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How do you use similar shapes to find missing lengths, areas and volumes?

Using similarity to find missing lengths in similar shapes, and applying the linear, area and volume scale factors between similar figures.

A focused answer to the SQA National 5 Mathematics similarity content, covering similar shapes and the linear scale factor, finding missing lengths in similar triangles, and the area and volume scale factors as the square and cube of the linear scale factor.

Generated by Claude Opus 4.89 min answer

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

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  1. What this dot point is asking
  2. Similar shapes and the linear scale factor
  3. Similar triangles
  4. Area and volume scale factors
  5. Examples in context
  6. Try this

What this dot point is asking

The SQA wants you to recognise similar shapes, use the linear scale factor to find missing lengths in similar figures (especially similar triangles), and apply the area scale factor (the square of the linear scale factor) and the volume scale factor (the cube) to find missing areas and volumes.

Similar shapes and the linear scale factor

Similar shapes have the same shape but not necessarily the same size: their corresponding angles are equal and their corresponding sides are in the same ratio. That common ratio is the linear scale factor.

Similar triangles

Triangles are similar if their angles match (for example, when a line is drawn parallel to one side, or in nested triangles sharing an angle). Set up the ratio of corresponding sides and solve.

Area and volume scale factors

When you enlarge a shape, area grows faster than length, and volume faster still, because area depends on two dimensions and volume on three.

These factors also work in reverse. If you are told the areas and asked for the linear scale factor, take the square root of the area scale factor; for volumes, take the cube root.

Examples in context

Scale factors explain why size changes have outsized effects. A model car at 120\frac{1}{20} scale has 1400\frac{1}{400} of the surface area to paint and 18000\frac{1}{8000} of the volume of material. In cooking, doubling every length of a tin multiplies its capacity by 88, so the recipe must be scaled accordingly. Map and plan reading uses the linear scale factor directly to convert between drawing and real distances.

Try this

Q1. Two similar triangles have bases 55 cm and 2020 cm. Find the scale factor. [1 mark]

  • Cue. 205=4\dfrac{20}{5} = 4.

Q2. Two similar shapes have lengths in ratio 1:21:2. The smaller has area 77 cm2^2. Find the larger area. [2 marks]

  • Cue. 7×22=287 \times 2^2 = 28 cm2^2.

Q3. Two similar solids have lengths in ratio 1:31:3. The smaller has volume 55 cm3^3. Find the larger volume. [2 marks]

  • Cue. 5×33=1355 \times 3^3 = 135 cm3^3.

Exam-style practice questions

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

SQA National 5 20183 marksTwo triangles are similar. The smaller has a base of 66 cm; the larger has a corresponding base of 1515 cm. If a side of the smaller is 88 cm, find the corresponding side of the larger.
Show worked answer →

Find the linear scale factor from the bases: 156=2.5\dfrac{15}{6} = 2.5 (1 mark). Multiply the known side by the scale factor: 8×2.5=208 \times 2.5 = 20 cm (2 marks). Markers reward the scale factor and the enlarged side. Always divide the larger by the smaller to enlarge.

SQA National 5 20223 marksTwo similar bottles have heights 1010 cm and 1515 cm. The smaller holds 400400 ml. Calculate the volume of the larger bottle.
Show worked answer →

The linear scale factor is 1510=1.5\dfrac{15}{10} = 1.5 (1 mark). Volume scales by the cube of the linear scale factor: 1.53=3.3751.5^3 = 3.375 (1 mark). Multiply: 400×3.375=1350400 \times 3.375 = 1350 ml (1 mark). Markers reward cubing the linear scale factor and applying it to the volume.

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