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
MathsUKwww.geekhero.co.uk
- 1.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?
- 2.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?
- 3.The angle between north and a footpath is 35°, measured clockwise from north. What is the three-figure bearing of the footpath?
- 4.A treasure map has a scale of 1 cm to 4 m. A path from the start to a rock is drawn as two straight sections, measuring 3 cm and 2.5 cm. Work out the total real distance from the start to the rock, in metres.
- 5.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?
- 6.A line segment is measured with a ruler and found to be 6.5 cm long. What is this length in millimetres?
- 7.A hiker walks on a bearing of 065°. She then turns clockwise through 90° and continues walking in a straight line. What bearing is she now walking on?
- 8.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?
- 9.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?
- 10.The bearing of a campsite B from a walker's position A is 070°. What is the bearing of A from B?
- 11.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?
- 12.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 13.A garden plan is drawn to a scale of 1 cm : 1.8 m. A path on the plan measures 4.5 cm. What is the real length of the path, in metres, to 1 decimal place?
- 14.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.
- 15.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?
Answer key
- (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) 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) 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.
- (b) 22 — Method: add the two drawn lengths together first, then apply the scale to the total. Working: 3 cm + 2.5 cm = 5.5 cm; 5.5 cm × 4 = 22 m. A student who answers 10 has only converted one of the two sections (2.5 cm × 4) and forgotten the other. A student who answers 5.5 has added the two drawn lengths but forgotten to apply the scale at all. A student who answers 44 has doubled the correct answer, effectively applying the scale twice. Answer: 22 m.
- (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) 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) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (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) 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.
- (a) 250° — The back bearing (the bearing of A from B) differs from the bearing of B from A by exactly 180°. Because the given bearing, 070°, is less than 180°, add 180°: 070 + 180 = 250°, so the bearing of A from B is 250°. Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 110° comes from subtracting 180° from 070° and dropping the negative sign (070 − 180 = −110) instead of adding 180°. Choosing 160° comes from adding only 90° instead of 180° (070 + 90 = 160).
- (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) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (a) 8.1 m — Multiply the length on the plan by the scale factor: 4.5 × 1.8 = 8.1, so the real length is 8.1 m. Choosing 6.3 m adds the two numbers instead of multiplying them (4.5 + 1.8 = 6.3). Choosing 2.5 m divides the plan length by the scale factor the wrong way round (4.5 ÷ 1.8 = 2.5) instead of multiplying. Choosing 9.0 m rounds the scale factor up to 2 before multiplying (4.5 × 2 = 9.0), losing the accuracy the 1.8 was giving.
- (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.
- (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.
Build your own mix at the worksheet builder.