Lesson video

Hello, my name's Miss Parnham.

In this lesson, we're going to learn how to calculate the area of compound shapes with circles.

The shaded area is made up of two distinct sections.

So working out the area of one of these regions and then doubling it, we'll find the shaded area.

So what we have here for one of the shaded areas is the area of a square subtract the area of a quarter circle.

The square has got side length 15 millimetres, and the radius of the quarter circle is also 15 millimetres.

So 15 squared subtract pi r squared over four, gives us 48.

3 millimetres squared, for just one of these shaded sections.

So multiplying that by two gives us a total shaded region on this shape of 96.

6 millimetre squared.

We're now asked what percentage of the shape is unshaded? So to find the area of the unshaded region, if we subtract the 96.

6 from 225, that gets us 128.

4 millimetres squared, percentage of the shape therefore is going to be this value we've just found divided by 225, multiplied by 100, which is 57.

1% to three significant figures.

An alternative way of working this out would have been to work out what percentage was shaded using 96.

6 divided by 225 times a hundred, but then we would need to remember to subtract that percentage from a hundred percent in order to get 57.

1.

Here's some questions for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

In part b the 48 is divided by three to get to diameter of 16 millimetres.

And therefore a radius of eight millimetres on each semicircle.

So you perhaps notice that six semi circles, makes three circles, so you could subtract three lots of 64 pi from the area of the rectangle and that might have got you to the answer quicker.

And the same with part c those quarter circles will form a complete circle with a radius of 1.

6 metres.

So this would have saved some time as well when finding out that shaded area.

Here's a further question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

You could have worked out the unshaded area by subtracting pi from four pi because you have an unshaded circle of radius two, split into two semi circles.

And the shaded circle of radius one, split into two semi circles, but you'd still get that same answer of three pi metres squared.

The shaded area in this example can be found by subtracting the area of two sectors from that of an isosceles triangle.

Let's first work out the area of a triangle.

This is height multiplied by base divided by two.

This gives us 0.

968 metres squared.

Both of these sectors are exactly the same area.

So let's work out the area of one of them and double it.

Area of one sector is pi r squared, multiplied by the fraction formed from 42 degrees over 360 degrees.

This gives us 0.

443 metres squared, to three significant figures.

Let's double that and take it away from 0.

968 to find the shaded region is 0.

0810 metres squared, to three significant figures.

If you had preferred, you could have worked out the area of both sectors together because they have the same radius of 1.

1.

So you could have subtracted pi r squared, multiplied by 84 over 360.

You would have still got to the same answer.

Here's a question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

In this example, you have two overlapping sectors with the same angle.

So one way of answering it would be to think about it as complete circles first.

So that's 100 pi subtract 49 pi giving us 51 pi, and then multiplying that 51 pi by a fraction made from 80 over 360.

You would still get exactly the same shaded area for your answer.

Here's another question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

No angle is marked here on the diagram, but the text tells us it's an equilateral triangle.

And so using our knowledge that it has an interior angle of 60 degrees will help us solve this problem.

That's all for this lesson.

Thank you for watching.