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.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?
- 2.A plan of a car park is drawn to a scale of 1 : 500. A parking bay is drawn as a rectangle measuring 1.2 cm by 0.6 cm. The real width of the bay corresponds to the 0.6 cm side. What is the real width of the bay, in metres?
- 3.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?
- 4.A line segment is measured with a ruler and found to be 6.5 cm long. What is this length in millimetres?
- 5.A scale drawing shows a room that is 5 cm wide. The real room is 15 m wide. Write the scale of the drawing in the form 1 : n.
- 6.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.
- 7.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?
- 8.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?
- 9.A submarine travels due south. What is the three-figure bearing for this direction?
- 10.The angle between north and a cycle path is 40°, but it is measured anticlockwise from north. What is the three-figure bearing of the cycle path?
- 11.A hiker walks due west. What is the three-figure bearing for this direction?
- 12.A plan of a school hall uses a scale of 1 : 250. A wall is drawn 3.4 cm long on the plan. Sam wants to know the wall's real length in metres. What is it?
- 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.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?
- 15.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?
Answer key
- (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.
- (b) 3 — The real width is 0.6 × 500 = 300 cm, which converts to 3 m by dividing by 100. A candidate who uses the wrong side of the rectangle, 1.2 cm, instead of the 0.6 cm width, gets 1.2 × 500 = 600 cm = 6 m. A candidate who multiplies correctly but converts the 300 cm to metres by dividing by 1000 instead of 100 gets 0.3 m. A candidate who converts by dividing by 10 instead of 100 gets 30 m. The real width of the bay is 3 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°.
- (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.
- (c) 1 : 300 — First convert both lengths to the same unit: 15 m = 1500 cm. The scale compares 5 cm on the drawing to 1500 cm in real life, so dividing both parts by 5 gives a scale of 1 : 300. A candidate who compares 5 to 15 without converting units gets 1 : 3. A candidate who converts 15 m to 150 cm, using the wrong conversion factor, gets 1 : 30. A candidate who converts 15 m to 15000 cm, again using the wrong conversion factor, gets 1 : 3000. The scale of the drawing is 1 : 300.
- (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) 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) 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.
- (d) 180 — Method: bearings are measured clockwise from north and written using three figures. Working: south is a half turn (180°) clockwise from north. A student who answers 090 has confused south with east. A student who answers 270 has confused south with west. A student who answers 018 has written the correct digits in the wrong order. Answer: 180.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (b) 270° — Bearings are measured clockwise from north (000°). Due north is 000°, due east is 090°, due south is 180° and due west is 270°. Choosing 090° gives the bearing for due east, not west. Choosing 180° gives the bearing for due south. Choosing 000° gives the bearing for due north.
- (b) 8.5 m — First apply the scale to convert the plan length to a real length in centimetres: 3.4 × 250 = 850 cm. Then convert centimetres to metres by dividing by 100: 850 ÷ 100 = 8.5, so the wall is 8.5 m long. Choosing 850 m applies the scale correctly but forgets to convert the answer from centimetres into metres. Choosing 0.85 m divides by 1000 instead of 100, confusing the centimetre-to-metre conversion with a metre-to-kilometre one. Choosing 3.4 m ignores the scale factor completely and just restates the plan length as if it were already the real length.
- (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).
- (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°.
- (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.
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