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G26 Circumference and area of a circle

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1 G26 Circumference and area of a circle
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle G26 Circumference and area of a circle This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation.

2 G26.1 Circumference of a circle
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle G26.1 Circumference of a circle

3 Circle circumference and diameter
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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).

4 Boardworks KS3 Maths 2009 G26 Circumference and area of a circle
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 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.

5 Approximations for the value of π
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle Approximations for the value of π When we are doing calculations involving the value π we have to use an approximation for the value. π For a rough approximation we can use 3. Better approximations are 3.14 or 7 22 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 of calculations. Most questions will tell you which approximation to use. When a calculation has lots of steps we write π as a symbol throughout and evaluate it at the end, if necessary.

6 The circumference of a circle
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle The circumference of a circle For any circle: π = circumference diameter or: π = d C We can rearrange this to make a formula to find the circumference of a circle given its diameter. Photo credit: © 2009 Jupiterimages Corporation Pupils should be asked to learn these formulae. C = πd

7 The circumference of a circle
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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.

8 Finding the circumference given the radius
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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

9 The circumference of a circle
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle The circumference of a circle Use π = 3.14 to find the circumference of the following circles: 4 cm C = πd 9 m C = 2πr = 3.14 × 4 = 2 × 3.14 × 9 = cm = m 23 mm C = πd 58 cm 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

10 Finding the radius given the circumference
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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 r = ? Link: A8 Changing the subject and deriving formulae 2 × 3.14 12 = = 1.91 cm (to 2 d.p.)

11 Find the perimeter of this shape
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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

12 Circumference problem
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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, 50 cm 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. The number of complete rotations = ÷ 157 ≈ 636

13 Boardworks KS3 Maths 2009 G26 Circumference and area of a circle
G26.2 Area of a circle

14 Boardworks KS3 Maths 2009 G26 Circumference and 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 and ask a volunteer to use the pen tool to label the length and the height in terms of r. Discuss how the diagram would look if the circle was divided into more pieces. Deduce from this that the area of a circle is r2.

15 Formula for the area of a circle
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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

16 The circumference of a circle
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle The circumference of a circle Use π = 3.14 to find the area of this circle. A = πr2 4 cm ≈ 3.14 × 4 × 4 = cm2

17 Finding the area given the diameter
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle Finding the area given the diameter The radius of a circle is half of its diameter, or r = 2 d 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

18 Boardworks KS3 Maths 2009 G26 Circumference and area of a circle
The area of a circle Use π = 3.14 to find the area of the following circles: 2 cm A = πr2 10 m A = πr2 = 3.14 × 22 = 3.14 × 52 = cm2 = 78.5 m2 A = πr2 A = πr2 23 mm 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 forget to half the diameter to find the radius. 78 cm = 3.14 × 232 = 3.14 × 392 = mm2 = cm2

19 Find the area of this shape
Boardworks KS3 Maths 2009 G26 Circumference and area of a circle 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 Compare this with slide 11, where we found the perimeter of the same shape. Area of rectangle = 6 × 13 = 78 cm2 Total area = = cm2

20 Boardworks KS3 Maths 2009 G26 Circumference and area of a circle
Area of a sector What is the area of this sector? Area of the sector 360° 72° × π × 52 = 72° 5 cm = × π × 52 5 1 = π × 5 Discuss how this area could be calculated before revealing the solution. The area of a sector is a fraction of the area of a full circle. We can find this fraction by dividing the angle at the centre by 360°. = 15.7 cm2 (to 1 d.p.) We can use this method to find the area of any sector.

21 Boardworks KS3 Maths 2009 G26 Circumference and area of a circle
Area problem Find the shaded area to 2 decimal places. Area of the square = 12 × 12 = 144 cm2 Area of sector = × π × 122 4 1 = 36π Discuss how this area could be calculated before revealing the solution. Shaded area = 144 – 36π 12 cm = cm2 (to 2 d.p.)


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