Circle: Circumference, arc length, area and sector. KS3 Mathematics The aim of this unit is to teach pupils to: Deduce and use formulae to calculate lengths, perimeters, areas and volumes in 2-D and 3-D shapes Material in this unit is linked to the Key Stage 3 Framework supplement of examples pp 234-241. Circle: Circumference, arc length, area and sector.

S8.5 Circumference of a circle Contents S8 Perimeter, area and volume S8.5 Circumference of a circle S8.6 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 = 3.141592653589793238462643383279502884197169 39937510582097494459230781640628620899862803482 53421170679821480865132823066470938446095505822 31725359408128481117450284102701938521105559644 62294895493038196 (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.

Approximations for the value of π When we are doing calculations involving the value π we have to use an approximation for the value. For a rough approximation we can use 3. Better approximations are 3.14 or . 22 7 We can also use the π button on a calculator. Most questions will tell you what approximations to use. 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.

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

The circumference of a circle Use π = 3.14 to find the circumference of this circle. C = πd 8 cm = 3.14 × 8 = 25.12 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.

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

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 = 12.56 cm = 56.52 m 23 mm 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. 58 cm C = πd C = 2πr = 3.14 × 23 = 2 × 3.14 × 58 = 72.22 mm = 364.24 cm

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.)

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 × 13 + 6 + 6 = 52.82 cm

Finding the length of an arc What is the length of arc AB? A B 6 cm The arc length is of the circumference of the circle. 1 4 This is because, 90° 360° = 1 4 So, Length of arc AB = × 2πr 1 4 = × 2π × 6 Length of arc AB = 9.42 cm (to 2 d.p.)

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 = 100 000 cm Explain that we can ignore any remainder when dividing 100000 by 157 because we are asked for the number of complete rotations. 50 cm The number of complete rotations = 100 000 ÷ 157 ≈ 636

Finding the length of an arc B r θ For any circle with radius r and angle at the centre θ, Remind pupils that in finding the length of an arc, we are finding a fraction of the circumference of the circle. Arc length AB = θ 360 × 2πr This is the circumference of the circle. Arc length AB = 2πrθ 360 = πrθ 180

The perimeter of shapes made from arcs Find the perimeter of these shapes on a cm square grid: 40° The perimeter of this shape is made from three semi-circles. 40° 360° = 1 9 1 2 × π × 6 + 1 9 × π × 12 + Perimeter = Perimeter = For the first example ask pupils to explain why the perimeter can be written as 6π cm. Similarly, in the second example, ask pupils to explain why the perimeter can be written as 2π + 6. 1 2 × π × 4 + 1 9 × π × 6 + 1 2 × π × 2 3 + 3 = 6π cm = 2π + 6 = 18.85 cm (to 2 d.p.) = 12.28 cm (to 2 d.p.)

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

The circumference of a circle Use π = 3.14 to find the area of this circle. 4 cm A = πr2 ≈ 3.14 × 4 × 4 = 50.24 cm2

Finding the area given the diameter The radius of a circle is half of its diameter, 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

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 = 12.56 cm2 = 78.5 m2 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 and to substitute this value into the formula. Pupils need to be careful not to make this mistake. 78 cm A = πr2 A = πr2 = 3.14 × 232 = 3.14 × 392 = 1661.06 mm2 = 4775.94 cm2

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 = 132.665 cm2 Compare this with slide 74, where we found the perimeter of the same shape. Area of rectangle = 6 × 13 = 78 cm2 Total area = 132.665 + 78 = 210.665 cm2

Area of a sector What is the area of this sector? Area of the sector = 72° 360° × π × 52 72° 5 cm 1 5 = × π × 52 = 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.

Find the shaded area to 2 decimal places. Area problem Find the shaded area to 2 decimal places. Area of the square = 12 × 12 = 144 cm2 1 4 Area of sector = × π × 122 = 36π Discuss how this area could be calculated before revealing the solution. Shaded area = 144 – 36π 12 cm = 30.96 cm2 (to 2 d.p.)

Finding the area of a sector B r θ O For any circle with radius r and angle at the centre θ, Remind pupils that in finding the area of a sector, we are finding a fraction of the area of the circle. Area of sector AOB = θ 360 × πr2 This is the area of the circle. Area of sector AOB = πr2θ 360

The area of shapes made from sectors Find the area of these shapes on a cm square grid: 40° 40° 360° = 1 9 Area = 1 2 × π × 32 1 2 × π × 12 + Area = 1 9 × π × 62 Explain that we can think of the first shape as a semi-circle of radius 3 + a semi-circle of radius 1 – a semi-circle of radius 2. By factorizing, we can write this as ½π(32 + 12 – 22) = ½π(9 + 1 – 4) = ½π × 6 = 3π 1 2 × π × 22 – 1 9 × π × 42 – = 3π cm2 = cm2 1 9 × π × 20 = 9.42 cm2 (to 2 d.p.) = 6.98 cm2 (to 2 d.p.)