Edexcel International GCSE Maths Geometric Proof Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Geometry and Trigonometry

Geometrical Reasoning and Proof: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every geometrical reasoning and proof question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 3 questions worth 9 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

13 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q15

Using algebra, prove that, given any 3 consecutive whole numbers, the sum of the square of the smallest number and the square of the largest number is always 2 more than twice the square of the middle number.

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  1. Let the numbers be n − 1, n and n + 1 (so n is the middle number).
  2. (n − 1)2 + (n + 1)2 = n2 − 2n + 1 + n2 + 2n + 1 = 2n2 + 2
  3. 2n2 + 2 is 2 more than twice the square of the middle number.

Answer: (n − 1)2 + (n + 1)2 = 2n2 + 2

23 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q17

Using algebra, prove that, given any 3 consecutive even numbers, the difference between the square of the largest number and the square of the smallest number is always 8 times the middle number.

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  1. Let the three consecutive even numbers be 2n, 2n + 2 and 2n + 4, where n is an integer.
  2. (2n + 4)2 − (2n)2 = 4n2 + 16n + 16 − 4n2 = 16n + 16
  3. Factorise: 16n + 16 = 8(2n + 2), and 2n + 2 is the middle number.

Answer: (2n + 4)2 − (2n)2 = 8(2n + 2), which is 8 × the middle number

33 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q17

Prove that, for any three numbers which are consecutive multiples of 4, the difference between the square of the largest number and the square of the smallest number is always a multiple of 64. Show clear algebraic working.

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  1. Let the three consecutive multiples of 4 be 4n, 4n + 4 and 4n + 8, where n is an integer.
  2. Square of the largest: (4n + 8)2 = 16n2 + 64n + 64
  3. Square of the smallest: (4n)2 = 16n2
  4. Difference: 16n2 + 64n + 64 − 16n2 = 64n + 64
  5. Factorise: 64n + 64 = 64(n + 1). Since n + 1 is an integer, this is always a multiple of 64.

Answer: (4n + 8)2 − (4n)2 = 64(n + 1), a multiple of 64

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