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GCSE Maths Revision Summary

Similar Shapes & Triangles — Key Notes & Worked Examples (3 Steps)

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Similar Shapes & Triangles — GCSE Maths Tutorial, Worked Examples & Practice

Revise Similar Shapes & Triangles (linear scale factors, missing lengths in similar polygons and overlapping similar triangles) within Geometry and Measures — 2D Shapes, aligned to Edexcel, AQA & OCR GCSE (9–1) and International GCSE Maths. The interactive practice tool at the top of this page includes 8 self-marking practice questions across 3 structured skill steps, with instant diagnostic feedback and full worked solutions.

Step 1 of 33 interactive practice questions

Linear Scale Factors in Similar Shapes

Finding the linear scale factor k and calculating missing corresponding lengths

Key Method & Revision Notes

  • Two 2D shapes are mathematically similar if all corresponding angles are equal and all corresponding side lengths are in the same ratio.
  • Step 1 — Find the linear scale factor ($k$): Choose a pair of corresponding known sides:
  • $$k = \frac{\text{Length on new (larger) shape}}{\text{Corresponding length on old (smaller) shape}}$$
  • Step 2 — Multiply or Divide: Multiply by $k$ to find a side on the larger shape; divide by $k$ to find a side on the smaller shape. Perimeters also scale by the exact same factor $k$.

Worked Example

Question:

Trapezia $A$ and $B$ are mathematically similar. Work out the length $x$.

Step-by-Step Solution:

1. Match the two known bases: $k = \dfrac{15}{6} = 2.5$.

2. Multiply the corresponding slanted side on $A$ by $k$: $x = 5 \times 2.5 = 12.5\text{ cm}$.

Ready to test this skill? Use the interactive tool at the top of the page to practise 3 self-marking exam-style questions (8 questions loaded in this topic module).

Step 2 of 33 interactive practice questions

Overlapping Similar Triangles (Parallel Line Inside)

Separating small and whole triangles when DE is parallel to BC

Key Method & Revision Notes

  • Whenever a line $DE$ is drawn parallel to the base $BC$ inside triangle $ABC$, corresponding angles are equal ($\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$), so the small top triangle $\triangle ADE$ is mathematically similar to the whole large triangle $\triangle ABC$.
  • Sketch the two triangles separately! Label the whole side of $\triangle ABC$ as $AB = AD + DB$ (and $AC = AE + EC$).
  • Never divide by the bottom strip $DB$ alone — $DB$ is part of a trapezium, not a side of a triangle!

Worked Example

Question:

In $\triangle ABC$, $DE$ is parallel to $BC$, with $AD = 4\text{ cm}$, $DB = 6\text{ cm}$, and $DE = 5\text{ cm}$. Work out the length of $BC$.

Step-by-Step Solution:

1. Whole side of $\triangle ABC$: $AB = 4 + 6 = 10\text{ cm}$ (do not use $6\text{ cm}$!).

2. Scale factor from $\triangle ADE$ to $\triangle ABC$: $k = \dfrac{10}{4} = 2.5$.

3. $BC = 5 \times 2.5 = 12.5\text{ cm}$.

Ready to test this skill? Use the interactive tool at the top of the page to practise 3 self-marking exam-style questions (8 questions loaded in this topic module).

Step 3 of 32 interactive practice questions

Alternate ('Bow-Tie') Similar Triangles & Proof

Matching alternate-angle vertices in hourglass triangles and writing formal similarity proofs

Key Method & Revision Notes

  • When two parallel lines $AB$ and $DE$ are joined by two crossing lines meeting at $C$ (a "bow-tie" / hourglass pair of triangles):
  • Match alternate ("Z") angles carefully! Along the straight line $ACE$, $\angle CAB = \angle CED$ (alternate angles), so side $AC$ corresponds to $EC$ (on the same straight line through $C$), not the side directly underneath it!
  • Proving two triangles are similar (AAA): State at least two pairs of equal angles with full geometric reasons (alternate angles and vertically opposite angles are equal).

Worked Example

Question:

In the diagram, $AB$ is parallel to $DE$ and $AE$ meets $BD$ at $C$. Work out the length of $CE$.

Step-by-Step Solution:

1. Parallel bases give scale factor $k = \dfrac{9}{6} = 1.5$.

2. Along straight line $ACE$, alternate angles match $AC$ ($7.5\text{ cm}$) with $EC$.

3. Going to the smaller triangle, divide by $1.5$: $CE = 7.5 \div 1.5 = 5\text{ cm}$.

Ready to test this skill? Use the interactive tool at the top of the page to practise 2 self-marking exam-style questions (8 questions loaded in this topic module).