Skip to main content
EnglandProduct Design and TechnologiesSyllabus dot point

How are scales, ratios and geometry used in technical drawing and to size products?

Working with scale and ratio in drawings and models, reading and using scales (for example 1:2, 1:5, 1:10 and enlargement scales), calculating areas and volumes for material estimation, using trigonometry and geometry to find lengths and angles, surface area and capacity calculations, and converting between units in a design context.

A focused answer to the Edexcel 9DT0 content on scale, ratio and geometry, covering reading and using drawing scales, ratio and proportion, area and volume for material estimation, trigonometry for lengths and angles, surface area and capacity, and unit conversion.

Generated by Claude Opus 4.813 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

Edexcel wants you to work with scale and ratio in drawings and models, read and use scales, calculate areas and volumes for material estimation, use trigonometry and geometry to find lengths and angles, work out surface area and capacity, and convert between units in a design context.

The answer

Scale and ratio in drawings

Ratio, proportion and scaling models

Ratio sets the relative sizes of parts (a 2 ⁣:12 \colon 1 ratio means one part is twice another) and is used in proportion and in scaling a whole model up or down by a single factor. Note that scaling a length by a factor kk scales area by k2k^2 and volume by k3k^3, which matters for material and capacity.

Geometry and trigonometry

Area, surface area, volume and capacity

Area estimates how much sheet material to buy and the surface area to finish (paint, coat); volume estimates the material mass (volume times density) and the capacity of containers. Convert to consistent units before calculating.

Converting units

Work in one set of units: 1010 mm =1= 1 cm, 100100 cm =1= 1 m, 10001000 mm =1= 1 m. For area and volume, the conversion factor is squared or cubed (11 cm squared =100= 100 mm squared; 11 cm cubed =1000= 1000 mm cubed). Getting units consistent first prevents most errors.

Examples in context

An architect or product designer draws a large item at 1 ⁣:101 \colon 10 or 1 ⁣:1001 \colon 100 to fit the page and a tiny electronic part at 5 ⁣:15 \colon 1 to detail it, reading real sizes off the scale. Material is estimated by area (sheet needed, surface to finish) and by volume (mass from volume times density, and the capacity of bottles and tanks in litres). Trigonometry sizes braces, slopes and frames. Scaling a prototype up reminds the designer that doubling the length multiplies the volume and material eightfold. Using scale and ratio correctly, calculating area, volume and capacity, and converting units consistently, are the maths-in-context skills Edexcel tests most often here.

Try this

Q1. A part is drawn at 1 ⁣:21 \colon 2 and measures 6060 mm on the drawing. What is its real length? [1 mark]

  • Cue. Real length =60×2=120= 60 \times 2 = 120 mm.

Q2. A rectangular sheet is 1.21.2 m by 0.80.8 m. Find its area in square metres. [1 mark]

  • Cue. A=1.2×0.8=0.96A = 1.2 \times 0.8 = 0.96 square metres.

Q3. A cylindrical bottle has radius 33 cm and height 2020 cm. Find its volume and its capacity in litres. [2 marks]

  • Cue. V=πr2h=π×9×20=180π=565V = \pi r^2 h = \pi \times 9 \times 20 = 180\pi = 565 cubic centimetres; capacity =565÷1000=0.57= 565 \div 1000 = 0.57 litres.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20204 marksA component is drawn at a scale of 1:5. On the drawing it measures 36 mm. Calculate the real length, and state the scale needed to draw a 4 mm electronic part clearly.
Show worked answer →

Award marks for the real length and a sensible enlargement scale.

At 1 ⁣:51 \colon 5, the real object is 55 times the drawing, so real length =36×5=180= 36 \times 5 = 180 mm.

A 44 mm part is too small to draw clearly at full size, so an enlargement scale such as 5 ⁣:15 \colon 1 or 10 ⁣:110 \colon 1 is needed: at 5 ⁣:15 \colon 1 the 44 mm part is drawn 2020 mm, and at 10 ⁣:110 \colon 1 it is drawn 4040 mm, large enough to detail.

Markers reward multiplying the drawing length by 5 (180180 mm) and naming an enlargement scale (for example 5 ⁣:15 \colon 1 or 10 ⁣:110 \colon 1) that makes a small part legible.

Edexcel 20226 marksA cylindrical container has a diameter of 80 mm and a height of 150 mm. Calculate its volume in cubic centimetres and its capacity in litres, and calculate the area of material needed for its curved surface.
Show worked answer →

Extended-response calculation marked on levels (correct formulae, working and units).

Radius =802=40= \dfrac{80}{2} = 40 mm =4.0= 4.0 cm; height =150= 150 mm =15= 15 cm.

Volume =πr2h=π×4.02×15=π×16×15=240π=754= \pi r^2 h = \pi \times 4.0^2 \times 15 = \pi \times 16 \times 15 = 240\pi = 754 cubic centimetres (to 3 s.f.).

Capacity: 11 litre =1000= 1000 cubic centimetres, so 754÷1000=0.75754 \div 1000 = 0.75 litres.

Curved surface area =2πrh=2π×4.0×15=120π=377= 2\pi r h = 2\pi \times 4.0 \times 15 = 120\pi = 377 square centimetres.

A strong answer converts units first, uses πr2h\pi r^2 h for volume and 2πrh2\pi r h for the curved surface, and gives capacity in litres (0.750.75 L), with correct units throughout.

Related dot points

Sources & how we know this