Printable · GCSE Foundation · ages 14-16
Measuring, scale drawings and bearings worksheet — GCSE Foundation
Fifteen questions on "measuring, scale drawings and bearings" — DfE statement G15. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Measuring, scale drawings and bearings worksheet — GCSE Foundation
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- 1.The angle between north and a footpath is 35°, measured clockwise from north. What is the three-figure bearing of the footpath?
- 2.A scale drawing of a rectangular lawn uses a scale of 1 : 125. On the drawing, the lawn measures 4 cm by 2.4 cm. Work out the perimeter of the real lawn, in metres.
- 3.A scale drawing of a bridge uses a scale of 1 : 25. A support beam on the drawing measures 4.4 cm. What is the real length of the beam, in metres?
- 4.A scale drawing of a park uses a scale of 1 : 2000. A path is drawn 8.5 cm long on the drawing. A cyclist rides the length of the path and then rides straight back again along the same path. How far does the cyclist travel in total, in metres?
- 5.A model car is built to a scale of 1 : 24. The real car is 4.32 m long. Work out the length of the model, in centimetres.
- 6.A map has a scale of 1 cm : 5 km. A road is drawn 3 cm long on the map. What is the real length of the road?
- 7.A line segment is measured with a ruler and found to be 6.5 cm long. What is this length in millimetres?
- 8.A bearing is described as being exactly halfway between north and east, measured clockwise. What is the three-figure bearing?
- 9.Two straight roads cross at a junction. The angle between them, measured with a protractor, is 118°. What is the size of the angle vertically opposite to it?
- 10.A scale drawing uses a scale of 1 : 60. A path on the drawing is measured as 9.8 cm long. What is the real length of the path, in metres, to 1 decimal place?
- 11.Two straight paths cross at a junction in a park. One of the angles between them, measured with a protractor, is 76°. What is the size of the angle immediately next to it on a straight line?
- 12.An angle is measured with a protractor as 47°. What type of angle is this?
- 13.A map has a scale of 1 : 20 000. The distance between two villages is 3.4 km in real life. How long is this distance on the map, in centimetres?
- 14.A ship sails 12 km on a bearing of 070° from a harbour to a buoy. It then sails a further 9 km, still on a bearing of 070°, from the buoy to a lighthouse. How far is the lighthouse from the harbour, in kilometres?
- 15.A furniture catalogue drawing uses a scale of 1 cm to 1.5 m. A sofa in the drawing is measured as 7 cm long. What is the real length of the sofa, in metres?
Answer key
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (d) 16 — Method: use the scale to find the real length and width separately, then use the perimeter formula. Working: real length = 4 cm × 125 = 500 cm = 5 m; real width = 2.4 cm × 125 = 300 cm = 3 m; perimeter = 2 × (5 + 3) = 16 m. A student who answers 8 has added the real length and width but forgotten to double the total for the perimeter. A student who answers 1600 has correctly worked out the perimeter in centimetres but forgotten to convert it to metres. A student who answers 500 has only converted the length to real centimetres and stopped there, ignoring the width and the perimeter step. Answer: 16 m.
- (d) 1.1 — Method: multiply the drawing length by the scale factor to get the real length, then convert to the units asked for. Working: 4.4 cm × 25 = 110 cm = 1.1 m. A student who answers 4.4 has forgotten to use the scale at all. A student who answers 110 has correctly worked out the real length in centimetres but forgotten to convert it to metres. A student who answers 11 has used a scale factor of 2.5 instead of 25 by misreading the scale. Answer: 1.1 m.
- (b) 340 — The real length of the path is 8.5 × 2000 = 17000 cm, which converts to 170 m by dividing by 100. The cyclist rides the path there and back, so the total distance is 170 × 2 = 340 m. A candidate who forgets the return journey gives only the one-way distance, 170 m. A candidate who halves the one-way distance instead of doubling it, reading “there and back” as splitting the journey, gets 85 m. A candidate who doubles the scale factor to 4000 by mistake, getting a one-way length of 340 m, and then doubles that for the return journey, gets 680 m. The total distance the cyclist travels is 340 m.
- (b) 18 — Method: convert the real length to centimetres, then divide by the scale factor to shrink it down to the model's size. Working: 4.32 m = 432 cm; 432 cm / 24 = 18 cm. A student who answers 432 has converted the units correctly but forgotten to divide by the scale factor at all. A student who answers 10368 has multiplied by the scale factor instead of dividing (432 x 24). A student who answers 1.8 has converted the metres to centimetres by multiplying by 10 instead of 100, getting 43.2 cm, and then divided by 24. Answer: 18 cm.
- (a) 15 km — Multiply the length drawn on the map by the scale: 3 × 5 = 15 km. Choosing 8 km adds the two scale numbers together (3 + 5 = 8) instead of multiplying them. Choosing 30 km comes from misreading the scale as 1 cm : 10 km and doubling the correct answer. Choosing 5 km simply repeats the scale's distance figure and ignores that the road is drawn 3 cm long, not 1 cm.
- (a) 65 — There are 10 millimetres in every centimetre, so to convert from cm to mm, multiply by 10: 6.5 × 10 = 65 mm. A candidate who forgets to convert at all writes down the original number, 6.5. A candidate who multiplies by 100 instead of 10, confusing cm-to-mm with m-to-cm, gets 650. A candidate who divides by 10 instead of multiplying gets 0.65. The correct length in millimetres is 65.
- (b) 045° — North is 000° and east is 090°; halfway between them, measured clockwise, is 045°. Choosing 090° gives east itself, not the midpoint between north and east. Choosing 135° gives the midpoint between east (090°) and south (180°) instead of between north and east. Choosing 225° is the back bearing of 045° (045° + 180°), not the bearing itself.
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (a) 5.9 — The real length is 9.8 × 60 = 588 cm. Converting to metres, by dividing by 100, gives 5.88 m, which rounds to 5.9 m to 1 decimal place. A candidate who rounds 5.88 down instead of up gets 5.8 m. A candidate who forgets to convert from centimetres to metres gets 58.8. A candidate who divides by 60 instead of multiplying gets 0.16, to 2 decimal places. The real length, to 1 decimal place, is 5.9 m.
- (b) 104 — Method: angles on a straight line add up to 180°. Working: 180° − 76° = 104°. A student who answers 76 has mistaken this for the vertically opposite angle, which is equal, rather than the adjacent angle on a straight line. A student who answers 90 has wrongly assumed the two paths must be perpendicular. A student who answers 14 has subtracted 76° from 90° instead of from 180°. Answer: 104°.
- (b) Acute — An acute angle is any angle less than 90°. Since 47° is less than 90°, it is acute. An obtuse angle is between 90° and 180°, which 47° is not. A reflex angle is greater than 180°, which is far too big for 47°. A right angle is exactly 90°, not 47°. The angle is acute.
- (a) 17 cm — Convert the real distance to centimetres first: 3.4 km = 340 000 cm. Then divide by the scale factor, 20 000, since the map is 20 000 times smaller than real life: 340 000 ÷ 20 000 = 17, so the map distance is 17 cm. Choosing 1700 cm divides by 200 instead of 20 000, losing two zeros from the scale factor (340 000 ÷ 200 = 1700). Choosing 170 cm divides by 2 000 instead of 20 000, losing one zero from the scale factor (340 000 ÷ 2 000 = 170). Choosing 0.17 cm divides by 2 000 000 instead of 20 000, adding two extra zeros to the scale factor (340 000 ÷ 2 000 000 = 0.17).
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
- (b) 10.5 — Method: multiply the drawing length by the scale factor to get the real length. Working: 7 cm × 1.5 = 10.5 m. A student who answers 8.5 has added the scale factor to the drawing length instead of multiplying (7 + 1.5). A student who answers 14 has rounded the scale factor up to 2 before multiplying. A student who answers 3.5 has divided the drawing length by 2 instead of multiplying it by 1.5. Answer: 10.5 m.
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