Lesson video

Hello, I'm Mrs. Lashley.

I'm looking forward to guiding you through your learning today.

Okay.

So in our lesson we're gonna use the knowledge of transformations to solve problems. So you've been working on transformations and we're now gonna try and apply them to problems. Words that you've met are transformation, which is the process that changes the size, the orientation or its position.

And the object is the starting figure before a transformation has been applied.

And once the transformation has been applied, then we call it an image, so the object and the image.

There's three parts to this lesson.

One is to do with inverse problems, one's to do with perimeter and area and the last one is coordinate problems. So we're gonna make a start on the first one to do with inverse problems. So you've worked with inverse many times before.

So for example, when you want to undo adding a number, we'd use its inverse operation and subtract the number and this would return you to the start.

So that's an example of where you've used inverse before.

So for example, if you did 125 plus 32, then the result is 157.

But if you didn't want to have added the 32, if you wanted to undo that adding, then the correct thing to do would be to subtract 32 back and you'd get back to the start number, the original number.

And so we're gonna use this premise for this part of the lesson, trying to undo what you just did.

So if the framed photo is moved to the right, how would you get it back to the start? Well, you would move at the same distance but to the left because the right and the left are opposite or inverse directions.

If the arrow is rotated 90 degrees clockwise, how could you return it back to its original orientation? You could continue another 270 degrees clockwise.

So you could continue in the clockwise direction, but now at 270 degrees to get your full turn.

Or you could do anticlockwise 90 degrees.

And the word anti, to understand that that means the opposite of clockwise.

If a shape had been enlarged by a scale factor of three from the coordinate two, three, what transformation could return the shape back to its original position? So because the lengths here have changed, it's been enlarged, then we need to do another enlargement to get back.

No other transformation will change the lengths.

But what would the scale factor be? Well, the scale factor would have to be a third because all of those lengths were originally multiplied by three and we need it to therefore divide by three and multiplying by a third is the same as dividing by three.

So our scale factor would have to be one third, but still from the same centre of enlargement.

So here we've got a problem.

Rectangle A is reflected in the x-axis to become rectangle B.

You can see that in the diagram.

Then rectangle B is then reflected in the y-axis to become rectangle C.

Again, hopefully you can follow that on the diagram.

Laura wants to know, is there a single transformation that maps rectangle C back onto rectangle A? So we did two transformations to get from A to C, but Laura wants to know is there one, is there a single transformation that could get her back from C to A? So she could translate rectangle C because the orientation looks to be the same.

She could translate rectangle C by 6, 4, 6 to the right four up would get her back to where she started.

She has also realised that you could rotate it 180 degrees about the origin and that would work as well.

So here you've got a check.

So triangle A has been rotated 90 degrees anticlockwise about the origin and the image is labelled B.

So A is your object and B is the image after a rotation.

So can you complete the two transformations that would map triangle B back onto triangle A? So fill in the blanks.

So the two transformations that would map triangle B back to triangle A.

So if we think about triangle B as the object this time and triangle A as the image is if we went 90 degrees anticlockwise about the origin, then we could go 90 degrees clockwise about the origin, that would get you back, undo.

Or because of the shape here, you could reflect it in the y-axis, they're both equidistant from the y-axis.

That y-axis would work as a line of reflection.

So you've got some practise, you are going to answer these three questions.

So pause the video and have a go at these three questions.

Read them carefully and then when you're done, come on back.

Question four, you need to fully describe a single transformation that would get C, shape C, back onto shape A.

So shape A is reflected in the x-axis and then shape B is reflected in the y-axis.

You need to come up with a single transformation that would get C back to A.

Pause the video and when you're ready come back for the answers.

So question one, you needed to undo a translation by three, two.

And so the vector that would get you back, well the original vector was three to the right.

So we need to go three to the left because that's the inverse direction.

And it was two up, so you need to go two down.

So negative three, negative two is the answer for question one.

Question two, shape P is reflected in the x-axis to create shape Q, what transformation would map shape Q back to shape P? Well you could just reflect it back in the x-axis.

Depending on the shape, it could be a translation or a rotation as well.

So depending on what shape the object was, there's actually a few answers.

But some objects would only have the one answer of a reflection back.

Question three, shape R is said to be enlarged by a scale factor of four from the origin.

So what in enlargement scale factor would you need to map back to shape R? So once again, if the lengths have changed because of an enlargement, a further enlargement would have to take place in order to change the lengths.

We don't wanna multiply it by a number greater than one because then the lengths just get longer.

So it needed to be multiplied by a fractional amount and that fraction is one quarter so that it would go back to its original.

And then question four, the single transformation that would get C back onto A is a rotation of 180 degrees about the origin.

You can't translate this shape because it's orientation has changed.

So you need to rotate it.

So we're now gonna move on to the second part of the lesson, which is still thinking about transformations but linked perimeter and area.

So the perimeter is the distance around a 2D shape.

You'll have been familiar with that I'm sure.

So here's an example, just in case.

So if we've got a rectangle whose width is 3.

7 centimetres and its length is 8.

2 centimetres, then for its perimeter we need to add up, find the total of all of those edges.

So 3.

7 plus 8.

2 plus 3.

7 plus 8.

2 is 23.

8 centimetres.

So shape A has been translated by the vector, negative two, three, to become shape B.

The transformation that has taken place as a translation and it's translated left by two and up by three.

What's the perimeter of shape A and shape B respectively? So the dimensions are two and three.

Because it's a translation, the object and the image are congruent.

So the dimensions of the image is two and three as well.

So the perimeter of A and B are both 10 units, two plus three, plus two plus three.

Here we've got another problem.

So shape C has been reflected in the y-axis to become shape D.

So shape C is your object shape D is your image, it's a reflection this time.

So what would the perimeter of shape C and shape D be? Well we need to know the dimensions.

We can find them by counting using the coordinate axis grid.

So it's three by four, the rectangle is three by four.

They're congruent to each other once again, so the perimeters will be the same.

They are both 14 units.

Three add four, add three, add four.

Maybe you've got a quicker way to find the perimeter than that.

So translation and reflection have not changed the perimeter and Lucas believes that that's gonna be the same for rotation as well.

So here is a diagram that shows a rotation.

So E is our object and F is our image.

The dimensions of the rectangle, the width is two and the length is four.

And F has a width of two and a length of four.

The orientation has changed but the shape is congruent to its object.

So the perimeter of E and the perimeter of F are in fact the same.

Lucas is correct.

So translation, reflection, rotation, maintain the lengths.

One of the properties of the shape that is invariant after a reflection, a rotation or a translation is length and that's why the perimeter is unchanged.

Alex has said "So the area would not change either." If they're congruent, the area's not gonna change.

Let's have a look at that.

So the areas the size of surface and states the number of unit squares needed to completely cover that surface.

So that's a definition of area that hopefully you feel familiar with anyway.

So we'll just go through an example.

So we've got our rectangle again.

So it's four centimetres by eight centimetres.

And so this area is 32 square centimetres 'cause there are four along the width and eight along the length.

And so if you've got four rows of eight or eight rows of four, then there are 32 units squares to cover up that rectangle.

So we've seen this example already, but last time we thought about the perimeter.

So this time, let's think about the area.

The dimensions are two and three.

So how many units squares cover up each shape? Well the shapes are the same.

So their areas are the same.

Hopefully you've written six square units or thought the answer is six square units.

Again, when we had shape C reflecting to become shape D, shape C was a rectangle of three by four.

Shape D is a rectangle of three by four.

The area is 12 square units, there are 12 unit squares covering up the surface of that 2D shape.

So shape E has been rotated 90 degrees clockwise about the origin to become the shape F.

So what is the area of shape E and what is the area of shape F? So again, if we look at their dimensions, E is a rectangle two by four units and F is a rectangle two by four units.

So they've both got eight square units covering up their surface.

So again, the areas are the same.

So a check for you.

If a shape with a perimeter of 52 millimetres is rotated, the image will have the same perimeter, true or false, and justify your answer.

So hopefully you went for true, they would have the same perimeter.

And that's because rotation maintains length, it is invariant property.

So Aisha's posed a question here.

What about enlargement? We've done reflection, we've done rotation, we've done translation.

What about enlargement? Does that change the perimeter and area? Andeep says the lengths are not maintained as the image is not congruent, so he thinks they must change.

So because the property, the length is not invariant after an enlargement, Andeep believes that they are going to change.

If the lengths change then the perimeter and the area must change also.

So shape H has been enlarged by scale factor of two to become shape I.

So H is your object, I is your image.

The scale factor is two.

So all lengths have been multiplied by two.

So if we now put the lengths on there for you and you can see that's happened.

So the 3.

2 corresponds with the 6.

4, it's been doubled.

So what's the perimeter of H? Well, the perimeter, we add them all up, we get 9.

8 centimetres and the perimeter of I is 19.

6 centimetres.

So Jun said "The perimeter of I is twice the perimeter of H because of that scale factor two." So because the lengths have all doubled, then it makes sense that the perimeter, the total of those lengths has also doubled.

So the Alex said, "Will the area be double as well then?" Just by looking at that, what do you think? Do you think Alex is correct? I is an enlargement by scale factor two of H.

So here we've got the three centimetre, which is the base, these are triangles.

We're gonna use the formula base times perpendicular height divided by two.

We've got the three centimetres, which then is six centimetres on the image because of the scale factor.

The perpendicular height is three centimetres.

It's not an edge but it is a length, so that has also doubled.

So the area of H is 4.

5 square centimetres, three times three divided by two.

And the area of I is 18 because six from six divided by two is 18.

So they're not doubled.

The area of the image is four times larger than the object and it fits within it four times.

And so you can see that on this diagram here.

So the object, the purple triangle, fits exactly four times over the image.

So check, fill in the blanks.

When an enlargement takes place, the perimeter changes.

If the enlargement scale factor is three, the perimeter would be three times as long.

The area changes but not by the scale factor.

And now we're onto the practise part of this section.

So pause the video, have a go and when you're ready, come on back.

Question two, you need to work out the perimeter and area of both the object and the image in the following enlargements.

And you can assume that one square is one centimetre.

Pause the video and when you're ready, come back for the answers.

So on question one, a shape has a perimeter of 10 centimetres.

If it's translated it still will be 10 centimetres.

If it is rotated, it'll still be 10 centimetres.

If it is reflected, it will still be 10 centimetres and that is because those three transformations maintain length.

Okay, so those three maintain length and therefore the perimeter is unchanged.

If it is enlarged by a scale factor of 0.

5, then the perimeter would be five centimetres because the total length would also be enlarged by 0.

5.

0.

5 is equivalent to one half and a half of 10 is five.

And question two, you needed to get the perimeter and the area of the, so perimeter of A is four centimetres and the area of A is one square centimetre, the perimeter of B is eight centimetres and the area is four centimetres squared.

Okay, so we're now gonna go onto the last part of this lesson.

And this part is gonna look at coordinate problems. So all to do with transformations as well as coordinates.

So here we've got a rectangle that is translated by two, one.

So the object is the purple one and the image is the green one.

And it's been translated by two one.

What would the image of the point A, B be? So if we've got that vertex and it's coordinate it's A, B, we've got no scales on the X and the y-axis, so it could be anything.

So if it's A, B and it's been translated by two, one, what would the coordinate of the image of A, B be? Well it'd be A plus two because the two in the vector tells us that it's increased by two, it's gone to the right by two.

And then it would be B plus one for the y-coordinate because the vector tells us it's increased by one, it's gone up by one.

So A plus two, B plus one.

Another example similar to this, but this time it's negative four, negative two.

So A, B is on the object.

And after a translation of negative four, negative two, what would the coordinate of that image point be? So it'd be A minus four for the x-coordinate 'cause it's gone to the left by four and B minus two for the y-coordinate because it has gone down by two.

If A and B had a numerical value after a translation of negative four, negative two, the x-coordinate would go down by four, you would subtract four from it and the y-coordinate would go down by two, you would subtract two from that one.

So a translation has taken place.

We're gonna look at it the other way round this time.

If you've got the coordinate A, B, that arrow there is saying it's mapped to.

So it's telling you that a transformation has taken place.

So A, B has mapped to this new image of A plus five, B minus two.

So which vector has it been translated? So on the x-coordinate, A has gone from A to A plus five.

So that means it's moved right by five units.

And the y-coordinate B, has gone from B to B minus two.

It is two below where it would've been, so it's gone down.

So the vector would've been five minus two, horizontally increased by five and vertically decreased by two.

So just to check on that one, which of these vectors has translated A, B to A plus three, B minus four.

So it would've been C.

The the x-coordinate has increased by three, it's gone to the right by three.

The y-coordinate has gone down by four, so minus four.

Another translation has taken place, but this time our object coordinate is not A, B, our objects coordinate is A plus four, five plus B.

So we're gonna do the same thing.

We're just gonna compare the x-coordinates of the object and the image and the y-coordinate of the object and the image.

So the x-coordinate was A plus four and is now A plus two.

So it has gone down by two, so minus two.

And the y-coordinate was five plus B and is now seven plus B.

So it is increased by two.

So the vector is minus two, two.

So if you think about that again, the top number in a vector is the horizontal displacement, that tells you what's happened to the x-coordinates.

And the x-coordinate has changed from A plus four to A plus two.

So within that algebraic expression you have taken away two and that is why that horizontal displacement is minus two.

So let's have a check with one that looks like that.

A plus 11, B plus six, translates to A plus eight, B plus two.

So which vector has it translated by? So compare the x-coordinates and then compare the y-coordinates.

So this one is negative three, negative four.

A plus 11 has subtracted three to get to A plus eight.

And B plus six has subtracted four to get to B plus two.

So it has moved, it's translated left by three and down by four.

So now we're gonna look at, not just a single point, but the vertices that make a triangle, so three of them.

So we've got a triangle with three vertices, A, which is three four, B, which is five, four and C, which is five, nine.

And we are told that it is transformed and we are given the coordinates of its image.

So A prime or the image of A is negative three, four.

The image of B is negative five, four.

And the image of C is negative five, nine.

And we need to figure out what transformation has taken place.

So this is quite a challenge with just coordinates.

So there is the option of drawing and sketching out a coordinate axis, plotting the points and looking to see what has happened.

That way we can think about lengths, angles, orientation, sense and position.

So Jun has noticed that none of the y-coordinates have changed.

And I think hopefully you notice that as well.

Jacob has said the x-coordinates are negative, so on the other side of the y-axis, so if you just imagine in your head or sketch down on a piece of paper an xy-axis, if you were at three four, where is minus three four, it's on the other side of the y-axis.

But if we just look at this one from its coordinates, then only the x-coordinates have changed and those x-coordinates haven't changed by the same amount.

So three to minus three is a drop of six, but five to minus five is a drop of 10.

This tells us it's not a translation because if it was a translation, the same horizontal displacement would've taken place on all three of the vertices and that hasn't happened here.

So it isn't a translation.

And with an image, it becomes a lot more clear what transformation has taken place.

And so a reflection in the y-axis, and that's why the y-coordinates didn't change, but the x-coordinates have become negative 'cause it has flipped over the y-axis into the negative second quadrant.

So can you complete the missing coordinate after a rotation by 180 degrees? Look for a pattern to support you.

So it should have been negative four, negative six.

The rotation of 180 degrees has taken it from that positive first quadrant where both X and Y values are positive and 180 degrees, a half turn has moved into the third quadrant where both the X and Y are negative.

So you're now gonna do a little bit of practise.

So the first question, you need to write down the coordinate of the image, these algebraic coordinates after it's been translated by negative five, three.

Pause the video and when you're ready, come back for the next question.

Question two, you need to figure out which transformation has occurred for both A and B.

They're both triangles, they're the same triangle, but a transformation has happened to it twice.

You may wish to plot the coordinates to support you, but you might wanna sit there and think about the coordinates before you plot it as well.

Pause the video and when you're ready, come back.

So the answers for the algebraic coordinates.

So all of the x-coordinates should have subtracted five.

So when it was A, it just became A subtract five.

On B, it was A plus one.

So if you take away five from the numerical part of that expression, you should have written A minus four.

So check your answers there.

And then for question two, part A was a translation by three zeros.

So by comparing the coordinates, you may have figured out it was a translation because the same thing happened to all three vertices.

But plotting it is another way that would've helped you see that it has moved.

And for part B, it was a reflection in a horizontal line.

So that one is much harder to do by just thinking about the coordinates.

But hopefully you did look at that the x-coordinates hadn't changed.

So there wasn't any sort of horizontal movement or reflection across an axis.

So this one was a reflection in a horizontal line.

And the horizontal line means that it has reflected vertically, which is why the x-coordinates didn't change and the y-coordinates did.

So during today's lesson, we've been looking at problem solving with transformations and really knowing what changes and what's invariant in every transformation will help you with your solving of problems. And when you come to trying to do a transformation or when you've got a problem to do with transformations, it's a good idea to either do it or sketch it and that will really help you visualise what has happened and help you to solve any problems. Well done today and I look forward to working with you again in the future.