OxfordAQA International GCSE Mathematics: Scale drawings, bearings, constructions and loci (9260) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover scale drawings, bearings, constructions and loci from the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards), references G12 and G13 in section 3.3.1. Both references are Core content, which is also assessable on the Extension papers, so no question here is Extension only. Have a ruler, protractor and compasses ready. Learn the methods in the study guide and the revision notes. See the course hub and the printable checklist. The free 10-minute diagnostics show where to focus.

Questions

1. (a) Write the three-figure bearing of north-west. [1] (b) A road runs from town A to town B on a bearing of 135°. State the compass point of B from A, and find the bearing of A from B. [2]

2. A map has a scale of 1 : 75 000. (a) Two villages are 8.6 cm apart on the map. How far apart are they on the ground? Give your answer in km. [2] (b) A canal is 12 km long. Find its length on the map in cm. [2] (c) On a second map the same 3 km is drawn as 5 cm. Write the scale of the second map as a ratio 1 : n. [2]

3. The bearing of a lighthouse L from a ship S is 251°. A buoy B has bearing 320° from S. (a) Find the bearing of S from L. [2] (b) Find the size of angle LSB. [2]

4. A boat leaves harbour H and sails 7.5 km on a bearing of 038° to point X. It then sails 10.5 km on a bearing of 112° to point Y. Use a scale of 1 cm to 1.5 km. (a) Make an accurate scale drawing of the journey. [4] (b) Measure HY and find the real distance from H to Y. [1] (c) Find the bearing of H from Y. [2]

5. Using ruler and compasses only, construct triangle ABC with AB = 7 cm, BC = 5.5 cm and AC = 4.5 cm. Leave in your construction arcs. Measure angle ABC. [5]

6. (a) Draw a line PQ of length 8 cm. Using ruler and compasses only, construct an angle of 60° at P. [3] (b) Using ruler and compasses only, bisect your angle. State the size of each part. [2]

7. Draw a line AB of length 8 cm. Point P is above AB with AP = 6 cm and BP = 5 cm. (a) Construct triangle ABP using ruler and compasses. [2] (b) Construct the perpendicular from P to AB. [2] (c) Measure the shortest distance from P to AB. [1]

8. X and Y are points 5 m apart. A path is the set of all points exactly 3 m from the line segment XY. (a) Describe the shape of the path. [2] (b) Calculate the length of the path, to 1 decimal place. [3]

9. ABCD is a rectangular playground with AB = 12 m and BC = 8 m. Draw it with 1 cm standing for 2 m. A prize is hidden at a point that is closer to A than to B, nearer to AB than to AD, and less than 7 m from A. (a) Draw the playground and shade the region where the prize may be. [7] (b) A point T is 5 m from AD and 4 m from AB. Is T in the region? Give a reason. [1]

10. A straight coast runs north to south through port P. A boat is 12 km from P on a bearing of 070°. Use a scale of 1 cm to 2 km to find the shortest distance from the boat to the coast. [5]

11. Port B is 18 km due east of port A. The bearing of a lighthouse L from A is 048° and the bearing of L from B is 312°. Use a scale of 1 cm to 3 km. (a) Draw the line AB and the two bearing lines, and mark L. [3] (b) Measure AL and write the real distance in km. [1] (c) Show that LA = LB. [2] (d) A radar at L detects points within 10 km. Is port A detected? Give a reason. [2] (e) Find the shortest distance from L to the line AB, by calculation. [2]

Answers

1. (a) North-west is 315° [1]. 315° (b) 135° is south-east, so B is south-east of A [1]. Back bearing: 135° + 180° = 315° [1]. [3] Examiner insight: Write every bearing with three figures, and name the compass point when the question asks for it.

2. (a) 8.6 × 75 000 = 645 000 cm [1] = 6.45 km [1]. (b) 12 km = 1 200 000 cm; 1 200 000 ÷ 75 000 [1] = 16 cm [1]. (c) 3 km = 300 000 cm [1]. 5 : 300 000 = 1 : 60 000 [1]. [6] Examiner insight: Show the conversion from cm to km as a separate line, so a slip in the power of ten is visible and the method can still be seen.

3. (a) 251° is more than 180°, so subtract: 251° − 180° [1] = 071° [1]. (b) Both lines start at S: 320° − 251° [1] = 69° [1]. [4] Examiner insight: For an angle between two bearings from the same point, subtract the bearings; do not add 180° unless you are switching to the other end of the line.

4. (a) Scale lengths: 7.5 ÷ 1.5 = 5 cm and 10.5 ÷ 1.5 = 7 cm [1]. Draw a north line at H, measure 038° clockwise and draw HX = 5 cm [1]. Draw a north line at X, measure 112° clockwise and draw XY = 7 cm [1]. Join H to Y [1]. (b) HY measures about 9.7 cm, so the distance is 9.7 × 1.5 ≈ 14.5 km [1] (accept 14.2 km to 14.8 km). (c) Bearing of Y from H is about 082° [1]. Add 180°: 262° (accept 260° to 264°) [1]. [7] Examiner insight: Keep your north lines on the page. A bearing of H from Y needs a north line drawn at Y.

5. Draw AB = 7 cm [1]. Set compasses to 4.5 cm, centre A, draw an arc [1]. Set to 5.5 cm, centre B, draw an arc crossing the first at C [1]. Join AC and BC, leaving the arcs [1]. Angle ABC measures 40° (accept 39° to 41°) [1]. [5] Examiner insight: A triangle made by trial and error gives no evidence of a construction, even if its sides are right, so keep the arcs.

6. (a) Draw PQ = 8 cm. Draw an arc centred at P cutting PQ at X [1]. With the same radius, draw an arc centred at X cutting the first arc at Z [1]. Join P to Z; angle ZPQ is 60° [1]. (b) Draw arcs from X and from Z with equal radius so that they cross [1]. Join P to the crossing point. Each part is 30° [1]. [5] Examiner insight: State the size of the angle you have made and the reason it is that size, instead of only measuring it.

7. (a) Draw AB = 8 cm [1]. Arc of radius 6 cm from A and arc of radius 5 cm from B cross at P [1]. (b) Draw an arc centred at P cutting AB at two points [1]. Arcs of equal radius from those points cross on the other side of AB; join to P [1]. (c) The perpendicular measures about 3.7 cm (accept 3.6 cm to 3.9 cm) [1]. [5] Examiner insight: The shortest distance is along the perpendicular, so a measurement from P to the nearer end of AB is not the shortest distance.

8. (a) Two straight lines [1], each 5 m long and 3 m from XY on either side, joined at each end by a semicircle of radius 3 m [1]. (b) The two straight parts have length 2 × 5 = 10 m [1]. The two semicircles make one circle: 2π × 3 = 18.85 m [1]. Total 10 + 18.85 = 28.85 (more exactly 28.8496), so 28.8 m to 1 d.p. [1]. [5] Examiner insight: Draw semicircles, not square corners, at the ends of the path.

9. (a) Draw rectangle 6 cm by 4 cm [1]. Construct the perpendicular bisector of AB (3 cm from A) [1]; the region is on the A side [1]. Construct the bisector of angle DAB, a line at 45° through A [1]; the region is below it, nearer AB [1]. Draw an arc of radius 3.5 cm centred at A [1]; the region is inside it. Shade the part that satisfies all three [1]. (b) T is 5 m from AD and 4 m from AB, so it is 2.5 cm across and 2 cm up: left of the perpendicular bisector, below the 45° line (4 < 5), and 6.4 m from A, which is less than 7 m. Yes, T is in the region [1]. [8] Examiner insight: Test a point against each boundary to decide which side to shade.

10. The boat is 12 ÷ 2 = 6 cm from P [1]. Draw the coast as a vertical line through P and a line of 6 cm at 070° clockwise from north [1]. Construct the perpendicular from the boat to the coast line [1]. It measures about 5.6 cm [1], so the distance is 5.6 × 2 = 11.3 km (accept 11.0 km to 11.6 km) [1]. [5] Examiner insight: Convert from cm back to km at the end and give units.

11. (a) Draw AB = 18 ÷ 3 = 6 cm [1]. At A draw a north line and measure 048° clockwise [1]. At B draw a north line and measure 312° clockwise; the lines cross at L [1]. (b) AL measures about 4.0 cm, so AL ≈ 12 km (calculation gives 12.1 km) [1]. (c) The bearing of B from A is 090°, so angle LAB = 90° − 48° = 42° [1]. At B, the bearing 312° is 48° west of north, so angle LBA = 90° − 48° = 42°. The base angles are equal, so triangle ALB is isosceles and LA = LB [1]. (d) AL ≈ 12 km [1], which is more than 10 km, so A is not detected [1]. (e) The perpendicular from L meets AB at its midpoint M, with AM = 9 km [1]. LM = 9 × tan 42° = 8.1 km [1]. [10] Examiner insight: “Show that” asks for each reason: state the bearings that make each base angle 42°, not just that the triangle looks isosceles.

Where marks are usually lost

• Measuring a bearing from the wrong north line, or anticlockwise.
• Not writing three figures for a bearing, such as 38° instead of 038°.
• Forgetting to convert km to cm, or to convert back, when using a scale.
• Using the wrong protractor scale, so an angle reads as 130° instead of 50°.
• Changing the compass width between two arcs of a construction.
• Leaving out the construction arcs, or rubbing them out.
• Measuring slanted lines as the shortest distance from a point to a line.
• Shading a region for only one or two of the loci conditions.

Next steps

• Re-read the revision notes and the study guide.
• Track coverage on the course hub and the printable checklist.
• Take one of the free 10-minute diagnostics to see what to revise next.
• Book a free trial class for one-to-one help with constructions.

Official syllabus

These questions follow the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA, references G12 and G13.