Lemmara logoLemmaraWorked solution

4024/12 May/Jun 2026, Question 17: Pythagoras and quadratics

Cambridge O Level Mathematics 4024 · Paper 12 (Non-calculator) · Non-calculator · 9 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

A right-angled triangle has base x cm, height x − 7 cm and hypotenuse x + 2 cm. Form an equation that simplifies to x² − 18x + 45 = 0, solve it by factorising, and find the triangle's area.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 4 marks

Form the equation and simplify it.

  1. 1Height = x − 7, hypotenuse = x + 2
  2. 2x² + (x − 7)² = (x + 2)²
  3. 3x² + x² − 14x + 49 = x² + 4x + 4
  4. 4x² − 18x + 45 = 0

Answerx² + (x − 7)² = (x + 2)², x² + x² − 14x + 49 = x² + 4x + 4, leading to x² − 18x + 45 = 0

Part (b) · 3 marks

Solve x² − 18x + 45 = 0 by factorising.

  1. 1Two numbers that multiply to 45 and add to −18: −3 and −15
  2. 2(x − 3)(x − 15) = 0
  3. 3x = 3 or x = 15

Answerx = 3 or x = 15

Part (c) · 2 marks

Find the area of the triangle.

  1. 1x = 3 gives height −4, so x = 15
  2. 2Base 15 cm, height 15 − 7 = 8 cm
  3. 3Area = 1/2 × 15 × 8 = 60 cm²

Answer60

Stuck on a step? Read the free lesson: Equations That Become Quadratics

Practise more pythagoras and quadratics questions

Real 4024 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on pythagoras and quadratics

Also searched as: 4024/12/M/J/26 · 4024_s26_qp_12 · 4024 s26 12 · question 17.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.