Lesson video

Hello, I'm Mrs. Lashley, and I'm gonna be working with you as we work through the lessons today.

The learning outcome today is to be able to find areas of compound shapes, including circles, in a variety of ways.

So we're gonna make a start at looking at doing that right now.

On the screen, there are some keywords that we're gonna use during the lesson.

I would encourage you to make sure you're familiar with them before we make a start.

So you might wish to pause the video whilst you read them.

And then when you are ready, press play.

So the first learning cycle is to look through the methods for finding area of compound shapes, and then the second learning cycle will look more closely at finding the area of compound shapes.

And remember, they'll all involve circles and sectors.

So let's make a start with the first one and looking back through methods that we can use to find the area.

Compound shapes are created from two or more basic shapes.

And so we've got three examples of compound shapes, and they've been split using the dash line into those basic shapes.

The importance of the basic shapes is that they have a formula or a standard way to find the area and one we want to find the area of a compound shape that's really useful.

I've split those shapes using the dash line as I say, but are there other ways that those compound shapes could have been split? And remember, you're splitting them into basic shapes.

So when it comes to finding the area, there are three standard methods, and certain shapes may lead you to use one of the methods over the other two.

And that's for efficiency.

So this shape can be split into two parts, and the areas can be totaled.

So can you see what the two parts could be and where would I split it? I can split it vertically, and I've now got two basic shapes.

A would be a quarter-circle, and the area for a quarter-circle, it's a sector of a circle, it is 1/4 of the area of the full circle with a radius of eight.

So 1/4 of pi r-squared where r is eight, that is 16 pi in its simplified form, and we're gonna leave it in terms of pi so that we have the most exact answer.

So that's the area of the basic shape A, which is part of this compound shape.

And B is a square.

So the area of that square would be 64.

So now we've got our two component parts.

To find the total area, we are going to sum them.

16 pi plus 64 square centimetres.

Once again, I'm gonna leave that in terms of pi so it's exact.

Whereas this shape is also a compound shape, but it lends itself to a method where we would complete the shape and subtract the additional area.

So rather than splitting the shape up and finding those areas, we're going to complete the shape.

So a rectangle would be used to complete the shape, and we've got the dimensions there of that rectangle.

And so the area would be 350 square centimetres because we multiplied the length and the width.

But that rectangle has completed the shape by including a semicircle and a quarter-circle.

So the area of the compound shape is actually less than the area of the rectangle.

So we're going to work out the areas of these additional areas.

So the first one is A, which is the semicircle.

A semicircle is half of the area of the circle with the same radius.

I know that the width of the rectangle is 14, and that would be the diameter.

So that I need to half it to get the radius of seven.

So half times pi times 7 squared would give me 49 over 2 pi.

Once again, I'm gonna leave that in terms of pi to keep it as exact as possible.

B, the other area that we've included in the rectangle that is not part of the compound shape is a quarter-circle.

What's the radius of that quarter-circle? Well, it's 14.

So we need to do 1/4 of the area of a circle with a radius of 14, and we evaluate that it is 49 pi.

So now, what is the area of the compound shape? Well, the compound shape is less than the area of the rectangle.

And how much less? Well, it's the additional area of the semicircle and the quarter-circle.

So we need to subtract them from it.

So that's 119.

09 or 119.

1 to one decimal place square centimetres.

So this is the second method for finding the area of a compound shape.

The third standard method is to rearrange the original compound shape to minimise or simplify the calculations.

So the configuration of this compound shape can be rearranged.

The quarter-circle on the right-hand side could be chopped or rearranged, moved into the quarter-circle that was missing at the other end.

And this now completes a rectangle.

The area would be equivalent to this rectangle because all we have done is rearrange the configuration.

So 12 times 8 is 96.

The area of that quite complicated-looking compound shape is as simple as finding the area of this rectangle.

So here is a check for you.

For which of these compound shapes would rearranging be the most efficient method to calculate the area? So pause the video whilst you make a decision.

And then when you want to check, press play.

A, there is a semicircle at the top of this shape that would fit very neatly into the bottom where there is a removal of a semicircle.

So that rearranging would mean that this has the same area as a rectangle.

That's a much more simplified calculation.

For this compound shape, which method for finding the area would be the most efficient? So our three standard methods.

So splitting the shape up and then finding the sum of those areas, completing the shape where we then have to normally subtract off some additional areas, or rearranging it to make it a simplified calculation.

Pause the video whilst you make a decision, and when you're ready to check, press play.

Splitting the shape up.

This would be split into a rectangle and two quarter-circles.

Those two quarter-circles could be thought of as one semicircle 'cause they would be equivalent to that.

So you could simplify the amount of calculations, but it wouldn't rearrange into a neat shape like a rectangle where there's just one calculation to do.

So on this first task of finding the area of compound shapes, remember they involve sectors or circles, question 1, these shapes are created by either adding or removing rectangles, quarter-circles and semicircle.

So you're not assuming that they are a quarter-circle.

They will have been created using quarter-circles or a semicircle, or a rectangle.

So by splitting the shapes into component parts, find the area of these compound shapes and do leave your answers in terms of pi.

So pause the video whilst you work out part A and part B, and then when you press play, we'll move on within the task.

Question 2, similarly, they are created using rectangles, semicircles, and quarter-circles, either by adding them or removing them.

And this time, I want you to find the area by completing the shape.

And again, leave your answers in terms of pi.

So two parts to do, by completing the shape, work out the area.

Pause the video whilst you're working on that question.

And when you press play, we'll move on to question 3.

So the last question of this task, same idea again, that we want to find the area of the compound shapes, which have been created using rectangles, semicircles, and quarter-circles, either by the addition of or a removal of.

But this time, find the area by using the third method of rearranging.

So rearrange the shapes to simplify the calculation and leave your answers in terms of pi, where appropriate.

Pause the video, and then when you press play, we're gonna go through the answers to all of the questions in task A.

Okay, so question 1, part A, you were splitting this up.

This is one that we had seen previously.

So we've got two quarter-circles and a rectangle.

So the area of the rectangle would be 18 by 6, which is 108.

The area of the quarter-circle, its six, would be the radius of that.

Quarter-circle, and so 1/4 times pi times 6 squared is 9 pi.

It may be that you treated two of them together as one semicircle.

And that way, you'd have got 18 pi as your answer because you would've used half rather than 1/4.

So the total area, we're gonna sum up our component parts.

We've got the rectangle of 108 and two quarter-circles, nine pi each.

So our answer in terms of pi is 108 plus 18 pi.

On part B, we split it up, and we've got two semicircles.

Those semicircles are not congruent, and we know they are not equal because of the dimensions of the rectangle.

So you couldn't treat it as a whole circle because they did have different radii.

So the area of the rectangle is 10 by 8, which is 80.

Then semicircle B would have a radius of five because we know it's diameter is 10.

And then that comes out as 25 over 2 pi.

The area of the semicircle C, which is the one that has a radius of four, when you work it out, it comes out as eight pi.

And so your total area is to sum all of those component parts.

Simplifying the two terms, you can collect the two terms in terms of pi together, and that's 41 over 2 pi.

So 80 plus 41 over 2 pi.

Moving on to question 2, where you are using the method of completing the shape.

So the first one, we can complete it by making a rectangle, and then we'll need to remove the additional areas, one of which was a semicircle, and one was a quarter-circle.

So you're gonna find the area of the rectangle and subtract those additional areas.

So the answer was 110 minus 75 over 4 pi.

Remember, you should have been leaving it in terms of pi.

B, completing the shape again and then the removal.

Area of the rectangle, 24 by 12.

So 288.

It may have been that you didn't notice that the additional area is actually one full circle.

And that came with a little bit of thinking because the semicircle has a diameter of 12, which means it has a radius of 6.

The two quarter-circles have a radius of 6.

So 2/4 plus a half equals 1.

So this was a total area of one full circle with a radius of 6.

The total area was then 288, which was the area of the rectangle, minus the area of the full circle.

So 288 minus 36 pi square centimetres.

And then the last question was to rearrange the shape.

So being able to sort of imagine slicing or chopping one part off and moving it into a different position.

So just rearranging the configuration, the area doesn't change.

So the semicircle was removed from the sort of left-hand side and placed at the right where there was a cutout of a semicircle, and that formed a rectangle.

So the area was 460.

And on part B, there were two quarter-circles that had radius of seven.

And that means if there's two quarter-circles that are congruent, that's the same as a semicircle.

And there was a cutout of a semicircle with a diameter of 14.

A diameter of 14 is the same as a radius of 7.

So this one again fit together to create a rectangle.

The area was 266.

So our next learning cycle is to continue to find the area of compound shapes.

We've re-familiarized ourselves with the methods that we could use, but we're gonna look at a little bit more complicated compound shapes and maybe a little bit more thought into what we do actually know.

So a compound shape is described as being made from a square and four congruent quarter-circles.

The radius of the quarter-circles is equal to the length of the square.

Three friends draw the compound shape that has been described.

So Jacob draws it like this.

So look at that.

Do you agree that he has met the description? Jun draws it like this.

Has June met the description of the compound shape? And lastly, Sam.

Sam has drawn it like this.

Have they met the description of the compound shape? Would you have drawn it the way the three friends did, or is there another way that you could have drawn a compound shape that has a square and four congruent quarter-circles where the radius of the quarter-circle is equal to the length of the square? Maybe have a go.

So if the dimensions of the square are the same in each arrangement, then despite the different arrangements that the three friends went for, the area of the compound shape would be the same.

And we saw that we can split a compound shape into component parts to find the area.

And because these three shapes all split into the same component parts, then their area would be equal, as long as the dimension of the square is the same.

So you can see there how they have arranged those five basic shapes.

To calculate the area, dividing it into the component parts is important.

And then you can consider thinking, well, actually, four congruent quarter-circles is equivalent to a full circle, and that would be much more efficient than working out one quarter-circles area and then multiplying it by four.

So I would like you to work out the area of this compound shape to one decimal place.

So pause the video, use your calculator.

And when you're ready to check, press play.

So this was another rearrangement or another configuration of four congruent quarter-circles and one square.

It was different to what the three friends had drawn, but the components were still the same.

So those area of the four quarter-circles, I'm just going to work out the area of a circle.

And the area of the square is length times width, but they are the same.

So we know it's length squared.

What you needed to do though was to work out what that length was.

So you were told that the maximum width of this compound shape was 15 centimetres.

We know that all three were the same length, so we can split it into three equal parts.

So that's why the radius is five, and the length of the square is five.

So the total area to one decimal place is 103.

5.

Really well done if you manage to get that.

So if we look at this very interesting compound shape, I sort of see it like a jester's hat, then a combination of the three methods actually might come in useful.

And so as we become more familiar with our three methods and when we look at shapes, it doesn't have to be one standalone method.

You want to try and be efficient with your working, and it might be that you need to combine some of them together.

So if we firstly split it into the component parts, then we've got a semicircle, we've got a triangle in that middle part, and then we've got these odd compound shapes themselves on the other two edges of the triangle.

So some of them are still not standard shapes.

So the ones I've just said were odd.

So to find the area of these parts, we need to consider the other two methods, completing the shape and rearranging.

So the part that's at the bottom, highlighted with the rectangle there, lends itself to be in a completed shape because we can work out the area of a semicircle, and we can work out the area of a rectangle.

But the shape that's made by removing a semicircle is more difficult to find.

So if I complete it by putting a semicircle in there, I can work out its area by doing rectangle, subtract the area of the semicircle.

The part on the other edge of the triangle also lends itself to completing the shape, so that it becomes a sector.

However, if the triangle is isosceles, then the other semicircle, which is a standard shape when we split the shape, could be rearranged to fit the missing space of a semicircle.

But that's only true if it was an isosceles triangle because then both of those lengths would be the same.

So that semicircle there could fit in the gap there.

Assuming that the isosceles triangle is also a right-angled triangle, then the angle in the sector would be 135 degrees, because you can only calculate the area of a sector if you have that interior angle.

And so without that further information, we wouldn't be able to work out the area.

So I have not put any dimensions on here.

We're just discussing the method and the process we'd go through.

But if you had knowledge of it being a right-angled isosceles triangle, that would mean the base angles that are equal to each other are 45 degrees, and then we could work out that the exterior angle is 135, which happens to be the angle within that sector.

Here is a check for you, another weird and wonderful-looking compound shape.

I hope you agree.

So given that this compound shape is created with sectors of a circle and a parallelogram, how do we know that none of the sectors are equal? So pause the video and really think about this one.

It might be that you want to sort of sketch the diagram, add on information that you do know.

If you've with somebody, talk to them, discuss it.

So how do we know for certain that none of the sectors are equal? Press play when you're ready to check.

So there are three sectors on here, and the largest one has a radius equal to the longest edge of a parallelogram.

So this was all really thinking about the fact that they were connected to a parallelogram.

And what are the properties of that? Well, the properties of a parallelogram are the opposite edges are equal in length.

So that first larger sector has a radius of the longest length of the parallelogram.

The medium sector has a diameter equal to the longest edge.

So one has a radius equal to the longest edge, and the other has a diameter, which means that they are different sizes.

The smallest sector has a diameter equal to the shorter edge.

So we need to think about, when we're met with a compound shape, what we know for certain and try not to assume the question and the content would need to give you enough information to move forward, or it will ask you to state the assumptions made.

So looking at another compound shape involving circles, well, this square is inscribed in a circle with length 14 centimetres.

What is the area of the shaded region? So the shaded region is on that sort of external part of the square.

The square is inscribed, which means its four vertices are exactly on the circumference of the circle.

So Lucas says, "The area of the square is 196 square centimetres." He's done 14 squared.

"But without the radius of the circle, it's impossible to work out the area of the shaded region." Well, we need to work out the area of the circle, and to work out the area of a circle, you need the radius or the diameter.

But Izzy said, "The diagonal of that square is the diameter of the circle." And that's because it is an inscribed square.

So because the vertices are on the circumference, then the diagonal from one vertex across the shape would be the length of the diameter.

Lucas said, "That's helpful.

If I think of it as an isosceles right-angled triangle, then I can use Pythagoras' theorem." Why is it isosceles? How do we know it's a right-angled triangle? Well, we know it's a right-angled triangle because it's a square.

So we know from our properties of a square that the interior angles are right angles.

How do we know it would be a right-angled isosceles? Because it's a square, and the edges are equal.

So we would have those two equal edges.

So it's 19.

798 centimetres.

Approximately, that's the diameter.

So the radius is 9.

89 or 9.

9 centimetres.

Using pi r-squared, the area of the circle would be 307.

9 roughly, which means that the shaded region is 111.

9 square centimetres to one decimal place.

And Izzy said, "That seems about right." So how did we get from the area of the circle to the shaded region? Lucas has subtracted the area of the square, which was 196.

So here's a check.

A rectangle fits exactly inside a circle.

So its four vertices are on the circumference.

Which of these are true? Diagonal of the rectangle is equal to the radius of the circle.

The diagonal of the rectangle is equal to the diameter of the circle.

The shorter edge of the rectangle is equal to the radius, or the longer edge of the rectangle is equal to the diameter.

So pause the video, and then when you're ready to check what you think is the correct answer, press play.

So it's B.

Again, if we find the diagonal of that rectangle, then it will be the diameter of the circle.

So here's another check.

A rectangle with width six centimetres and length 12 centimetres fits exactly inside a circle.

Find the length of the radius of the circle.

So pause the video whilst you're working through to get to the radius.

And when you're ready to check, press play.

So we know from the previous check that the diameter is equal to the diagonal.

So we need to get that length d that's marked on the diagram.

Because it's a rectangle, we know that we have right-angled triangles when we split it via its diagonal.

And so therefore, we can use Pythagoras' theorem.

We know the two shorter edge lengths because it was given in the question, the dimensions of the rectangle.

And so the diameter is 6 root 5 in its exact form, which is approximately 13.

4.

The question actually asked you for the radius.

So you needed to go through the relationship between diameter and radius, which is that the radius is half of the diameter.

So we can divide it by two.

And so in its exact form, it's 3 root 5, and that's approximately 6.

7 centimetres.

So we're now onto the final task for the lesson, which is to still find the area of compound shape, but sort of maybe we have a little bit more involvement and thinking about combinations of methods as opposed to just one fixed method.

So question 1, a rectangle is inscribed in a circle, find the area of the shaded region.

So press pause whilst you're going through that, and then when you're ready to move on, press play.

Okay, so question 2, find the area of these compound shapes, stating any assumptions that you have made.

Leave your answer in surd form where appropriate.

So once again, there are some that have been split to show you what they've been created from.

Any assumptions you make, you should be writing that down as well.

So press pause whilst you're working through those three parts.

When you press play, you've got one more question on this task.

So question 3, here is a compound shape.

Part A, I'd like you to complete the description of the shapes that have made it.

Part B, if the rectangle has a length of 18 centimetres and a width of six centimetres, find the area of the compound shape.

And part C, if the rectangle had a length of x centimetres and a width of y centimetres, find an expression for the area of the compound shape.

So pause the video whilst you're working through question 3.

And when you are ready to press play, we're gonna go through the answers to task B.

Question 1, find the area of the shade of region.

So it was going through the steps of knowing that the diagonal of that rectangle is the diameter of the circle.

Using Pythagoras' theorem to get the diameter.

From the diameter, finding the radius, which was 8.

5, which allowed you then to get the area of the circle, pi r-squared, and you could work out the area of the rectangle from the beginning.

And then you are going to do a subtraction 'cause it's like completing the area method.

You've found the area of the circle, and you've subtracted the rectangle.

You filled in the rectangle.

So the shaded area to three significant figures is 107 square centimetres.

Question 2, remember there are three parts.

So this is part A.

The assumption that I've made is that it's a rectangle and four semicircles.

So the area of the rectangle is 12 by 16, which is 192.

And now there are two circles.

Why are there two circles? Well, that's because there are two congruent semicircles.

How do I know that they are two congruent semicircles? Because they're on the opposite edges of a rectangle.

And therefore, their diameter and their radius would be equal.

So I can think of them as just one circle as opposed to doing two semicircles.

So the diameter of 16 means the radius is 8.

So the area of that circle or the two semicircles is 64 pi.

The area of the circles with diameter 12, which would mean that's a radius of 6, is 36 pi.

And so the total area of this compound shape, which I've assumed is made from rectangle and four semicircles, is 192 plus a 100 pi.

Part B, I'm assuming it's an isosceles trapezium and semicircles.

The area of the trapezium could be calculated from the given information, and that's 128 square centimetres.

I then needed the diameter to then be able to get the radius of the semicircle to get the area of those.

And that's where I'm gonna need to use Pythagoras' theorem.

By assuming it's an isosceles trapezium and then saying I've got two congruent semicircles, but also it tells me that this trapezium has got a line of symmetry.

And therefore, I can work out the six is the base of the triangle.

I've got the perpendicular height of eight, which was given in the trapezium.

But because of the symmetry, I can subtract 10 from 22 and half it.

By being an isosceles trapezium, I know there is symmetry.

Pythagoras' theorem then tells me that what's marked as d is 10, and that is the diameter of these semicircles.

So I use the radius of five.

Adding them together, I get 128 plus 25 pi.

Part C, assuming that this shape is a kite with a semicircular cutout, then I need to work out the area of it by doing the area of the kite, subtract the area of the semicircle.

However, I need to do a little bit of work to be able to get the area of the kite.

So I'm gonna split the area of the kite using its own diagonals, which will give me two pairs of congruent right-angled triangles because the diagonals of a kite are perpendicular to each other.

So I've got my right-angled triangles.

So there on the screen, you can see my triangle A, and I'm working out the vertical or the perpendicular height, which I've labelled lowercase a, and that is 12.

On triangle B, I'm working out again the sort of vertical perpendicular height of b, and that is five.

So, so far, I've now got the total length or the total height of the kite.

I'm gonna move on to the next slide to continue this solution.

So the area of the kite is a half times 10 times 12 plus a half times 10 times 5.

Where's that come from? Well, that's come from the fact that I've got two triangle As, which I can think of as one isosceles triangle.

I now know that it's base is 10.

I've got five and five that's given.

And we worked out that the sort of vertical height of triangle A is 12.

So my isosceles has got a perpendicular height of 12 and a base of 10.

That's the bottom two A triangles.

Then I'm gonna add on the top two triangles, which I've labelled as B.

And that's another isosceles, again, using the definition of a kite that we've got adjacent edges being equal, that then I've got a base of 10 and a perpendicular height of five.

I calculated that on the previous slide.

So I've now got my full area of the kite by splitting it up into two isosceles triangles.

Then I need to do the area of the semicircle that has been removed.

The diameter would be 5 root 2.

And I know that again because adjacent edges on a kite are equal.

And so I want a semicircle.

So it's a half of the full circle with the same radius.

That was the diameter.

So the radius, I'd have to half it.

And we end up with 25 over 4 pi for the area of the semicircle.

So the area of my compound shape is 85 minus 25 over 4 pi.

It might be that you want to go back through that work, go through the working out and check you're following that along nicely.

It was quite a complicated one, so really well done if you manage that independently.

The last question had three parts.

So first, I wanted you to complete the description.

So a rectangle adjoined to two semicircles and two quarter-circles.

If you'd stopped there, that is a fair description, but I've gone on further to discuss the sort of size and the dimensions of them.

So one semicircle has a diameter of the shortest edge of the rectangle, and the other has a diameter of the longest edge of the rectangle.

So they are not congruent semicircles.

They're on opposing sides of the rectangle.

And similarly with the quarter-circles, we've got one that has a radius of the short edge and one that has a radius with the longest edge of the rectangle.

Part B, we need to find the area of the compound shape, given the dimensions of the rectangle.

So I've got five sections, five component parts.

I've split it up.

So I've got A, which is a quarter-circle with a radius of six, B, which is a quarter-circle with a radius of 18, C, which is a semicircle with a radius of three, and D, which is a semicircle with a radius of nine, leaving them all in terms of pi, keeping them as exact as possible.

And finally, E is a rectangle.

So we've done length times width, and that's 108.

The area of the compound shape is then the sum of all five component areas, and that's simplifies to 135 pi plus 108 square centimetres.

Part C was then more general.

If you had a length of x and a width of y, what would be an expression for the area? And it's going through the same steps that you just did using numbers.

So we need the area of A, which is a quarter-circle with a radius of y.

Then B is a quarter-circle with a radius of x.

C is a semicircle with a radius of y divided by 2 because we don't want the diameter, we want the radius.

D is a semicircle with a radius of x over 2 or half x, and the rectangle is xy.

Adding those together and simplifying the fractions, we end up with this wonderful-looking expression of 3, open brackets, x-squared plus y-squared, close bracket.

So that could be an expanded form.

You might have it as 3x-squared plus 3y-squared divided by 8, lots of pi, plus xy.

So that's all of the circular sector areas plus the area of the rectangle.

So to summarise today's lesson, which was about area of compound shapes, which included circles and sectors, there are multiple methods to find them.

We went through the three standard methods, but the features of the shape may lead you to one method over the other because it's more efficient.

However, you might need to do a combination of those methods for certain shapes.

Really well done today.

I hope you enjoyed it, and I look forward to working with you again in the future.