Every scale drawings and bearings question from 14 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 4.3. 14 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/21/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Geometry past paper questions.
0580 Scale Drawings and Bearings past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q11
The scale diagram shows the positions of town A and town B. The scale is 1 cm represents 15 km.
(a) Measure the bearing of town B from town A. (b) Find the actual distance AB.
Show solution
- Put the protractor’s centre on A with 0° on the north line, and measure clockwise round to the line AB.
- Bearing = about 124° (122° to 126° is accepted)
- Measure AB with a ruler: about 7.6 cm.
- 1 cm represents 15 km, so 7.6 cm represents 7.6 × 15 = 114 km.
- Distance = about 114 km (111 to 117 km is accepted)
May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q8
The scale drawing shows the position of a town, T. The scale is 1 centimetre represents 20 kilometres. A village, V, is 94 km from T on a bearing of 112°. Mark the position of V on the map.
Show solution
- Change the real distance to a map distance: 1 cm stands for 20 km, so 94 km is 94 ÷ 20 = 4.7 cm.
- Put the protractor’s centre on T with 0° on the north line.
- Measure 112° clockwise from north and draw a faint line.
- Measure 4.7 cm along that line and mark the point V, 4.7 cm from T on a bearing of 112°
May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q9
The scale drawing shows the positions of three towns, A, B and C. Town B is directly south of town A. The scale is 1 cm represents 40 km.
(a) Measure the bearing of town C from town B.
(b) Work out the actual distance between town A and town B.
(c) Town D is 300 km from town A. Town D is 240 km from town B. Using a ruler and compasses only, construct the position of town D on the scale drawing.
Show solution
- Draw a north line at B. Put the protractor’s centre on B with 0° on north and measure clockwise to the line BC.
- Bearing = about 125°
- Measure AB with a ruler: about 5.5 cm.
- 1 cm stands for 40 km, so 5.5 cm stands for 5.5 × 40 = about 220 km
- Change the real distances to the scale: 300 ÷ 40 = 7.5 cm and 240 ÷ 40 = 6 cm.
- Set your compasses to 7.5 cm and draw an arc centred on A. Every point on this arc is 300 km from A.
- Set your compasses to 6 cm and draw an arc centred on B. Every point on this arc is 240 km from B.
- D is where the 7.5 cm arc centred on A crosses the 6 cm arc centred on B
Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q10
The scale drawing shows the positions of two schools, A and B. The scale is 1 centimetre represents 200 metres.
(a) Measure the bearing of B from A. (b) Write the scale in the form 1 : n.
(c) On a second scale drawing, A and B are four times as far apart as on the first scale drawing. Write down the scale of the second scale drawing in the form 1 : n.
Show solution
- Put the protractor’s centre on A with 0° on the north line, and measure clockwise to the line AB.
- Bearing of B from A = 078°
- Both sides of the ratio must be in the same units. 200 m = 200 × 100 = 20 000 cm.
- So 1 cm represents 20 000 cm: 1 : 20 000
- On the new drawing the same real distance is shown 4 times longer, so 1 cm stands for a quarter of the real distance.
- 20 000 ÷ 4 = 5000: 1 : 5000
Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q13
The scale drawing shows the positions of town A and town B. The scale is 1 centimetre represents 10 kilometres.
(a) Work out the actual distance from town A to town B.
(b) Measure the bearing of town B from town A.
(c) Town C is 45 km from town A on a bearing of 310°. Mark the position of town C.
Show solution
- Measure AB with a ruler: 8 cm.
- 1 cm represents 10 km, so 8 cm represents 8 × 10 = 80 km
- Put the protractor’s centre on A with 0° on the north line, and measure clockwise round to the line AB.
- Bearing of B from A = 032°
- Tip: Bearings always have three figures, so 32° is written 032°.
- Change the distance to the drawing: 45 km ÷ 10 = 4.5 cm.
- A bearing of 310° is 360° − 310° = 50° anticlockwise from north.
- Measure this angle at A, draw a line and mark C 4.5 cm from A on a bearing of 310°
Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q24
(a) The scale on a map is 1 cm represents 12 km. Write this in the ratio 1 : n.
(b) The bearing of B from A is 070°. Find the bearing of A from B.
Show solution
- Both sides of a ratio must be in the same units. Change 12 km to cm.
- 12 km = 12 × 1000 m = 12 000 m, and 12 000 m = 12 000 × 100 cm = 1 200 000 cm.
- Scale = 1 : 1 200 000
- The bearing back the other way is 180° different.
- 70 is less than 180, so add 180: 70 + 180 = 250°
Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q15
The scale drawing shows the position of ship A. The scale is 1 centimetre represents 40 kilometres. Ship B is 220 km from ship A on a bearing of 140°. Mark the position of ship B.
Show solution
- Change the real distance to a map distance: 1 cm stands for 40 km, so 220 km is 220 ÷ 40 = 5.5 cm.
- Put the protractor’s centre on A with 0° on the north line.
- Measure 140° clockwise from north and draw a faint line.
- Measure 5.5 cm along the line and mark B, 5.5 cm from A on a bearing of 140°
0580 Scale Drawings and Bearings past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q7
Ship Q is 40 km due east of ship R. Ship P is 40 km due north of ship Q. Work out the bearing of ship R from ship P.
Show solution
- Draw a sketch: R, then Q 40 km east of R, then P 40 km north of Q. Triangle PQR has a right angle at Q.
- QR = QP = 40 km, so the triangle is isosceles and the other two angles are each (180 − 90) ÷ 2 = 45°.
- Stand at P and face north. Due south is 180°. R is south of P, turned 45° further towards the west.
- Bearing of R from P = 180 + 45 = 225°
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q7
The diagram shows the position of two towns, A and B. The scale is 1 cm represents 2 km.
(a) Work out the actual distance between A and B. (b) Find the bearing of B from A.
Show solution
- Measure AB on the diagram with a ruler: about 9.4 cm.
- 1 cm represents 2 km, so 9.4 cm represents 9.4 × 2 = 18.8 km.
- Distance = about 18.8 km (18.4 to 19.2 km is accepted)
- Put the protractor’s centre on A with 0° on the north line.
- Measure clockwise all the way round to the line AB. It is more than 180°, so measure the angle anticlockwise from north (about 103°) and subtract from 360°.
- Bearing = about 257° (255° to 259° is accepted)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q20(a)
The diagram shows a field ABCD with a straight footpath from A to C. At B, the angle measured anticlockwise from north to the line BA is 103°. Find the bearing of A from B.
Show solution
- Bearings are always measured clockwise from north.
- The 103° angle is measured anticlockwise, so the clockwise angle is 360 − 103 = 257°
Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q10
The bearing of A from B is (2x + 50)°. The bearing of B from A is 060°. Find the value of x.
Show solution
- A back bearing (the bearing in the opposite direction) differs by 180°.
- The bearing of B from A is 060°, so the bearing of A from B = 60 + 180 = 240°.
- Set up the equation: 2x + 50 = 240.
- Subtract 50: 2x = 190. Divide by 2: x = 95
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q5(c)
Triangle ABC, with AB = 8 cm, is a scale drawing of a field. (i) The scale is 1 : 10 000. Find the actual distance from A to B in kilometres. (ii) B is due east of A. Find the bearing of A from B.
Show solution
- A scale of 1 : 10 000 means 1 cm on the drawing is 10 000 cm in real life.
- AB = 8 cm, so the real distance is 8 × 10 000 = 80 000 cm.
- Change to metres (÷ 100): 80 000 cm = 800 m.
- Change to kilometres (÷ 1000): 800 m = 0.8 km
- B is due east of A, so A is due west of B.
- Bearings are measured clockwise from north: north = 000°, east = 090°, south = 180°, west = 270°.
- The bearing of A from B is 270°
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q4
(a) Town B is 12 km from town A on a bearing of 080°. Using a scale of 1 cm represents 2 km, mark the position of town B on the diagram.
(b) The bearing of C from D is 130°. Work out the bearing of D from C.
Show solution
- Change the real distance to the scale: 1 cm stands for 2 km, so 12 km is 12 ÷ 2 = 6 cm.
- Put the protractor’s centre on A with 0° on the north line. Measure 80° clockwise and draw a faint line.
- Measure 6 cm along the line and mark B, 6 cm from A on a bearing of 080°
- The bearing back the other way differs by 180°.
- 130° is less than 180°, so add 180: 130 + 180 = 310°
May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q2
The scale drawing shows the positions of two villages, P and Q. The scale is 1 cm represents 0.5 km.
(a) Find the actual distance between village P and village Q.
(b) Measure the bearing of village Q from village P.
Show solution
- Measure PQ on the drawing with a ruler: 9 cm.
- 1 cm represents 0.5 km, so 9 cm represents 9 × 0.5 = 4.5 km
- Put the protractor’s centre on P with 0° on the north line, and measure clockwise round to the line PQ.
- Bearing of Q from P = 110°
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