IGCSE · Practice questions · Angles in polygons

One full turn.

Six original Cambridge-style questions on interior and exterior angles of polygons, ending with a tiling problem.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Exterior first, then interior.

Angle sum: \((n - 2) \times 180^\circ\).
Exterior sum: always \(360^\circ\).
Regular: exterior \(= 360^\circ \div n\), interior \(= 180^\circ - \text{exterior}\).
Number of sides: \(n = 360^\circ \div \text{exterior}\); it must be a whole number.
Shapes at a point: interior angles add to \(360^\circ\).

Recall

[2 marks]

(a) Find the sum of the interior angles of an octagon. (b) Find each interior angle of a regular octagon.

(a) \((8 - 2) \times 180^\circ = 1080^\circ\). ✓
(b) \(1080^\circ \div 8 = 135^\circ\) (or \(180^\circ - 360^\circ \div 8 = 135^\circ\)). ✓

Calculation

[3 marks]

Each exterior angle of a regular polygon is \(20^\circ\). Find (a) the number of sides (b) each interior angle (c) the sum of the interior angles.

(a) \(360 \div 20 = 18\) sides. ✓
(b) \(180^\circ - 20^\circ = 160^\circ\). ✓
(c) \((18 - 2) \times 180^\circ = 2880^\circ\) (check: \(18 \times 160 = 2880\)). ✓

Analysis

[3 marks]

Show that a regular polygon can have interior angles of \(140^\circ\) but not \(145^\circ\).

\(140^\circ\): exterior \(= 40^\circ\), \(n = 360 \div 40 = 9\), a whole number, so a regular nonagon works. ✓
\(145^\circ\): exterior \(= 35^\circ\), \(n = 360 \div 35 = 10.28...\), not a whole number, so it is impossible. ✓

Calculation

[3 marks]

(Extended) The interior angles of a hexagon are \(130^\circ\), \(125^\circ\), \(140^\circ\), \(85^\circ\), \(x\) and \(2x\). Find the largest angle.

Hexagon angle sum \(= 4 \times 180^\circ = 720^\circ\). ✓
\(130 + 125 + 140 + 85 + 3x = 720\), so \(480 + 3x = 720\) and \(x = 80\). ✓
The angles \(x\) and \(2x\) are \(80^\circ\) and \(160^\circ\); the largest is \(160^\circ\). ✓

Proof

[3 marks]

Explain why the interior angles of a polygon with \(n\) sides add up to \((n - 2) \times 180^\circ\).

Join one vertex to every other vertex that is not next to it. ✓
This splits the polygon into \(n - 2\) triangles (each side except the two at the chosen vertex forms one triangle). ✓
The triangles' angles together make up exactly the polygon's interior angles, and each triangle contributes \(180^\circ\), so the sum is \((n - 2) \times 180^\circ\). ✓

Analysis

[4 marks]

A floor is tiled with regular octagons and squares, with two octagons and one square meeting at every corner.

(a) Find the interior angle of a regular octagon. (b) Show that two octagons and one square fit exactly round a point. (c) Explain why regular pentagons alone cannot tile a floor.

(a) Exterior \(= 360 \div 8 = 45^\circ\), so interior \(= 135^\circ\). ✓
(b) \(135^\circ + 135^\circ + 90^\circ = 360^\circ\), exactly the angles at a point, so there is no gap or overlap. ✓
(c) Each interior angle of a regular pentagon is \(108^\circ\). \(360 \div 108 = 3.33...\) is not a whole number, so pentagons leave a gap (\(36^\circ\) with three) or overlap (with four). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.