G2: Constructions
Use the standard ruler and compass constructions; perpendicular bisector, angle bisector, constructing triangles
📋 Key Concepts
Definition: Constructions are accurate drawings made using only a ruler (for straight lines) and a compass (for circles and arcs).
Essential Equipment
- Sharp pencil - for accuracy
- Ruler - for measuring and drawing straight lines
- Compass - for drawing circles and arcs
- Protractor - for measuring angles (not always needed for constructions)
📝 Perpendicular Bisector
Definition: A perpendicular bisector is a line that cuts another line segment exactly in half at 90°.
Construction Steps
To construct the perpendicular bisector of line AB:
- Set your compass to more than half the length of AB
- With the compass point at A, draw an arc above and below the line
- With the compass point at B, draw arcs that intersect the first two
- Draw a straight line through the two points of intersection
Result: This line is the perpendicular bisector of AB. It passes through the midpoint at 90°.
📝 Angle Bisector
Definition: An angle bisector is a line that divides an angle exactly in half.
Construction Steps
To construct the bisector of angle ABC:
- With compass point at B, draw an arc that crosses both BA and BC
- Label the crossing points P and Q
- With compass point at P, draw an arc inside the angle
- With the same radius, draw an arc from Q that intersects the first arc
- Draw a line from B through the intersection point
Result: This line bisects the angle, creating two equal angles.
📝 Constructing Triangles
Methods: Triangles can be constructed given: SSS (three sides), SAS (two sides and included angle), ASA (two angles and included side).
Example 1: SSS Construction
Construct triangle ABC where AB = 6 cm, BC = 5 cm, AC = 4 cm
Solution:
- Draw line AB = 6 cm
- With compass set to 5 cm, draw an arc from point B
- With compass set to 4 cm, draw an arc from point A
- Point C is where the arcs intersect
- Complete the triangle by joining AC and BC
Example 2: SAS Construction
Construct triangle PQR where PQ = 5 cm, angle PQR = 60°, QR = 4 cm
Solution:
- Draw line PQ = 5 cm
- At Q, construct angle 60° using a protractor or construction
- From Q, measure 4 cm along the angle line to point R
- Join PR to complete the triangle
📝 Perpendicular from a Point
Two cases: Constructing a perpendicular from a point on a line, or from a point to a line.
From Point to Line
To construct a perpendicular from point P to line AB:
- With compass point at P, draw an arc that cuts AB at two points
- Label these points C and D
- With compass wider, draw arcs from C and D that intersect
- Draw line from P through the intersection point
🧮 Worked Examples — Exam Style
Constructions are practical marks: compass arcs must be VISIBLE. Perpendicular bisector, angle bisector, and SSS triangles are the three core builds.
Construct a 60° angle at a point P (describe the steps).
1) Draw a ray from P. 2) Arc centred at P crossing the ray. 3) Same radius, arc centred at that crossing point. 4) Line from P to where the two arcs intersect → 60° (the two radii form an equilateral triangle).
M1 for both arcs, A1 for the reason (equilateral). Keeping the same radius is the trick — that IS the 60°.
Exam wording to expect: "Construct …" (leave all arcs visible), "the locus of points equidistant from …".
❓ Practice Questions
Q1: What equipment is needed for geometric constructions?
Q2: Describe how to construct a perpendicular bisector of a 7 cm line segment.
Q3: What is an angle bisector?
Q4: What does SSS stand for in triangle construction?
Q5: Why must the compass be set to more than half the line length when constructing a perpendicular bisector?
Q6: Construct a triangle with sides 5 cm, 6 cm and 7 cm (describe the steps)
Q7: What does the perpendicular bisector of AB tell you about any point on it?
✅ Answers
- Sharp pencil, ruler, and compass (protractor sometimes)
- 1) Set compass > 3.5 cm, 2) Draw arcs from both endpoints, 3) Draw line through intersection points
- A line that divides an angle exactly in half
- Side-Side-Side (constructing a triangle given three sides)
- So the arcs from both endpoints will intersect each other
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. Sharp pencil, ruler and compasses (protractor only when the question allows angle-measuring)
Q2. Compass > 3.5 cm; arcs from BOTH endpoints (same radius); rule through the two arc intersections — the line meets AB at 90° at its midpoint
Q3. The line cutting the angle into two equal halves
Q4. Side-Side-Side — the three given side lengths
Q5. Arcs must intersect; with radius ≤ half, the arcs never meet
Q6. 1) Draw the 7 cm base. 2) Arc of radius 5 cm centred at one end, arc of radius 6 cm centred at the other (same side). 3) Join the intersection to both ends ∎
Q7. Every point on it is equidistant from A and B — that is its locus meaning
📌 Key Facts to Memorise
- Perpendicular bisector: two equal-radius arc sets from each endpoint
- Angle bisector: equal-radius arcs from the vertex arms, then intersect the two intermediate arcs
- SSS triangle: draw one side, arc two lengths, join the intersection
- 60° = equilateral construction; 90° = perpendicular bisector; 45° = bisect 90°
- Loci: distance-from-point = circle · distance-from-line = parallel pair · from two points = perpendicular bisector
🗺️ In the Exam
Constructions: 2–4 marks on Paper 1 (real equipment). Arcs are working — rubbing them out destroys the method marks.
Links to: loci (G2), geometric notation (G1), measuring (G15).
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Keep your pencil sharp for accurate constructions
- Leave all construction arcs visible - you get marks for showing your method
- Never rub out construction lines
- Check your compass doesn't slip while drawing
- Practise each construction several times before the exam
- Make sure your compass pencil and compass point are level
🧠 Problem-Solving Strategies
Problem-Solving
Use a compass for all constructions — never just estimate. For a perpendicular bisector: set compass to more than half the line length, draw arcs from both ends. For an angle bisector: draw an arc from the vertex, then arcs from where the first arc crosses each arm. For constructing triangles, always draw the longest side first.
Multi-Step Problem
Construct a triangle with sides 7 cm, 5 cm and 4 cm. Then construct the perpendicular bisector of the longest side. Does the bisector pass through the opposite vertex?
Solution: Draw the 7 cm base. From each end, draw arcs with radii 5 cm and 4 cm to locate the third vertex. Construct the perpendicular bisector of the 7 cm side. It does NOT pass through the opposite vertex because the triangle is scalene (only in isosceles triangles does the perpendicular bisector of the base pass through the opposite vertex).
⚠️ Common Errors
Watch Out!
1. Wrong: Setting the compass radius too small when constructing a perpendicular bisector Correct: The compass must be set to MORE than half the length of the line, otherwise the arcs will not intersect.
2. Wrong: Drawing only one arc when constructing a perpendicular bisector Correct: You must draw arcs from BOTH endpoints. The intersection of the two pairs of arcs determines the bisector line.
3. Wrong: When constructing a 60° angle, measuring with a protractor instead of using compass construction Correct: A 60° angle is constructed by drawing an equilateral triangle with compass arcs, not by measuring with a protractor. Constructions must use only compass and straightedge.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: (a) Construct triangle ABC where AB = 8 cm, AC = 6 cm and ∠BAC = 45°. (b) Construct the angle bisector of ∠BAC. (c) The bisector meets BC at point D. Measure BD and DC. Is BD = DC? Explain your observation.
(a) Draw AB = 8 cm. At A, construct a 45° angle (bisect a right angle). Mark C at 6 cm along this line. Join BC.
(b) From A, draw an arc cutting both arms. From each intersection, draw arcs that cross. Draw the bisector from A through this crossing point to meet BC at D.
(c) BD ≈ 4.8 cm, DC ≈ 3.7 cm. BD ≠ DC because the angle bisector divides the opposite side in the ratio of the adjacent sides (AB:AC = 8:6 = 4:3), not equally. Only in isosceles triangles does the angle bisector bisect the opposite side.
Mark scheme: M1 for construction of 45°, A1 correct triangle, M1 for angle bisector construction, A1 correct bisector, M1 measurement, A1 explanation with ratio
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A student wants to find the centre of a circular plate but cannot see the whole circle.
(a) Describe a construction method to find the centre of a circle.
(b) The student draws two chords and constructs their perpendicular bisectors. Why must the intersection point be the centre?
(c) Explain why using only one chord's perpendicular bisector is not sufficient.
Answers: (a) Draw any two chords. Construct the perpendicular bisector of each chord. The intersection of the two bisectors is the centre. (b) The perpendicular bisector of any chord passes through the centre. Two bisectors give two lines that both pass through the centre, so their unique intersection point must be the centre. (c) A single perpendicular bisector gives a line of possible centre points. Without a second bisector, the exact centre position along this line is unknown — infinitely many points could be the centre.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Mathematics exams?
Get the best revision books and guides to boost your grades.