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MathidiaMathematics Academy
Similarity, congruence and scale
Lesson 50 of 75
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Similarity modelling

Objective. Solve multi-stage geometric scale problems.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Similarity modelling uses chains of similar shapes to solve problems with multiple stages, such as scaling a model up to real life then finding an area or volume, or linking several similar triangles together. To solve these, break the problem into stages: find the scale factor between each pair of similar shapes, then apply ratios or powers (squared for area, cubed for volume) at each stage in turn. For a model car with scale 1:20, a length of 8cm on the model corresponds to 8×20=160cm in real life, and a model surface area would need multiplying by 20²=400 to find the real area. When problems involve several similar shapes in sequence, multiply the scale factors together to find the overall relationship between the first and last shape. Always track units carefully across every stage of a multi-step scaling problem.

Worked example

A model of a statue is built at a scale of 1:25. The real statue has a volume of 50,000 cm³. Find the volume of the model.

125500
Bar chart of the values 1, 25, 50, 0
  1. 1Linear scale factor from real to model is 1/25
  2. 2Volume scale factor = (1/25)³ = 1/15625
  3. 3Model volume = 50,000 ÷ 15625
  4. 4Calculate: 50,000 ÷ 15625 = 3.2 cm³

Answer: 3.2 cm³

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Common mistake. Pupils apply a linear scale factor to a volume calculation in a multi-stage problem without cubing it at the correct stage, mixing up when to use the factor directly and when to cube it. For a scale factor of 5 applied to volume, they forget to cube it, using 5 instead of 125. Always decide whether each stage concerns length, area or volume before applying the factor.

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