Year 11 · Measurement

M1Applications of measurement 0 / 22

YR 11

A1 A2 M1 M2 F1 S1 S2

YR 12

A4 M6 M7 F4 F5 S4 S5 N2 N3

💡 Quick tips — memory joggers

Converting areas and volumes? Multiply or divide by the conversion factor to the power of the unit:
- twice for area
- three times for volume
Volume and capacity conversions:
- $1\ \text{cm}^{3} = 1\ \text{mL}$
- $\begin{aligned}1\ \text{m}^{3} &= 1\ \text{kL} \\ &= 1000\ \text{L}\end{aligned}$
(not on reference sheet)
Trapezoidal rule: Provides an estimate area of an irregular boundary.
$A \approx \dfrac{h}{2}(d_f + d_l)$

(on reference sheet)

Length units

Conversions for any unit

Units & conversions

Length: e.g. convert $5\text{ m}^{2}$ to $\text{cm}^{2}$ — square the conversion factor:$\begin{aligned}1\text{ m}&=100\text{ cm}\\ 1\text{ m}^{2}&=(100)^{2}=10\,000\text{ cm}^{2}\\ 5\text{ m}^{2}&=5\times 10\,000=50\,000\text{ cm}^{2}\end{aligned}$
Capacity: $1\ \text{cm}^{3}=1\ \text{mL}$ $\begin{aligned}1\ \text{m}^{3}&=1\ \text{kL}\\&=1000\ \text{L}\end{aligned}$
Mass & energy: $1000\ \text{g}=1\ \text{kg}$, $1000\ \text{kg}=1\ \text{t}$; $1000\ \text{J}=1\ \text{kJ}$, $1\ \text{Cal}\approx 4.184\ \text{kJ}$.

Accuracy & error

Precision (limit of reading): Smallest unit of measure. (not on reference sheet) e.g.
$\begin{aligned}8\text{ cm} &\rightarrow 1\text{ cm} \\ 0.6\text{ cm} &\rightarrow 0.1\text{ cm} \\ 0.234\text{ m} &\rightarrow 0.001\text{ m}\end{aligned}$
Absolute error $=\dfrac{1}{2}\times\text{precision}$. (on reference sheet)
Upper bound $=\text{measurement}+\text{absolute error}$. (on reference sheet)
Lower bound $=\text{measurement}-\text{absolute error}$. (on reference sheet)
Percentage error $=\dfrac{\text{absolute error}}{\text{measurement}}\times 100$. (not on reference sheet)

Perimeter & area

Circle: $C=2\pi r$, $A=\pi r^{2}$. (not on reference sheet)
Sector: arc length $\ell=\dfrac{\theta}{360}\times 2\pi r$, area $A=\dfrac{\theta}{360}\times \pi r^{2}$. (on reference sheet)
Trapezium: $A=\dfrac{h}{2}(a+b)$. (on reference sheet)
Pythagoras: $c^{2}=a^{2}+b^{2}$. (not on reference sheet)
Triangle: $A=\dfrac{1}{2}bh$. (not on reference sheet)
Parallelogram: $A=bh$. (not on reference sheet)
Rhombus: $A=\dfrac{1}{2}xy$ (diagonals $x,y$). (not on reference sheet)

Surface area, volume & capacity

Surface area: cylinder $=2\pi r^{2}+2\pi r h$. (on reference sheet) Be careful with one end open or both ends open — the formula needs adjusting (count only the circular ends that are actually there).Both ends closed: $A=2\pi r^{2}+2\pi rh$
One end open: $A=\pi r^{2}+2\pi rh$
Both ends open (tube): $A=2\pi rh$
Surface area: sphere $=4\pi r^{2}$. (on reference sheet)
Volume: prism/cylinder $V=Ah$. (not on reference sheet)
Volume: pyramid/cone $V=\dfrac{1}{3}Ah$. (on reference sheet)
Volume: sphere $V=\dfrac{4}{3}\pi r^{3}$. (on reference sheet)

Volume is how much space a shape holds; capacity is how much liquid it holds.

Trapezoidal rule

$A\approx\dfrac{h}{2}(d_{f}+d_{l})$ $h$ is the distance between the parallel lines
$d_{f}$ is the first parallel line
$d_{l}$ is the second parallel line e.g. the area of a field beside a river: $\begin{aligned}A &\approx \tfrac{400}{2}(200+300) \\ &= 200\times 500 \\ &= 100\,000 \text{ m}^{2}\end{aligned}$

Power and energy

Electricity cost $=\text{power (kW)}\times\text{time (h)}\times\text{price per kWh}$.

Level 1 · Fluency

Convert 3.5 metres to centimetres.

Level 2 · Application

Convert 2.4 m² to square centimetres.
Level 3 · Further Application
A container has a volume of 0.75 m³.
1. (a) Convert this to cubic centimetres. (1 mark)
2. (b) Convert it to litres (1000 cm³ = 1 L). (2 marks)
750 litres

(a) 1 m³ = 1 000 000 cm³, so 0.75 m³ = 750 000 cm³.
(b) $750\,000 \div 1000 =$ 750 litres.
A drum holds 1.2 m³ of liquid.
1. (a) Convert this to cubic centimetres. (1 mark)
2. (b) Convert it to litres (1000 cm³ = 1 L). (2 marks)
1200 litres

(a) 1 m³ = 1 000 000 cm³, so 1.2 m³ = 1 200 000 cm³.
(b) $1\,200\,000 \div 1000 =$ 1200 litres.
A jug is marked as 2500 mL.
1. (a) Convert this to litres. (1 mark)
2. (b) Convert it to cubic centimetres (1 mL = 1 cm³). (2 marks)
2.5 L; 2500 cm³

(a) $2500 \div 1000 =$ 2.5 L.
(b) 1 mL = 1 cm³, so 2500 mL = 2500 cm³.
Level 1 · Fluency

A length is measured to the nearest centimetre. What is the absolute (maximum) error?

Level 2 · Application

A mass is measured as 4.2 kg to one decimal place. State the absolute error.
Level 3 · Further Application
A plank is measured as 2.35 m, correct to the nearest centimetre.
1. (a) State the absolute error. (1 mark)
2. (b) Write the range of possible true lengths. (2 marks)
(a) $\tfrac{1}{2} \times 0.01 = 0.005$ m.
(b) $2.35 \pm 0.005$ m, i.e. from 2.345 m to 2.355 m.
A bolt is measured as 46 mm, correct to the nearest millimetre.
1. (a) State the absolute error. (1 mark)
2. (b) Write the range of possible true lengths. (2 marks)
0.5 mm; 45.5 mm to 46.5 mm

(a) $\tfrac{1}{2} \times 1 =$ 0.5 mm.
(b) $46 \pm 0.5$ mm, i.e. from 45.5 mm to 46.5 mm.
A parcel is measured as 3.5 kg, correct to the nearest 0.1 kg.
1. (a) State the absolute error. (1 mark)
2. (b) Write the range of possible true masses. (2 marks)
0.05 kg; 3.45 kg to 3.55 kg

(a) $\tfrac{1}{2} \times 0.1 =$ 0.05 kg.
(b) $3.5 \pm 0.05$ kg, i.e. from 3.45 kg to 3.55 kg.
Level 1 · Fluency

A length is 20 cm to the nearest cm. State its lower bound.

Level 2 · Application

A time is recorded as 9.6 s to one decimal place. State the upper and lower bounds.
Level 3 · Further Application
A rectangular tile is measured as 30 cm by 20 cm, each to the nearest cm.
1. (a) State the bounds for each side. (1 mark)
2. (b) Find the largest possible area of the tile. (2 marks)
625.25 cm²

(a) Length: 29.5–30.5 cm; width: 19.5–20.5 cm.
(b) Largest area = $30.5 \times 20.5 =$ 625.25 cm².
A rectangular card is measured as 15 cm by 8 cm, each to the nearest cm.
1. (a) State the bounds for each side. (1 mark)
2. (b) Find the smallest possible area of the card. (2 marks)
108.75 cm²

(a) Length: 14.5–15.5 cm; width: 7.5–8.5 cm.
(b) Smallest area = $14.5 \times 7.5 =$ 108.75 cm².
A rectangular garden bed is measured as 12 m by 5 m, each to the nearest metre.
1. (a) State the bounds for each side. (1 mark)
2. (b) Find the largest possible area of the bed. (2 marks)
68.75 m²

(a) Length: 11.5–12.5 m; width: 4.5–5.5 m.
(b) Largest area = $12.5 \times 5.5 =$ 68.75 m².
Level 1 · Fluency

A scale that always reads 0.2 kg too high shows what type of error?

Level 2 · Application

Distinguish between a systematic error and a random error, giving one example of each.
Level 3 · Further Application
A student times an event five times and gets 12.1, 12.3, 12.0, 12.2, 12.2 s using a stopwatch by hand.
1. (a) Identify the main type of error in these readings and justify. (2 marks)
2. (b) Suggest one way to reduce it. (1 mark)
(a) Random error (reaction-time variation) — the readings scatter around a central value rather than all being offset the same way.
(b) Take more readings and average them (or use electronic timing gates).
A thermometer is later found to read 2°C too low. A student had recorded 18, 18, 19, 18, 18°C for the same object.
1. (a) Identify the main type of error and justify. (2 marks)
2. (b) State the true temperature of the object. (1 mark)
(a) Systematic error — every reading is offset by the same 2°C, so averaging will not remove it.
(b) The readings sit near 18°C, so the true temperature is about 20°C (add back the 2°C).
Two students each measure the same doorway. Student A gets 2.03, 2.05, 2.04 m; Student B (using a tape that starts at the 10 cm mark) gets 2.14, 2.14, 2.15 m.
1. (a) Identify the type of error affecting Student B and justify. (2 marks)
2. (b) State one way Student B could correct their results. (1 mark)
(a) Systematic error — every reading is about 0.10 m too large because the tape is read from the 10 cm mark rather than 0.
(b) Subtract 0.10 m from each reading (or re-measure from the 0 mark).
Level 1 · Fluency

Percentage error = $(\text{absolute error} \div \text{measurement}) \times 100$. Find it for an absolute error of 0.5 cm on a 50 cm length.

Level 2 · Application

A metal rod is measured as 12.5 cm, correct to one decimal place. Calculate the percentage error.
Level 3 · Further Application
A room’s length is measured as 4.5 m to the nearest 0.1 m.
1. (a) State the absolute error and calculate the percentage error. (2 marks)
2. (b) Explain why a longer measurement generally has a smaller percentage error for the same instrument. (1 mark)
1.11%

(a) Absolute error 0.05 m; % error = $\dfrac{0.05}{4.5} \times 100 =$ 1.11%.
(b) The absolute error stays the same ($\tfrac{1}{2}$ a unit), but dividing it by a larger measurement gives a smaller ratio, so the percentage error falls.
A pipe’s length is measured as 3.2 m to the nearest 0.1 m.
1. (a) State the absolute error and calculate the percentage error, to two decimal places. (2 marks)
2. (b) State what happens to the percentage error if a shorter pipe is measured with the same ruler. (1 mark)
1.56%

(a) Absolute error 0.05 m; % error = $\dfrac{0.05}{3.2} \times 100 = 1.5625\% \approx$ 1.56%.
(b) The percentage error increases, because the same 0.05 m error is divided by a smaller measurement.
A short wire is measured as 0.8 m to the nearest 0.1 m.
1. (a) Calculate the percentage error. (2 marks)
2. (b) Compare it with the percentage error of an 8 m wire measured to the same precision. (1 mark)
6.25%; the 8 m wire is 0.625%

(a) Absolute error 0.05 m; % error = $\dfrac{0.05}{0.8} \times 100 =$ 6.25%.
(b) For 8 m: $\dfrac{0.05}{8} \times 100 = 0.625\%$, which is ten times smaller — longer measurements give smaller percentage errors.
Level 1 · Fluency

Write 4 500 000 in standard form.

Level 2 · Application

Write 0.002073 in standard form, correct to two significant figures.
Level 3 · Further Application
A cell measures 0.0000042 m across.
1. (a) Write this in standard form. (1 mark)
2. (b) Write it correct to two significant figures using a suitable metric prefix (micrometres). (2 marks)
4.2 μm

(a) $4.2 \times 10^{-6}$ m.
(b) 1 μm = $10^{-6}$ m, so $4.2 \times 10^{-6}$ m = 4.2 μm.
A bacterium measures 0.0000015 m across.
1. (a) Write this in standard form. (1 mark)
2. (b) Write it using micrometres. (2 marks)
1.5 μm

(a) $1.5 \times 10^{-6}$ m.
(b) 1 μm = $10^{-6}$ m, so $1.5 \times 10^{-6}$ m = 1.5 μm.
A virus measures 0.000000025 m across.
1. (a) Write this in standard form. (1 mark)
2. (b) Write it in nanometres (1 nm = $10^{-9}$ m). (2 marks)
25 nm

(a) $2.5 \times 10^{-8}$ m.
(b) 1 nm = $10^{-9}$ m, so $2.5 \times 10^{-8} = 25 \times 10^{-9}$ m = 25 nm.
Level 1 · Fluency

What power of ten does the prefix “kilo” represent?

Level 2 · Application

Order these prefixes from smallest to largest: milli, mega, micro, kilo.
Level 3 · Further Application
A file is 3.2 gigabytes.
1. (a) State how many bytes this is in standard form (1 GB = $10^{9}$ bytes). (1 mark)
2. (b) Express 3.2 GB in megabytes. (2 marks)
3200 MB

(a) $3.2 \times 10^{9}$ bytes.
(b) 1 GB = 1000 MB, so 3.2 GB = 3200 MB.
A hard drive holds 2.5 terabytes.
1. (a) State how many bytes this is in standard form (1 TB = $10^{12}$ bytes). (1 mark)
2. (b) Express 2.5 TB in gigabytes. (2 marks)
2500 GB

(a) $2.5 \times 10^{12}$ bytes.
(b) 1 TB = 1000 GB, so 2.5 TB = 2500 GB.
A beam of light has a wavelength of 650 nanometres.
1. (a) Write this length in metres in standard form (1 nm = $10^{-9}$ m). (1 mark)
2. (b) Express 650 nm in micrometres. (2 marks)
0.65 μm

(a) $650 \times 10^{-9} = 6.5 \times 10^{-7}$ m.
(b) 1 μm = 1000 nm, so $650 \div 1000 =$ 0.65 μm.
Level 1 · Fluency

Find the area of a rectangle 8 cm by 5 cm.

Level 2 · Application

Calculate the area of a circle of radius 6 cm, correct to two decimal places.
Level 3 · Further Application
A shape is a rectangle 10 m by 6 m with a semicircle of diameter 6 m attached to one short end.
1. (a) Calculate the total area, to one decimal place. (2 marks)
2. (b) Calculate the perimeter of the composite shape, to one decimal place. (1 mark)
74.1 m²; 35.4 m

(a) Rectangle 60 m² + semicircle $\tfrac{1}{2}\pi(3)^{2} = 14.14$ m² → total 74.1 m².
(b) Perimeter = 10 + 6 + 10 + half-circumference $\pi \times 3 = 26 + 9.42 =$ 35.4 m.
A shape is a rectangle 8 m by 5 m with a semicircle of diameter 5 m attached to one short end.
1. (a) Calculate the total area, to one decimal place. (2 marks)
2. (b) Calculate the perimeter of the composite shape, to one decimal place. (1 mark)
49.8 m²; 28.9 m

(a) Rectangle 40 m² + semicircle $\tfrac{1}{2}\pi(2.5)^{2} = 9.82$ m² → total 49.8 m².
(b) Perimeter = 8 + 5 + 8 + half-circumference $\pi \times 2.5 = 21 + 7.85 =$ 28.9 m.
A washer is a flat ring: a circle of radius 10 cm with a concentric circle of radius 6 cm removed.
1. (a) Calculate the area of the ring, to two decimal places. (2 marks)
2. (b) Calculate the total perimeter (both circle edges), to two decimal places. (1 mark)
201.06 cm²; 100.53 cm

(a) $A = \pi(10)^{2} - \pi(6)^{2} = 100\pi - 36\pi = 64\pi =$ 201.06 cm².
(b) $2\pi(10) + 2\pi(6) = 20\pi + 12\pi = 32\pi =$ 100.53 cm.
Level 1 · Fluency

A right triangle has short sides 3 cm and 4 cm. Find the hypotenuse.

Level 2 · Application

A right-angled triangle has legs 5 cm and 12 cm. Find the hypotenuse.
Level 3 · Further Application
A ladder reaches 5.2 m up a wall, with its foot 1.8 m from the wall.
1. (a) Find the length of the ladder, to one decimal place. (2 marks)
2. (b) State whether it is safe if the maximum safe length is 5.4 m. (1 mark)
5.5 m

(a) $\sqrt{5.2^{2} + 1.8^{2}} = \sqrt{30.28} =$ 5.5 m.
(b) $5.5\ \text{m} \le 5.4\ \text{m}$, so it is within the safe length.
A rectangular gate is 3 m wide and 1.4 m tall. A straight diagonal brace runs corner to corner.
1. (a) Find the length of the brace, to one decimal place. (2 marks)
2. (b) State whether a single 3.5 m length of timber is long enough. (1 mark)
3.3 m; yes

(a) $\sqrt{3^{2} + 1.4^{2}} = \sqrt{10.96} =$ 3.3 m.
(b) $3.3\ \text{m} \le 3.5\ \text{m}$, so a single length is long enough.
A wheelchair ramp rises 1.5 m over a horizontal run of 3.6 m.
1. (a) Find the length of the sloping ramp surface, to one decimal place. (2 marks)
2. (b) A handrail is fitted along both sides of the slope. Find the total length of handrail needed. (1 mark)
3.9 m; 7.8 m

(a) $\sqrt{1.5^{2} + 3.6^{2}} = \sqrt{15.21} =$ 3.9 m.
(b) Two sides: $2 \times 3.9 =$ 7.8 m.
Level 1 · Fluency

Two similar rectangles have a scale factor of 3. If a side of the small one is 4 cm, find the matching side of the large one.

Level 2 · Application

A model car is built to a scale of 1 : 20. The model is 22 cm long. Find the real length of the car in metres.
Level 3 · Further Application
Two similar triangles have a scale factor of 2.5 (large : small).
1. (a) A side of the small triangle is 6 cm. Find the matching side of the large triangle. (1 mark)
2. (b) State the ratio of their areas. (2 marks)
6.25 : 1

(a) $6 \times 2.5 = 15$ cm.
(b) Area scale factor = $(2.5)^{2} = 6.25$, so the areas are in the ratio 6.25 : 1.
Two similar rectangles have a scale factor of 3 (large : small).
1. (a) A side of the small rectangle is 8 cm. Find the matching side of the large rectangle. (1 mark)
2. (b) State the ratio of their areas. (2 marks)
9 : 1

(a) $8 \times 3 = 24$ cm.
(b) Area scale factor = $3^{2} = 9$, so the areas are in the ratio 9 : 1.
Two similar figures have areas in the ratio 16 : 1 (large : small).
1. (a) Find the length scale factor. (1 mark)
2. (b) The small figure has a perimeter of 10 cm. Find the perimeter of the large figure. (2 marks)
Scale factor 4; 40 cm

(a) Length scale factor = $\sqrt{16} =$ 4.
(b) Perimeter scales by the length factor: $10 \times 4 =$ 40 cm.
Level 1 · Fluency

Find the surface area of a cube with side 5 cm.

Level 2 · Application

Calculate the surface area of a closed cylinder with radius 4 cm and height 10 cm, correct to two decimal places. ($SA = 2\pi r^{2} + 2\pi rh$.)
Level 3 · Further Application
A solid is a cube of side 6 cm with a triangular prism (two triangular faces of base 6 cm and height 5 cm, and length 6 cm) sitting exactly on top.
1. (a) Explain which cube face is hidden and does not count toward the surface area. (1 mark)
2. (b) Calculate the combined area of the two triangular faces. (2 marks)
30 cm²

(a) The top face of the cube is covered by the prism, so it is not part of the outer surface.
(b) Two triangles: $2 \times (\tfrac{1}{2} \times 6 \times 5) = 2 \times 15 =$ 30 cm².
A closed box is a rectangular prism 12 cm $\times$ 8 cm $\times$ 5 cm.
1. (a) Calculate its total surface area. (2 marks)
2. (b) If painting costs $0.02 per cm², find the cost to paint the whole outside. (1 mark)
392 cm²; $7.84

(a) $SA = 2(12{\times}8 + 12{\times}5 + 8{\times}5) = 2(96 + 60 + 40) = 2 \times 196 =$ 392 cm².
(b) $392 \times 0.02 =$ $7.84.
An open-top cylindrical tin (no lid) has radius 6 cm and height 15 cm.
1. (a) Calculate the total surface area, to two decimal places. (2 marks)
2. (b) Explain why only one circle is included in the calculation. (1 mark)
678.58 cm²

(a) $SA = \pi r^{2} + 2\pi rh = \pi(36) + 2\pi(6)(15) = 36\pi + 180\pi = 216\pi =$ 678.58 cm².
(b) The tin is open at the top, so it has only one circular end (the base) instead of two.
Level 1 · Fluency

Find the volume of a rectangular prism 4 cm $\times$ 3 cm $\times$ 10 cm.

Level 2 · Application

A tank is the bottom half of a sphere of radius 3 m. Calculate its volume, to one decimal place. ($V$ of sphere = $\tfrac{4}{3}\pi r^{3}$.)
Level 3 · Further Application
A cylindrical drink container has radius 5 cm and height 20 cm.
1. (a) Calculate its volume, to the nearest cm³. (2 marks)
2. (b) State its capacity in litres (1000 cm³ = 1 L), to two decimal places. (1 mark)
1571 cm³; 1.57 L

(a) $V = \pi r^{2}h = \pi \times 25 \times 20 =$ 1571 cm³.
(b) $1571 \div 1000 =$ 1.57 L.
A rectangular fish tank measures 50 cm $\times$ 30 cm $\times$ 40 cm.
1. (a) Calculate its volume in cubic centimetres. (2 marks)
2. (b) State its capacity in litres (1000 cm³ = 1 L). (1 mark)
60 000 cm³; 60 L

(a) $V = 50 \times 30 \times 40 =$ 60 000 cm³.
(b) $60\,000 \div 1000 =$ 60 L.
A spherical ball has radius 9 cm.
1. (a) Calculate its volume, to the nearest cm³. ($V = \tfrac{4}{3}\pi r^{3}$.) (2 marks)
2. (b) State its capacity in litres (1000 cm³ = 1 L), to two decimal places. (1 mark)
3054 cm³; 3.05 L

(a) $V = \tfrac{4}{3}\pi \times 9^{3} = \tfrac{4}{3}\pi \times 729 = 972\pi =$ 3054 cm³.
(b) $3054 \div 1000 =$ 3.05 L.
Level 1 · Fluency

How many millilitres are in 1 litre?

Level 2 · Application

A container holds 3.5 L. Convert this to cubic centimetres (1 L = 1000 cm³).
Level 3 · Further Application
A fish tank measures 40 cm $\times$ 25 cm $\times$ 30 cm (internal).
1. (a) Calculate its volume in cubic centimetres. (1 mark)
2. (b) State its capacity in litres. (2 marks)
30 litres

(a) $40 \times 25 \times 30 = 30\,000$ cm³.
(b) $30\,000 \div 1000 =$ 30 litres.
A storage bin measures 60 cm $\times$ 40 cm $\times$ 50 cm (internal).
1. (a) Calculate its volume in cubic centimetres. (1 mark)
2. (b) State its capacity in litres. (2 marks)
120 litres

(a) $60 \times 40 \times 50 = 120\,000$ cm³.
(b) $120\,000 \div 1000 =$ 120 litres.
A water trough measures 1.5 m $\times$ 0.8 m $\times$ 0.5 m (internal).
1. (a) Calculate its volume in cubic metres. (1 mark)
2. (b) State its capacity in litres (1 m³ = 1000 L). (2 marks)
0.6 m³; 600 litres

(a) $1.5 \times 0.8 \times 0.5 =$ 0.6 m³.
(b) $0.6 \times 1000 =$ 600 litres.
Level 1 · Fluency

To find the area of an L-shaped block, what is a useful first step?

Level 2 · Application

An L-shaped block is a 20 m $\times$ 15 m rectangle with a 8 m $\times$ 6 m rectangle removed from one corner. Find its area.
Level 3 · Further Application
A block is made of a 30 m $\times$ 18 m rectangle with a right-angled triangular section (base 30 m, height 8 m) attached along one long side.
1. (a) Find the total area. (2 marks)
2. (b) If turf costs $9/m², find the cost to turf the whole block. (1 mark)
660 m²; $5940

(a) Rectangle 540 m² + triangle $\tfrac{1}{2} \times 30 \times 8 = 120$ m² → 660 m².
(b) $660 \times 9 =$ $5940.
A garden bed is a trapezium with parallel sides 20 m and 14 m and a perpendicular width of 10 m.
1. (a) Find its area. (2 marks)
2. (b) If turf costs $12/m², find the cost to turf the bed. (1 mark)
170 m²; $2040

(a) $A = \tfrac{1}{2}(20+14) \times 10 = \tfrac{1}{2} \times 34 \times 10 =$ 170 m².
(b) $170 \times 12 =$ $2040.
A block is a 25 m $\times$ 12 m rectangle with a right-angled triangle (base 12 m, height 9 m) cut off one corner.
1. (a) Find the remaining area. (2 marks)
2. (b) If turf costs $8/m², find the cost to turf the block. (1 mark)
246 m²; $1968

(a) Rectangle $25 \times 12 = 300$ m²; triangle $\tfrac{1}{2} \times 12 \times 9 = 54$ m²; remaining $300 - 54 =$ 246 m².
(b) $246 \times 8 =$ $1968.
Level 1 · Fluency

The trapezoidal rule for one application is $A \approx \tfrac{h}{2}(a + b)$. What shape is being used to approximate the area?

Level 2 · Application

Explain how the area of a trapezium leads to the rule $A \approx \tfrac{h}{2}(d_{l} + d_{r})$.
Level 3 · Further Application
A river’s width is measured as 12 m at one point and 18 m at another point 25 m downstream.
1. (a) Use one application of the trapezoidal rule to estimate the surface area of that stretch. (2 marks)
2. (b) State one assumption this estimate makes. (1 mark)
375 m²

(a) $A \approx \tfrac{25}{2}(12 + 18) = 12.5 \times 30 =$ 375 m².
(b) It assumes the banks are straight (change uniformly) between the two measurements.
A field boundary is measured as 20 m at one point and 30 m at a point 16 m further along.
1. (a) Use one application of the trapezoidal rule to estimate the area of the strip. (2 marks)
2. (b) State one assumption this estimate makes. (1 mark)
400 m²

(a) $A \approx \tfrac{16}{2}(20 + 30) = 8 \times 50 =$ 400 m².
(b) It assumes the boundary runs straight between the two measured offsets.
A plot has parallel boundaries of 45 m and 55 m, taken 24 m apart.
1. (a) Use one application of the trapezoidal rule to estimate the area. (2 marks)
2. (b) If the true side boundaries actually curve inward, state whether the estimate is too large or too small. (1 mark)
1200 m²; too large

(a) $A \approx \tfrac{24}{2}(45 + 55) = 12 \times 100 =$ 1200 m².
(b) If the boundaries curve inward, the straight-line estimate encloses extra area, so it is too large (an over-estimate).
Level 1 · Fluency

For two applications with equal strip width $h$, the rule is $A \approx \tfrac{h}{2}(d_{0} + 2d_{1} + d_{2})$. Which measurement is doubled?

Level 2 · Application

A garden has offsets 30 m, 24 m and 18 m, each 15 m apart. Use two applications of the trapezoidal rule to estimate its area.
Level 3 · Further Application
A lake is surveyed with offsets 0 m, 14 m, 22 m, 16 m and 0 m taken every 10 m across its length.
1. (a) Use the trapezoidal rule (four applications) to estimate the surface area. (2 marks)
2. (b) State whether the estimate is likely to be an over- or under-estimate if the shoreline curves outward. (1 mark)
520 m²

(a) $A \approx \tfrac{10}{2}(0 + 2\times 14 + 2\times 22 + 2\times 16 + 0) = 5 \times (28 + 44 + 32) = 5 \times 104 =$ 520 m².
(b) If the shoreline bulges outward beyond the straight segments, the rule under-estimates the true area.
A pond is surveyed with offsets 0 m, 6 m, 10 m, 8 m and 0 m taken every 5 m across its length.
1. (a) Use the trapezoidal rule (four applications) to estimate the surface area. (2 marks)
2. (b) State whether the estimate is an over- or under-estimate if the shoreline curves outward. (1 mark)
120 m²

(a) $A \approx \tfrac{5}{2}(0 + 2\times 6 + 2\times 10 + 2\times 8 + 0) = 2.5 \times (12 + 20 + 16) = 2.5 \times 48 =$ 120 m².
(b) If the shoreline bulges outward, the rule under-estimates the true area.
A pool is 4 m wide throughout. Its depth is measured every 3 m along its length as 1.0 m, 1.4 m, 1.8 m and 1.2 m.
1. (a) Use the trapezoidal rule to estimate the area of the vertical cross-section. (2 marks)
2. (b) Estimate the volume of water by multiplying by the width. (1 mark)
12.9 m²; 51.6 m³

(a) $A \approx \tfrac{3}{2}(1.0 + 2\times 1.4 + 2\times 1.8 + 1.2) = 1.5 \times 8.6 =$ 12.9 m².
(b) Volume $= 12.9 \times 4 =$ 51.6 m³.
Level 1 · Fluency

A cube has side 10 cm. State both its volume and its surface area.

Level 2 · Application

A cylindrical water tank has radius 1.5 m and height 2 m. Calculate its capacity in litres, to the nearest litre. (1 m³ = 1000 L.)
Level 3 · Further Application
A rectangular pool is 8 m $\times$ 4 m with a uniform depth of 1.5 m.
1. (a) Calculate the volume of water when full, in cubic metres. (1 mark)
2. (b) If water costs $2.30 per kilolitre (1 kL = 1 m³), find the cost to fill the pool. (2 marks)
$110.40

(a) $8 \times 4 \times 1.5 = 48$ m³.
(b) 48 kL $\times$ $2.30 = $110.40.
A rectangular pool is 10 m $\times$ 5 m with a uniform depth of 1.2 m.
1. (a) Calculate the volume of water when full, in cubic metres. (1 mark)
2. (b) If water costs $2.10 per kilolitre (1 kL = 1 m³), find the cost to fill the pool. (2 marks)
$126.00

(a) $10 \times 5 \times 1.2 = 60$ m³.
(b) 60 kL $\times$ $2.10 = $126.00.
A cylindrical water tank has radius 2 m and height 3 m.
1. (a) Calculate its volume, to one decimal place. (2 marks)
2. (b) State its capacity in kilolitres (1 kL = 1 m³). (1 mark)
37.7 m³; 37.7 kL

(a) $V = \pi \times 2^{2} \times 3 = 12\pi =$ 37.7 m³.
(b) 1 kL = 1 m³, so the capacity is 37.7 kL.
Level 1 · Fluency

Convert 2500 g to kilograms.

Level 2 · Application

A truck carries 3.4 tonnes of gravel. Convert this to kilograms.
Level 3 · Further Application
A pallet holds 48 bags of cement, each 25 kg.
1. (a) Find the total mass in kilograms. (1 mark)
2. (b) Express the total mass in tonnes. (2 marks)
1.2 tonnes

(a) $48 \times 25 = 1200$ kg.
(b) $1200 \div 1000 =$ 1.2 tonnes.
A truck is loaded with 60 bags of sand, each 40 kg.
1. (a) Find the total mass in kilograms. (1 mark)
2. (b) Express the total mass in tonnes. (2 marks)
2.4 tonnes

(a) $60 \times 40 = 2400$ kg.
(b) $2400 \div 1000 =$ 2.4 tonnes.
A delivery of soil weighs 3.5 tonnes.
1. (a) Convert the total mass to kilograms. (1 mark)
2. (b) Find how many 50 kg sacks it fills. (2 marks)
3500 kg; 70 sacks

(a) $3.5 \times 1000 =$ 3500 kg.
(b) $3500 \div 50 =$ 70 sacks.
Level 1 · Fluency

How many joules are in 1 kilojoule?

Level 2 · Application

A snack contains 220 Calories (kilocalories). Convert this to kilojoules (1 Cal $\approx$ 4.184 kJ), to the nearest kJ.
Level 3 · Further Application
A drink is labelled 630 kJ.
1. (a) Convert this to Calories (kilocalories), to the nearest Calorie (1 Cal $\approx$ 4.184 kJ). (2 marks)
2. (b) A person’s daily intake is about 8700 kJ. State this drink as a percentage of that intake, to one decimal place. (1 mark)
151 Cal; 7.2%

(a) $630 \div 4.184 =$ 151 Cal.
(b) $\dfrac{630}{8700} \times 100 =$ 7.2%.
A snack is labelled 850 kJ.
1. (a) Convert this to Calories (kilocalories), to the nearest Calorie (1 Cal $\approx$ 4.184 kJ). (2 marks)
2. (b) State this snack as a percentage of an 8700 kJ daily intake, to one decimal place. (1 mark)
203 Cal; 9.8%

(a) $850 \div 4.184 = 203.2 \approx$ 203 Cal.
(b) $\dfrac{850}{8700} \times 100 =$ 9.8%.
A meal is labelled 2100 kJ.
1. (a) Convert this to Calories (kilocalories), to the nearest Calorie (1 Cal $\approx$ 4.184 kJ). (2 marks)
2. (b) State this meal as a percentage of an 8700 kJ daily intake, to one decimal place. (1 mark)
502 Cal; 24.1%

(a) $2100 \div 4.184 = 501.9 \approx$ 502 Cal.
(b) $\dfrac{2100}{8700} \times 100 =$ 24.1%.
Level 1 · Fluency

Food A has 500 kJ per serve and food B has 800 kJ per serve. Which is more energy-dense per serve?

Level 2 · Application

A cereal provides 1600 kJ per 100 g. Calculate the energy in a 45 g serve.
Level 3 · Further Application
Two snacks are compared: Snack A is 1500 kJ per 100 g; Snack B is 1900 kJ per 100 g.
1. (a) Find the energy in a 30 g serve of each. (2 marks)
2. (b) State which snack is lower in energy for the same serve size. (1 mark)
(a) A: $1500 \times 0.30 = 450$ kJ. B: $1900 \times 0.30 = 570$ kJ.
(b) Snack A is lower in energy per 30 g serve.
A cereal is 1500 kJ per 100 g. A serve is 40 g eaten with 125 mL of milk that provides 250 kJ.
1. (a) Find the energy from the cereal alone. (2 marks)
2. (b) Find the total energy of the serve including the milk. (1 mark)
600 kJ; 850 kJ

(a) $1500 \times \dfrac{40}{100} =$ 600 kJ.
(b) $600 + 250 =$ 850 kJ.
Two drinks are compared: Drink A is 180 kJ per 100 mL; Drink B is 150 kJ per 100 mL.
1. (a) Find the energy in a 375 mL can of each. (2 marks)
2. (b) State which drink is lower in energy for the same can size. (1 mark)
A: 675 kJ, B: 562.5 kJ; Drink B is lower

(a) A: $180 \times \dfrac{375}{100} = 675$ kJ; B: $150 \times \dfrac{375}{100} = 562.5$ kJ.
(b) Drink B is lower in energy per can.
Level 1 · Fluency

Walking uses about 20 kJ per minute. How much energy is used in 15 minutes?

Level 2 · Application

Cycling uses about 30 kJ per minute. Calculate the energy used in a 40-minute ride.
Level 3 · Further Application
A person burns about 25 kJ per minute jogging.
1. (a) Calculate the energy used in a 35-minute jog. (1 mark)
2. (b) A muesli bar contains 700 kJ. Determine how long they must jog to burn off one bar, to the nearest minute. (2 marks)
28 minutes

(a) $25 \times 35 = 875$ kJ.
(b) $700 \div 25 =$ 28 minutes.
A person burns about 40 kJ per minute skipping.
1. (a) Calculate the energy used in a 20-minute session. (1 mark)
2. (b) A chocolate bar contains 1000 kJ. Determine how long they must skip to burn off one bar, to the nearest minute. (2 marks)
25 minutes

(a) $40 \times 20 = 800$ kJ.
(b) $1000 \div 40 =$ 25 minutes.
A person burns about 24 kJ per minute walking briskly.
1. (a) Calculate the energy used in a 30-minute walk. (1 mark)
2. (b) A muffin contains 600 kJ. Determine how long they must walk to burn off the muffin, to the nearest minute. (2 marks)
25 minutes

(a) $24 \times 30 = 720$ kJ.
(b) $600 \div 24 =$ 25 minutes.
Level 1 · Fluency

A 2 kW heater runs for 3 hours. How many kilowatt-hours does it use?

Level 2 · Application

An electricity bill shows 1240 kWh used over a 91-day quarter at 26.4c/kWh, plus a 98c per day supply charge. Calculate the total bill.
Level 3 · Further Application
A household runs a 1.5 kW appliance for 4 hours a day.
1. (a) Calculate the energy used over 30 days, in kWh. (1 mark)
2. (b) At 28c/kWh, calculate the cost for the 30 days. (2 marks)
$50.40

(a) $1.5 \times 4 \times 30 = 180$ kWh.
(b) $180 \times 0.28 =$ $50.40.
A household runs a 2 kW heater for 5 hours a day.
1. (a) Calculate the energy used over 30 days, in kWh. (1 mark)
2. (b) At 32c/kWh, calculate the cost for the 30 days. (2 marks)
$96.00

(a) $2 \times 5 \times 30 = 300$ kWh.
(b) $300 \times 0.32 =$ $96.00.
A household runs a 0.9 kW appliance for 6 hours a day.
1. (a) Calculate the energy used over 30 days, in kWh. (1 mark)
2. (b) At 27c/kWh, calculate the cost for the 30 days. (2 marks)
$43.74

(a) $0.9 \times 6 \times 30 = 162$ kWh.
(b) $162 \times 0.27 =$ $43.74.