me find the area of one of those triangles.

So I'm looking at the area of this triangle here.

The area is base multiplied by height divided by two.

I know that the base is four.

The height is r, I called it r, you've called it anything else.

I don't know what the radius of that circle is.

So the area is 1/2 multiplied by four multiplied by r, which gives me 2r centimetre squared.

Half of four is two.

So that gives me the area of one of those triangles.

Now, I'm going to work out the height of the equilateral triangle.

Now the height of the equilateral triangle, let me use the highlighter here to show you where it is.

So I am looking at, I want to work at this height here of the equilateral triangle.

I want this.

Now, we have done Pythagoras before, so we should be able to use Pythagoras to find the length of the height.

Okay? Now, this line here crossed at the midpoint.

So this side must be two centimetres.

The whole thing, all of this is four.

All of this is four centimetres.

And we had a line coming down that equilateral triangle.

That is our line of symmetry really, isn't it? It cuts the triangle in half.

It gives us two centimetres on each end.

So that was the midpoint, right here is the midpoint.

So now I have two centimetres here.

I know that this side is four.

I should be able to calculate the height of that equilateral triangle.

So to find the height of the equilateral triangle, according to Pythagoras it's four squared minus two squared equals the height squared.

So if I substitute that, the height is the square root of 12.

Good job if you had this correctly.

Now, what do I need to do next? Really good.

Now, I can find the area of the equilateral triangle.

I know the base of the equilateral triangle, and I've just worked out the perpendicular height of it so I can work out the area of it.

So the area of the equilateral triangle is the base, which is four, multiplied by the height, square root of two, divide by two.

Now, four multiplied by square root of 12 divide by two is going to give me four lots of square root three centimetre squared.

Okay? Really good.

Now I know the area of the equilateral triangle, I also know the area of one triangle.

Okay.

The bigger triangle, the equilateral triangle is made out of three smaller triangle, isn't it? So if I have this here, I can divide it by three.

It should give me the area of one of the smaller triangles.

It should give me the area of this here.

So let's do that.

If I divide this by three, what would it give me? Four root three divide by three.

I can leave it as a fraction, I don't have to do anything fancy with it at the moment.

So now I have two things that represent the area of one of those triangles.

I have four root three out of three, and I have 2r.

Can they both represent the area of one of these triangles? So I can now write an equation and say, well, 2r is equal to four root three out of three.

Okay.

Now, if two lots of r is equal to this, r must be, half of that, right? So if I want to divide four root three out of three by two, I get two root three divide by three.

Now I know what the radius is.

So I know the radius of the area, I can calculate.

Sorry, I know what the radius of the circle is, so I can calculate the area of the circle.

The area of the circle is pi r squared.

So that's pi multiplied by that radius and I squared it altogether.

Okay now, what does this equal to? It's equal to pi four out of three, and I can leave it in pi format.

The question didn't ask me to write it to one decimal place or to significant figures, I can leave it in terms of pi.

And I can say that the area of this circle is 4/3 of pi centimetre squared.

Wasn't that easy question, so well done for tackling it.

You should be really, really proud.

Now the second one, I think is a lot easier.

We have been given an equilateral triangle.

It's the same equilateral triangle we know that this side here is equal to four, and we want to really know what's the radius of the circle or the diameter of the circle is, in order for us to calculate the area of that circle.

Now we've done this before in the previous question, we know that this line here, the hypotenuse of this triangle, we know that it's equal to square root of 12.

Really good.

So if we know that the diameter is equal to root 12, what is the radius? It's half of that.

So what's half of root 12? Square root of three.

Excellent.

And now to find the area of the circle, we need pi multiplied by r squared, pi multiplied by square root of three, all squared.

So the answer is three pi centimetre squared, and we are done.

Well done on all the fantastic learning that you have done in today's lesson and for all the resilience that you are showing in tackling those difficult questions, difficult and challenging questions.

So, well done.

Please do not forget to complete the exit quiz, enjoy the rest of your learning for today, and I will see you in the next lesson.

Bye.