OCR GCSE Mathematics: Ruler-and-compass constructions and loci (J560) – 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 follow section 8.02 (8.02a to 8.02d) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). All of the content is Foundation and Higher; no part of it is restricted to the Higher tier. You need a ruler, compasses, a protractor and a sharp pencil. Learn the methods first from the study guide and the revision notes. The OCR GCSE Mathematics hub, the checklist and the 10-minute diagnostics are linked for planning.

Questions

1. Draw a line segment AB of length 8.4 cm. Construct the perpendicular bisector of AB, mark the midpoint M, and write down the length of AM. [3]

2. Use a protractor to draw an angle of 76°. Construct the bisector of this angle and state the size of each of the two equal angles. [3]

3. Draw a horizontal line l. Mark a point P exactly 4 cm above l. Construct the perpendicular from P to l, and label the foot F. Mark a point X on l, 3 cm from F. Find the length PX by calculation, and state which is shorter, PF or PX. [4]

4. Draw a horizontal line AB of length 8 cm. Construct the perpendicular to AB at A. Mark C on this perpendicular so that AC = 6 cm. Join BC and find its length. [4]

5. (a) Explain what “equidistant” means. [1] (b) State the locus of points equidistant from two fixed points A and B. [1] (c) State the locus of points equidistant from two straight lines that cross each other. [1]

6. The line segment CD is 6 cm long. (a) Draw the locus of all points exactly 2 cm from CD. [3] (b) Work out the length of this locus, giving your answer to 1 decimal place. [2]

7. The points A, B and P have coordinates A(1, 2), B(5, 2) and P(3, 5). Show that P is equidistant from A and B, and say what this tells you about where P lies. [3]

8. A point P is 2.4 cm from a straight line. F is the foot of the perpendicular from P to the line. X is a point on the line with FX = 1.8 cm. Work out PX, and explain why the perpendicular distance is the shortest distance from P to the line. [3]

9. A rectangular lawn ABCD has AB = 12 m and AD = 8 m. A sprinkler at D waters every point within 6 m of D. Use a scale of 1 cm to 2 m. (a) Draw the lawn to scale. [1] (b) Construct the bisector of angle ADC. [2] (c) Draw the boundary of the watered area and shade the part of the lawn that is watered and nearer to DA than to DC. [1] (d) A flower bed F is on the bisector, 5 m from D. Is it watered? [1]

10. Draw a rectangle ABCD with A at the bottom left, AB = 10 cm and AD = 6 cm. Shade the region inside the rectangle that is nearer to AD than to BC, nearer to AB than to DC, and more than 4 cm from A. Use constructions to show your boundaries. Then state how far from AD the boundary arc meets the line that is equidistant from AB and DC. Give your answer to 1 decimal place. [6]

11. Two straight roads meet at a junction J at an angle of 70°. A wind turbine W must be the same distance from both roads and 5 km from J. Use a scale of 1 cm to 1 km. (a) Draw the roads from J (make each at least 7 cm long) and construct the line of points equidistant from them. [3] (b) Mark the position of W. [2] (c) Construct the shortest distance from W to one of the roads and measure it, giving the real distance in km to 1 decimal place. [3]

Answers

1. Set the compasses wider than 4.2 cm and draw arcs from A above and below AB [1]. Repeat from B with the same radius, then join the crossing points with a ruler [1]. AM = 8.4 ÷ 2 = 4.2 cm [1]. [3] Examiner insight: Leave your construction arcs on the page. They are the evidence that you used compasses, and a line drawn by eye or by measuring may not earn credit for the construction.

2. Draw the 76° angle with a protractor [1]. Construct the bisector with an arc from the vertex, then equal arcs from the two points where it meets the lines [1]. Each angle is 76° ÷ 2 = 38° [1]. [3] Examiner insight: Check your bisector with the protractor before moving on. If the two parts differ by more than a degree or so, one arc setting probably changed.

3. Arc from P cutting l twice, then equal arcs from those two points crossing below l, and the line joining P to the crossing point [1]. Mark F where this line meets l; PF = 4 cm [1]. PX² = 4² + 3² = 25, so PX = 5 cm [1]. PF is shorter, because 4 cm is less than 5 cm [1]. [4] Examiner insight: “Find by calculation” asks for working from Pythagoras. A length read off your drawing is a check, not a substitute.

4. Arcs from A cutting AB at equal distances either side of A (extend AB to the left if needed) [1]. Wider equal arcs from those two points crossing above A, and the line through A [1]. Mark C 6 cm from A on this line, then join BC [1]. BC² = 6² + 8² = 100, so BC = 10 cm [1]. [4] Examiner insight: A set square or a protractor does not show the construction. Use arcs, then confirm with the protractor if you wish.

5. (a) Equidistant means the same distance from two or more things [1]. (b) The perpendicular bisector of AB [1]. (c) The angle bisector of the angle between the lines [1]. [3] Examiner insight: Name the line and say what it bisects. “The middle” is too vague to count as a locus.

6. (a) Draw CD 6 cm long [1]. Draw two lines parallel to CD, 2 cm above and 2 cm below, each 6 cm long [1]. Join them with a semicircle of radius 2 cm centred on C and another centred on D [1]. (b) Length = 2 × 6 + 2 × π × 2 = 12 + 4π [1] = 24.6 cm [1]. [5] Examiner insight: Round only at the end. The two semicircles together make one full circle, so use 2πr once, not twice.

7. PA² = (3 − 1)² + (5 − 2)² = 4 + 9 = 13 [1]. PB² = (5 − 3)² + (5 − 2)² = 4 + 9 = 13 [1]. So PA = PB = √13, and P lies on the perpendicular bisector of AB (the line x = 3) [1]. [3] Examiner insight: In a “show that” question, write both distances out in full and finish with a conclusion in words. The statement alone is not enough.

8. PX² = 2.4² + 1.8² = 5.76 + 3.24 = 9, so PX = 3 cm [1]. PX is the hypotenuse of right-angled triangle PFX [1]. The hypotenuse is the longest side, so PX is longer than PF, and the perpendicular PF is the shortest distance [1]. [3] Examiner insight: Give a reason that applies to any point on the line, not only to the one given. Refer to the right-angled triangle.

9. (a) Rectangle 6 cm by 4 cm [1]. (b) Arcs from D cutting DA and DC, then equal arcs from those points crossing inside the angle, and a line from D through the crossing point [1]. This line makes 45° with each side [1]. (c) Arc of radius 3 cm centred on D (6 m is 3 cm), with the lawn shaded between the bisector, side DA and the arc, on the DA side [1]. (d) 5 m is 2.5 cm, which is less than 3 cm, so F is watered [1]. [5] Examiner insight: Convert the 6 m to 3 cm before you set the compasses. Writing the scale conversion beside the drawing shows you are using the scale the right way round.

10. Draw the rectangle 10 cm by 6 cm [1]. Construct the perpendicular bisector of AB, which is 5 cm from AD [1]. Construct the perpendicular bisector of AD, which is 3 cm from AB [1]. Draw an arc of radius 4 cm centred on A [1]. Shade the region left of the first line, below the second line and outside the arc [1]. The arc meets the 3 cm line where x² + 3² = 4², so x = √7 = 2.6 cm from AD [1]. [6] Examiner insight: Use a different colour or hatching for the final region, and shade only the part that meets all three conditions.

11. (a) Draw a 70° angle at J with both arms at least 7 cm long [1]. Arc from J cutting both arms, then equal arcs from those points crossing [1]. Draw the bisector through J and the crossing point [1]. (b) Set the compasses to 5 cm, centre J [1]. The point where this arc cuts the bisector is W [1]. (c) Perpendicular from W to one road: arc from W cutting the road twice, equal arcs from those points crossing, and a line from W [1]. Measure the length of this line on the drawing [1]. The distance is about 2.9 cm, which represents 2.9 km [1]. [8] Examiner insight: Your measurement may differ from 2.9 by a tenth or so, which is normal for hand drawing. Take care with the pencil and show the perpendicular as a construction.

Where marks are usually lost

• Arcs from A and from B drawn with different radii.
• Compasses opened to less than half of AB, so the arcs do not cross.
• Measuring the distance to a road along a slanted line instead of the perpendicular.
• Forgetting to convert real lengths to drawing lengths before setting the compasses.
• Drawing a square end instead of a semicircle at each end of a locus around a segment.
• Shading a region that meets only some of the conditions.
• Rubbing out the arcs that show the construction.
• Giving the answer to the wrong accuracy, or with no units.

Next steps

• Review the revision notes and the study guide.
• Go to the OCR GCSE Mathematics hub and the checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class: /trial/.

Official syllabus

OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), section 8.02, published by OCR.