Lesson video

Hello there.

You made a great choice with today's lesson.

It's gonna be a gooden.

My name is Dr.

Rowlandson, and I'm gonna be supporting you through it.

Let's get started.

Welcome to today's lesson from the unit of 2D and 3D shape with compound shapes.

This lesson is called problem solving with complex 2D shapes.

And by the end of today's lesson, we'll be able to use our understanding of 2D shapes to solve problems. Here are some previous keywords that will be useful again during today's lesson.

So you may want to pause the video, if you want to remind yourself what any of these words mean before pressing play to continue.

The lesson contains two learning cycles.

In the first learning cycle, we're going to focus on some abstract problems that involve complex 2D shapes.

And then the second learning cycle, we'll apply what we've learned to some contextual problems. But let's start off with abstract problems involving complex 2D shapes.

Here we have a shape that is constructed by using the arcs of three 60 degree sectors of a circle.

Two of those arcs are curved outwards.

Those are on the left and the right.

The centre of those sectors are in the opposite bottom left or bottom right corner of the shape.

And the third arc is curved inwards, that's at the bottom.

Now you can even think of that in one of two ways.

You can either think of it as being the arc of a sector where the centre of that sector is somewhere below the shape, or you can think of it as being the arc of a sector where the centre is at the top of the shape, but the arc has been inverted.

Either way, does it matter? We've got the same shape, and each sector has a radius of 12 centimetres.

And what we need to do is find the area of this shape.

Now Andeep is going to walk us through this problem in a moment, but before he does, pause the video and think about how you might approach this problem yourself.

In particular, think about how you might break this problem down, or even what your first step might be.

And then press play when you are ready to continue.

Okay, let's hear from Andeep.

He says, "With geometry problems, it can sometimes be helpful to add extra lines and information on the diagram.

For example, I could draw the radii for one of the sectors." Like this.

Now we've done that, we can visualise this particular sector more clearly, and it may give us some clues for what we might do next.

Andeep then says, "With geometry problems, it can also sometimes be helpful to cover up some parts of the diagram while you focus on other parts.

For example, I could just focus on one sector." Like this.

Now just think what we could do with that sector.

Andeep says, "I could then find the area of that sector." And this calculation contains pi times 12 squared, which would be the area of a full circle.

And before it, we have the fraction 60 over 360, which is what fraction of the circle we have with this sector.

A 60 degree angle out of a total possible 360 degrees around a point.

So that calculation will give us the area of the sector, which is 24 pi centimetres squared.

Now we could write that as a decimal.

Well, we've gotta be very careful about any numbers that we use.

We don't round until we get our final answer.

Otherwise, the final answer may not be completely accurate.

So it might be more accurate for the time being to keep this answer in terms of pi.

Anyway, back to our problem.

'Cause the original problem wasn't just to find the area of this sector, there was more to it than that.

Andeep then says, "If I gradually reveal the parts of the diagram that I previously covered, I can see there is a region that needs to be added to the area so far, as well as a region that needs to be subtracted from the area." But I wonder if you notice something about those two regions.

He says, "The areas of these two regions are equal, so the area of the compound shape is equal to the area of the sector." Therefore, this calculation gives us the answer to the area of this complex 2D shape.

So let's check what we've learned then.

The shape below is constructed using the arcs of semicircles.

And we can see it has a couple of measurements on a 6 centimetres and a 12 centimetres.

Jacob draws a line segment on the shape below to split it into three regions.

And that's what we can see on the right.

We have three statements labelled a, b, and c.

And what you need to do is decide which of those statements is true.

Pause the video while you do that and press play when you're ready for an answer.

The answer is c, the area of region A is equal to the area of region B.

They are both semicircles with a diameter of 6 centimetres.

Jacob transforms a shape on the left into a semicircle that has the same area using the steps below.

So based on this, what you need to do is use Jacob's working so far to calculate the area.

Pause the video while you do that, and press play when you are ready for an answer.

Okay, let's take a look.

The answer is either 18 pi centimetres squared, if you write it in terms of pi, or if you write it as a decimal, you can have 56.

5 centimetres squared if it's rounded to one decimal place.

Here we have a different abstract problem now.

The diagram shows a square inscribed inside a circle.

The length of the square is 10 centimetres, and what we need to do is find the total area of the shaded regions.

Now Jun's gonna talk us through this problem shortly, but before he does, perhaps pause the video while you think about how you might approach this problem yourself, or even how you might get started with this, or even just do a tiny part of this problem.

And then press play when you're ready to continue.

Let's now hear from Jun.

Jun says, "When a problem requires multiple steps, it can sometimes be helpful to start by writing a plan." Here's Jun's plan.

First, he's going to find the area of the square, and then he'll find the radius of the circle, and then you'll find the area of the circle.

And then finally, he'll find the area of the shade of region by doing the area of the circle, subtract the area of the square.

It's almost like we are cutting the square outta the circle.

And what this plan does is it takes this complex problem and breaks it down into four separate smaller problems where each problem is more manageable than the original problem.

Jun says, "I can now just focus on one step at a time." So let's now work through this plan together.

And let's start off with the area of the square, because Jun says, "I always like to do the easiest step first." Here we have a square, and we know the length of the square is 10 centimetres.

So the area would be 10 squared, which is 100.

Step two was to find the radius of the circle, because if we know the radius, we can find the area.

But how can we use our information about the length of the square to find the radius of the circle.

Well, Jun says, "The diameter of the circle is equal to the diagonal of the square." Because those two vertices on the square are both on the circumference.

Now if we look at the diagram with that diameter drawn on it, what we can see inside the square is we have a right-angled triangle.

Jun says, "I could find this diameter by using Pythagoras' theorem." Because we have a right-angled triangle where we know two lengths, 10 centimetres and 10 centimetres.

And what we're trying to find is the hypotenuse of this right-angled triangle.

So we can do 10 squared plus 10 squared, and then square root it to get 10 root 2.

Once again, we could write out as a decimal, but we would need to be careful not to round in it until a final solution.

So it'd be more accurate for now to use 10 root 2.

So if the diameter is 10 root 2, we can divide that by 2 to get the radius, and that'll give us 5 root 2.

With that, we can then find the area of the circle.

We could do pi times the square of 5 root 2.

And that would give 50 pi centimetres squared.

We then just have one last step, and that is to find the area of the shaded region.

We know the area of the circle and the area of the square.

So to find the area of the shaded region, we can do the area of the circle subtract the area of the square, and that would be 50 pi subtract 100, which we could put as our final answer with some brackets around it and centimetres squared.

Or we could write it as a decimal, which is 57.

1 centimetres squared rounded to one decimal place.

Here's Jun's solution now in full where each calculation is labelled with a part of Jun's plan.

And by doing this, we've taken a complex problem, which requires quite a few calculations and broken it down into five separate questions, where each question is much easier and more manageable than the problem as a whole.

So let's check what we've learned.

Here we have a diagram that shows a square inscribed inside a circle.

And what you need to do is pause the video and find the area of the square, and then press play when you're ready for the answer.

The answer is 8 squared, which is 64 centimetres squared.

So now find the diameter of the circle.

And remember, the diameter of the circle is equal to the diagonal of the square.

Pause the video while you do this, and press play when you are ready to continue.

To do that, you'll do 8 squared plus 8 squared, and then square root it, and that'll give you 8 root 2 centimetres squared.

You may have written it as a decimal, but your rounded answer would not be as accurate as it is in third form here.

So now we know the diameter of this circle, find its area.

Pause the video while you do that, and press play when you're ready for an answer.

There, the circle is 32 pi centimetres squared, which you get from doing pi times 4 root 2 squared.

So let's put all those pieces together now to solve this more complex problem.

The diagram shows the square inscribed inside the circle.

We know the area of the square is 64 centimetres squared.

We know the area of the circle is 32 pi centimetres squared.

And what you need to do is find the total area of the shaded regions.

Give your answer to one decimal place.

Pause the video while you do that, and press play when you're ready for the answer.

Well, you previously did all the hard work of this question when you found the areas of the separate things.

This final part is just a subtraction, 32 pi subtract 64, and that is equal to 36.

5 centimetres squared when rounded to one decimal place.

And what you've done with this series of check for understanding questions is you've taken a complex problem and treat it as separate smaller problems. So it's over to you now for task A.

This task contains two questions, and here is question one.

Pause the video while you work through this, and press play when you are ready for Question 2.

Here's Question 2, pause the video while you do that, and press play when you're ready to go through some answers.

Okay, let's see how we got on.

For part A, your answer is 50.

3 centimetres squared.

In part B, your answer is 18.

3 centimetres squared, which you can get if you subtract the right-angled triangle from part A.

But part C, your answer is 13.

7 centimetres squared, which you can get by subtracting your answer from part B from the right-angled triangle you worked out in part B.

And then in part D, your answer is 36.

5 centimetres squared, which you can get from either doubling your answers from part B, or you could subtract 2 lots of part C from a square.

And there are other ways to do it as well.

Now it's worth noting that if you did use your answers from parts A, B, and C to get your answer for part D, and you used the rounded versions of your answer, your answer part D might be a little bit different.

But if you went back and got the unrounded versions for parts A, B, and C, and use those to get your answer for part D, you should get 36.

5 centimetres squared.

And then Question 2.

In part A, you've given the length of the square, so you would need to find the damage of the circle and the radius of the circle.

And then finally, you'd get an area of 82.

2 centimetres squared for your shaded regions.

Part B is a bit different, because you're given the diameter of the circle, so that means you know the radius of the circle, but you need to use Pythagoras in a different way to get the length of the square, so you can get the area of it.

But your final answer should be 41.

1 centimetre squared.

Fantastic work so far.

Now let's move on to the second part of today's lesson where we're gonna apply what we've learned to solve some contextual problems. Here we have Izzy.

Izzy is calculating the cost to decorate the room shown in the diagram.

She wants to tile the floor and to paint the walls and ceiling the same colour.

She looks up some information about the materials she'll need, and she finds these details.

The paint is sold in 2.

5 litre tins.

Each paint tin costs 22 pounds.

One litre of paint covers 10 square metres.

There's also a door in that room, which Izzy will not be painting, and that door is 0.

9 metres by 2 metres.

Each floor tile costs 14 pounds.

And floor tiles are squares of length 30 centimetres.

So that's a lot of information, but all of it will be useful here.

How could Izzy calculate the total cost to decorate the room? Maybe pause the video while you think about this.

And in particular, think about how you could break this big problem down into lots of smaller problems, and maybe even write a plan.

And then press play when you're ready to continue.

Izzy's now going to talk us through the solution.

She says, "Before I start though, I'll write a plan for my solution." And here is Izzy's plan.

First, she's going to find the area of the ceiling, and then she's going to find the area of the walls, and then she'll find the area of the door, and then the cost of the paint by using parts one, two, and three.

After that, she'll work out the number of floor tiles she needs, and then the cost of the floor tiles, until finally, working out the total cost of all the materials altogether.

Quite a few steps to this plan, but let's start off with step one.

Izzy says, "I'll start by finding the area of the ceiling, and there are multiple ways I could do this." Here's one way.

We could split this compound shape into two rectangles.

We could find the area of rectangle A by doing 2.

4 times 3, which is 7.

2 metres squared.

And then the area of rectangle B, which is 1.

5 times 1.

8, which is 2.

7 metres squared.

And then the total layer of the ceiling will be the sum of those two answers, which is 9.

9 metres squared.

Izzy then says, let's find the total area of the walls.

I could do this by multiplying the perimeter of the room by the height of the walls.

If you're not sure why that works, what you could imagine is taking each of those separate walls, and then lining them up into one long rectangle.

The height of that rectangle will be 2.

4 metres.

That's the height of the walls.

And the length of the rectangle will be the total length around the edge of the room, because that's the sum of the base for each of those separate rectangles.

For your own sure, you could always find the area of each wall and then add them together.

But a quicker way would be to do what Izzy suggests, multiplying the perimeter by the height of the walls.

The perimeter of the room can be calculated by adding together the length of all the sides, which is 13.

8 metres.

And if we multiply that by 2.

4, we would get 33.

12 metres squared.

Izzy then says, "Next, I'll find the area of the door." Can we see why we need to find the area of the door? We have the area of the walls that we're going to paint, but part of one of the walls is taken up by the door, which we will not be painting.

So we need to find the area of the door so we know what to subtract from the total area of the walls.

The dimensions of the door is 0.

9 metres by 2 metres, so the area of the door is 0.

9 times 2, which is 1.

8 metres squared.

Izzy says, "I'm using the same paint to cover the walls and the ceiling, but I'm not painting the door." So we know the area of the walls, the area of the ceiling, and the area of the door.

To find the total area to be painted, we'll add together the areas for the walls and ceiling, and subtract the area the door.

And that would give 41.

22 metres squared.

So now we know how much area will be painted, we need to work out how many tins of paint we need.

1 litre of paint covers 10 square metres.

And paint is sold in 2.

5 litre tins, where each paint tin costs 22 pounds.

And we know the total area to be painted is 41.

22 square metres.

Now there are multiple different ways we could work out the number of paint tins we need.

But one way could be to start by finding the amount of paint required.

We would do that by dividing 41.

22 by 10, because each litre covers 10 square metres.

And when we do that division, we get 4.

122, which means we need 4.

122 litres to cover all the surfaces that need the painting.

So then we can work out how many tins we need by dividing 4.

122, which is how many litres we want, by 2.

5, 'cause each tin contains 2.

5 litres, and that would give 1.

6488.

But hang on a minute, we can't buy 1.

6488 tins.

We would need to buy 2 tins.

So with each tin costing 22 pounds, the total cost would be 2 lots of 22, which is 44 pounds.

So that is the walls and the ceiling covered in terms of cost.

But what about the floor? Izzy says, "I also need to work out how many floor tiles are required." And we know the length of each square floor tile is 30 centimetres.

She says, "I'll start by working out how many fit along the 3 metre length." That's the vertical length on the left-hand side.

Now the length of the floor tile is given in centimetres, and the length of the wall is given in metres.

Let's make the units the same.

The length of each tile would be 0.

3 metres.

So to work out the number of tiles to cover that 3 metre long edge of the room, we could do 3 divided by 0.

3 to get 10.

We would need 10 tiles to go along that edge of the floor.

She then says, "I'll now work out how many tiles fit across the 2.

4 metre length." We can do a similar thing.

We could divide 2.

4 by 0.

3 and that would give 8, which means we need 8 tiles to go along that part of the floor.

Izzy then says, "I'll now work out how many tiles are needed to cover the left part of the room." That left part of the room is a rectangle.

We know we can fit 10 tiles going vertically, upwards, and 8 tiles going across.

So the total number of tiles would be 8 times 10, which would be 80.

We need 80 tiles to cover that part of the room.

Izzy says, "I can then use the same process to find the number of tiles required for the rest of the room," so we have another rectangle.

We could work out how many tiles cover 1.

5 metres by doing 1.

5 divided by 0.

3, that's 5.

So it's 5 tiles across.

And then the number of tiles to cover 1.

8 metres by doing 1.

8 divided by 0.

3, that's 6.

So it's 6 tiles down.

We have a rectangle that is 5 tiles by 6 tiles, which means the tiles needed that part of the room would be 5 times 6, which is 30.

Izzy says, "Once I work out the total number of tiles required, I can calculate their cost." We know we need 80 tiles for one part of the room, 30 tiles for the other part.

Altogether, we need 110 tiles.

Each tile costs 14 pounds, so the total cost of the floor tiles is 110 times 14, which is 1,540 pounds.

So we know the cost of the paint and we know the cost of the floor tiles.

Izzy says, "I can now calculate the total cost to decorate the room." By adding those two numbers together, which is 1,584 pounds.

Now at the start of this, Izzy wrote a plan, and she says, "My original plan can also be used as a breakdown of the costs." Here it is.

And this can be useful in a couple of ways.

For Izzy, it provides a way for it to go back and check each of the numbers, and see where her final answer comes from.

But also you may sometimes see breakdowns of numbers like this when you pay for some kind of service, such as the work of a trades person.

For example, if Izzy had paid for a decorator to come and decorate the room.

When the decorator provides Izzy with a price, not only will a decorator provide the total cost of everything, but often they provide a breakdown such as what you can see here.

So let's check what we've learned.

Here, we have a compound shape.

Which calculations would find the area of the shape? Pause a video while you choose, and might be more than one answer, and then press play when you're ready to continue.

The answers are a, b, and c.

This is the shape of a room.

The perimeter of the room is 20 metres, the walls are 3 metres high.

What is the total area of the walls? Pause the video while you work it out, and press play when you're ready for an answer.

The total area of the walls is equal to the perimeter multiplied by the height, which is 60 square metres.

Here we have Sofia who is using the same paint for the walls and ceiling.

1 litre of paint covers 10 square metres of surface.

The area of the walls is 60 square metres.

The area of the ceiling is 20.

24 square metres.

How many litres of paint is required? Pause the video while you write down your answer, and press play when you're ready to continue.

We can get our answer by adding our two areas together, and then divide them by 10 to get 8.

024 litres.

Let's now think about the number of tins required.

Sofia needs 8.

024 litres of paint.

The paint is sold in 2.

5 litre tins.

Therefore, Sofia needs to buy 3 tins of paint.

Is that true or is it false? Pause the video why you choose and also choose a justification, and then press play when you're ready for an answer.

The answer is false.

And when it comes to the reason why, yes, when we divide 8.

024 by 2.

5, we do get 3.

2096, which does round down to 3.

But if Sofia only buys 3 tins, she will not have 8.

024 litres.

She have less than that.

So 3 tins wouldn't be enough, she would need to buy 4 tins, and have some paint left over.

Let's take a look at the design of the room again, and one part of the room is highlighted here.

Sofia wants to cover the floor with tiles in the highlighted section of the room.

Each tile is a square with length 20 centimetres.

How many tiles would be needed? Pause the video while you work this out, and press play when you're ready for an answer.

Sofia would need 336 tiles.

Let's now think about a different context that uses 2D shapes, because some problems require a combination of knowledge from different areas of maths to solve.

For example, when drawing multiple pie charts, different size circles can be used to represent the differences in the sample size for each dataset.

And in these cases, the areas of the pie charts are proportional to the total frequency for each dataset.

And doing this requires applying knowledge about multiple aspects of mathematics, pie charts, angles, areas of circles, and proportional reasoning.

Let's take a look at an example.

Andeep surveys two companies about how the staff travel to work.

We have company A and company B, and the different modes of transport they take.

And one thing you might notice about these two companies is that company B is much larger than company A in terms of its staff, or at least a larger sample was taken from company B than company A.

Andeep plots a pie chart for company A with a radius of 6 centimetres.

And he wants to plot a pie chart for company B, but he wants to make it so that these sizes of these two pie charts differ, so that they reflect the different sizes of the samples.

So with that in mind, what radius should he use so that the areas of the pie charts are in the same proportions as the total frequencies? Now we're going to work through his problems shortly together, but before we do, what you might wanna do is pause the video to give yourself a bit of time to think about what this problem is trying to achieve and any steps you might need to take along the way in order to solve it.

And then press play when you're ready to continue.

Let's work through this together now.

Sometimes it can be helpful to present information in a table.

So what we can see in this table is we have the total frequency for company A and the total frequency for company B.

And if we think about those two numbers as a ratio 40 to 90, we could simplify that ratio to 4 to 9.

And the bottom row of this table is for the areas of each circle for the pie charts, and we want those areas to be in the same ratio 4 to 9.

Now we know the radius of the pie chart for company A is 6 centimetres.

So what we could work out at this point is the area of that pie chart.

We could do it by doing pi times 6 squared, which is 36 pi centimetre squared and fill it in our table.

Now if we want the areas for those two pi charts to be in the same ratio 4 to 9, what will be helpful to work out is a multiplier between these two ratios.

We can do it by dividing 36 pi by 4 to get a multiplier of 9 pi, which means if we take the ratio 4 to 9 and multiply both parts by 9 pi, we'd get the ratio for the areas of the circles, 36 pi, and we'll do 9 times 9 pi, which is 81 pi.

So now we know what area we want that pie chart to be, we can work out its radius.

We could set up an equation like this, pi r squared equals 81 pi.

We could simplify the equation by dividing both sides by pi and solve it by square rooting both sides to get a radius of 9 centimetres.

Now we know the radius, we can plot the pie chart.

Here are Andeep's two pie charts, one for company A and one for company B, where the areas of those two pie charts are in the same ratio as the total frequencies for each company.

So let's check what we've learned by taking a similar problem and breaking it down to small steps.

Laura is drawing a pie chart to represent data about two groups.

She wants the areas of the pie chart to be in proportion with the total frequencies of each group.

The table shows the total frequency for each group, and what you need to do is write the ratio for the total frequencies in its simplest form.

Pause while you do that and press play to continue.

The ratio 60 to 45 simplifies to 4 to 3.

Laura draws a pie chart for group A using a radius of 8 centimetres.

What is the area of the pie chart for group A? Pause the video while you work it out, and press play to continue.

You would do pi times 8 squared, which is 64 pi centimetres squared.

So with that in mind, what is the area for the pie chart for group B? Pause while you work it out and press play to continue.

The answer is 48 pi centimetres squared.

So now you know the area, what is the radius of the pie chart for group B? And give your answer to one decimal place.

The answer is 6.

9 centimetres, and there's the working to get to it.

Okay, it's over to you now for task B.

This task contains two questions, and here is Question 1.

Pause the video while you do it, and press play when you're ready for Question 2.

And here is Question 2.

Pause the video while you work through these, and then press play when you're ready for some answers.

Let's see how we got on.

For Question 1, your final answer should be 1,854 pounds.

Now, if you didn't get that, don't worry about it too much, here's a breakdown of all the numbers you should get along the way.

So if your answer was different, compare your work into what you can see here.

Find where your answer is different, rework out that part, and just fix anything else you need to fix before recalculating the final answer.

And then Question 2, your pie chart for company A should look like this.

And then for part B, your pie chart should look something like this where your radius is 8.

9 centimetres, and the breakdown of how you get to that on the screen.

Wonderful work today.

Now let's summarise what we've learned.

When solving problems, consider whether any elements of that problem are familiar from other aspects of mathematics.

Problems don't just fit neatly into geometry, algebra, or number and so on, sometimes you might have to use different things from different parts of the curriculum.

But it's always worth trying to keep the goal in mind throughout the problem, because it's very easy to get distracted by lots of different pieces of information along the way.

Therefore, it can sometimes be helpful to write a plan before you start.

That way, you can treat the big problem as lots of small problems and worry about how to solve the whole thing when you're ready for it.

And with diagrams, it can be beneficial to add information to them such as drawing extra lines or writing it on anything you work out.

And then when it comes to things like pie charts, the size of a pie chart can be scaled to reflect the population or sample size of the data by using some of the things you know about 2D shapes.

Thank you very much today.

Have a great day.