1. Shape and Space Circles The aim of this unit is to teach pupils to:
Identify and use the geometric properties of triangles, quadrilaterals and other polygons to solve problems; explain and justify inferences and deductions using mathematical reasoning
Understand congruence and similarity
Identify and use the properties of circles
Material in this unit is linked to the Key Stage 3 Framework supplement of examples pp
Circles

2. The value of π
For any circle the circumference is always just over three times bigger than the radius.
The exact number is called π (pi).
We use the symbol π because the number cannot be written exactly.
π =
(to 200 decimal places)!
Explain that pi is just a number. We call it pi because it is not possible to write the number exactly. Even written to 200 decimal places, although extremely accurate, is an approximation.

3. Approximations for the value of π
When we are doing calculations involving the value π we have to use an approximation for the value.
Generally, we use the approximation 3.14
We can also use the π button on a calculator.
It is useful to approximate pi to a value of 3 when approximating the answers to calculations.
When a calculation has lots of steps we write π as a symbol throughout and evaluate it at the end, if necessary.

4. The circumference of a circle
For any circle,
π =
circumference
diameter
or,
π = C d
We can rearrange this to make a formula to find the circumference of a circle given its diameter.
Pupils should be asked to learn these formulae.
C = πd

5. Circle circumference and diameter
Drag the circle to different sizes and ask pupils to record the length of the diameter and the circumference for each one. Challenge them to find a link between these two numbers.
Tell pupils that the ratio of the diameter to the circumference of a circle is a fixed amount equal to just over 3. This means that the circumference of a circle is always just over three times the diameter.
Once pupils know the link between the diameter and the circumference, hide the circumference and ask them to find it given the diameter (by multiplying the diameter by 3.14).
Also hide the diameter and ask them how we can find it given the circumference (by dividing by 3.14).

6. The circumference of a circle
Use π = 3.14 to find the circumference of this circle.
C = πd
8 cm
= 3.14 × 8
= cm
Tell pupils that when solving a problem like this they should always start by writing down the formula that they are using.
This will minimize the risk of using the radius instead of the diameter, for example.

7. Finding the circumference given the radius
The diameter of a circle is two times its radius, or
d = 2r
We can substitute this into the formula
C = πd
to give us a formula to find the circumference of a circle given its radius.
C = 2πr

8. The circumference of a circle
Use π = 3.14 to find the circumference of the following circles:
4 cm
9 m
C = πd
C = 2πr
= 3.14 × 4
= 2 × 3.14 × 9
= cm
= m
23 mm
58 cm
C = πd
C = 2πr
For each one, start by asking pupils what formula we have to use.
Estimate each answer first using  = 3, or use this to check the answer.
= 3.14 × 23
= 2 × 3.14 × 58
= mm
= cm

9. Finding the radius given the circumference
Use π = 3.14 to find the radius of this circle.
C = 2πr
12 cm
How can we rearrange this to make r the subject of the formula? C 2π
r = ?
Link:
A3 Formulae – changing the subject of a formula 12
2 × 3.14 =
= 1.91 cm (to 2 d.p.)

10. Find the perimeter of this shape
Use π = 3.14 to find perimeter of this shape.
The perimeter of this shape is made up of the circumference of a circle of diameter 13 cm and two lines of length 6 cm.
13 cm
6 cm
Perimeter = 3.14 ×
= cm

11. Circumference problem
The diameter of a bicycle wheel is 50 cm. How many complete rotations does it make over a distance of 1 km?
Using C = πd and π = 3.14,
The circumference of the wheel
= 3.14 × 50
= 157 cm
1 km = cm
Explain that we can ignore any remainder when dividing by 157 because we are asked for the number of complete rotations.
50 cm
The number of complete rotations
= ÷ 157
= 637

12. Formula for the area of a circle
We can find the area of a circle using the formula
Area of a circle = π × r × r
radius or
Area of a circle = πr2

13. Area of a circle
This animation shows how the area of a circle can be approximated to the area of a parallelogram of base length r and height r.
Watch the circle pieces rearrange into an approximate parallelogram an ask a volunteer to use the pen tool to label the length and the height in terms of r.
Deduce from this that the area of a circle is r2.

14. The circumference of a circle
Use π = 3.14 to find the area of this circle.
4 cm
A = πr2
= 3.14 × 4 × 4
= cm2

15. Finding the area given the diameter
The radius of a circle is half of its radius, or
r = d 2
We can substitute this into the formula
A = πr2
to give us a formula to find the area of a circle given its diameter.
A =
πd2 4

16. The area of a circle
Use π = 3.14 to find the area of the following circles:
2 cm
10 m
A = πr2
A = πr2
= 3.14 × 22
= 3.14 × 52
= cm2
= 78.5 m2
23 mm
78 cm
A = πr2
A = πr2
Explain that rather than use the formula on the previous slide, it is usually easier to halve the diameter mentally to give the radius, before substituting it into the formula.
The most common error is to neglect to half the diameter to find the radius and to substitute this value into the formula. Ensure that pupils do not make this mistake.
= 3.14 × 232
= 3.14 × 392
= mm2
= cm2

17. Find the area of this shape
Use π = 3.14 to find area of this shape.
The area of this shape is made up of the area of a circle of diameter 13 cm and the area of a rectangle of width 6 cm and length 13 cm.
Area of circle = 3.14 × 6.52
13 cm
6 cm
= cm2
Area of rectangle = 6 × 13
Compare this with slide 74, which finds the perimeter of the same shape.
= 78 cm2
Total area =
= cm2