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

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

Polygons: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every polygons question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 9 questions worth 40 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.

14 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q3

AB and BC are two sides of a regular polygon with n sides. At B there is a third line; the angle between this line and BA is 148° and the angle between this line and BC is 50°. Work out the value of n. Show your working clearly.

50°148°BAC
Show solution
  1. Angles around a point add up to 360°, so the interior angle ABC = 360° − 148° − 50° = 162°
  2. Each exterior angle of the polygon = 180° − 162° = 18°
  3. The exterior angles of any polygon add up to 360°, so n = 360 ÷ 18

Answer: n = 20

24 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q4

The diagram shows parts of three regular polygons, A, B and C, meeting at a point. The interior angle of B is 7x°, of C is 3x° and of A is 8x°. Polygon B has n sides. Work out the value of n.

7x°3x°8x°BCA
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  1. The three angles meet at a point, so they add up to 360°: 7x + 3x + 8x = 360
  2. 18x = 360, so x = 20
  3. Interior angle of B = 7 × 20 = 140°, so its exterior angle = 180° − 140° = 40°
  4. The exterior angles of a polygon add up to 360°, so n = 360 ÷ 40

Answer: n = 9

33 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q4

The diagram shows a pentagon with angles x°, 119°, 67°, 135° and a right angle. Work out the value of x.

Show solution
  1. Sum of the interior angles of a pentagon = (5 − 2) × 180° = 540°
  2. The right angle is 90°: x + 119 + 67 + 135 + 90 = 540
  3. x + 411 = 540

Answer: x = 129

44 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q5

In the diagram, ABCDE is a regular pentagon. Angle AEF = 96°. Work out the size of the obtuse angle FED. Show your working clearly.

96°AEDCBF
Show solution
  1. First find the interior angle of a regular pentagon. Sum of interior angles = (5 − 2) × 180° = 540°
  2. All 5 angles are equal, so each interior angle = 540° ÷ 5 = 108°. So angle AED = 108°.
  3. Angles around the point E add up to 360°. The three angles at E are AEF, FED (the obtuse one) and AED.
  4. Angle FED = 360° − 96° − 108°

Answer: 156°

54 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q8

Here is a 10-sided polygon. Nine of its interior angles are 148°, 150°, 150°, 168°, 168°, 134°, 134°, 125° and 125°. At the tenth vertex, the angle x° is marked on the outside of the polygon. Work out the value of x.

148°150°150°168°168°134°134°125°125°x°
Show solution
  1. Sum of interior angles of a 10-sided polygon = (10 − 2) × 180° = 1440°
  2. The nine known angles add to 148 + 2 × 150 + 2 × 168 + 2 × 134 + 2 × 125 = 1302°
  3. The tenth interior angle = 1440° − 1302° = 138°
  4. x is marked outside the polygon, so it is the reflex angle at that vertex: 360° − 138°

Answer: x = 222

Tip: Look carefully at where the angle arc is drawn. Here it is on the outside, so the answer is not 138.
65 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q8

The diagram shows a regular pentagon ABCDE, a regular hexagon CFGHID, and a quadrilateral EDIJ. AEJ and HIJ are straight lines. Work out the size of the angle marked x (angle EJI). Show your working clearly.

ABCDEFGHIJx
Show solution
  1. Interior angle of a regular pentagon = (5 − 2) × 180° ÷ 5 = 108°. Interior angle of a regular hexagon = (6 − 2) × 180° ÷ 6 = 120°.
  2. At D: angles around a point add up to 360°: angle EDI = 360° − 108° − 120° = 132°
  3. At E: AEJ is straight, so angle DEJ = 180° − 108° = 72°
  4. At I: HIJ is straight, so angle DIJ = 180° − 120° = 60°
  5. Angles in the quadrilateral EDIJ add up to 360°: x = 360° − 132° − 72° − 60°

Answer: x = 96°

74 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q15

BCDEFG is a hexagon. Angle GBC = 90°, angle CDE = 57°, angle DEF = 104°, angle EFG = 149° and angle FGB = 95°. AB, BC and CD are three sides of a regular n-sided polygon. Calculate the value of n. Show your working clearly.

ABCDEFG57°104°149°95°
Show solution
  1. Sum of interior angles of a hexagon = (6 − 2) × 180° = 720°
  2. The hexagon’s angle at C = 720° − (90° + 57° + 104° + 149° + 95°) = 720° − 495° = 225°. This is a reflex angle.
  3. The interior angle of the regular polygon at C is the angle on the other side: 360° − 225° = 135°
  4. Exterior angle = 180° − 135° = 45°
  5. n = 360° ÷ 45°

Answer: n = 8

86 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q23

A polygon has n sides, where n > 5. When arranged in order of size, starting with the largest number, the sizes of the interior angles of the polygon, in degrees, are the terms of an arithmetic sequence. The first five terms of this sequence are 177, 175, 173, 171, 169. Find the value of n. Show clear algebraic working.

Show solution
  1. The angles form an arithmetic sequence with a = 177 and d = −2.
  2. Sum of n terms = n⁄2[2 × 177 + (n − 1)(−2)] = n⁄2(356 − 2n) = n(178 − n)
  3. The interior angles of an n-sided polygon add up to 180(n − 2).
  4. So 178n − n2 = 180n − 360, which gives n2 + 2n − 360 = 0
  5. Factorise: (n + 20)(n − 18) = 0, so n = 18 (the number of sides cannot be negative)

Answer: n = 18

96 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q24

A polygon has n sides, where n > 5. The interior angles of the polygon form an arithmetic sequence. The smallest angle of the polygon is 84°. The common difference of the sequence is 4°. Work out the sum of the interior angles of the polygon. Show clear algebraic working.

Show solution
  1. The sum of the interior angles can be found in two ways, so set them equal.
  2. Using the arithmetic series formula with a = 84 and d = 4: S = n⁄2[2 × 84 + (n − 1) × 4] = n⁄2(164 + 4n) = 82n + 2n2
  3. Using the polygon rule: sum of interior angles = 180(n − 2) = 180n − 360
  4. So 2n2 + 82n = 180n − 360, which gives 2n2 − 98n + 360 = 0
  5. Divide by 2: n2 − 49n + 180 = 0, which factorises as (n − 4)(n − 45) = 0
  6. n = 4 or n = 45. We are told n > 5, so n = 45.
  7. Sum = 180 × (45 − 2) = 180 × 43

Answer: 7740°

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