✏️ Constructions

Grade 8 · Cambridge Lower Secondary Stage 8

What you'll learn

Construct triangles (SSS, SAS, ASA, AAS)
Perpendicular bisector of a line segment
Angle bisector using a compass
Construct special angles (60°, 90°, 45°, 30°)
Loci — sets of points satisfying given conditions

📖 Learn

1. Constructing Triangles

SSS (3 sides given): Draw base, set compass to each side length, draw arcs from each end, cross = third vertex

SAS (2 sides + included angle): Draw one side, measure angle, draw second side along that direction

ASA (angle-side-angle): Draw the base, measure angle at each end, lines cross = third vertex

AAS (angle-angle-side): Uses the fact that third angle = 180 − other two

💡 Always use a ruler and compass for constructions — never use a protractor for "proper" construction, but it is acceptable in exams when not specifically a "construction" task.

⚠️ Leave all construction arcs visible — examiners must see your method.

2. Perpendicular Bisector

The perpendicular bisector of AB is the line that cuts AB at 90° at its midpoint.

Step 1: Open compass to more than half of AB

Step 2: Draw arc from A (above and below AB)

Step 3: Same compass width, draw arc from B

Step 4: Join the two crossing points — this is the perp. bisector

💡 Every point on the perpendicular bisector is equidistant from A and B.

3. Angle Bisector

The angle bisector divides an angle exactly in half.

Step 1: Draw an arc centred on the vertex, cutting both arms of the angle

Step 2: From each intersection, draw equal arcs that cross inside the angle

Step 3: Draw a straight line from the vertex through the crossing — this bisects the angle

💡 Every point on the angle bisector is equidistant from both arms of the angle.

4. Special Angles

60°: Draw equilateral triangle — each angle is 60°. Or: open compass to full radius, mark 6 equal arcs around a circle.

90°: Perpendicular bisector at a point on a line

45°: Construct 90°, then bisect it

30°: Construct 60°, then bisect it

5. Loci

A locus (plural: loci) is the set of all points satisfying a given condition.

Distance r from point P: circle radius r centred on P

Equidistant from A and B: perpendicular bisector of AB

Equidistant from two lines: angle bisector of the angle between the lines

Distance d from a line: two parallel lines at distance d on each side

Combined loci: region satisfying multiple conditions = intersection of regions

✏️ Worked Examples

Example 1 – Constructing an Equilateral Triangle (SSS)

Construct an equilateral triangle with side 5 cm.

Step 1: Draw base AB = 5 cm

Step 2: Set compass to 5 cm. Draw arc from A.

Step 3: Same compass width. Draw arc from B. Mark crossing point C.

Step 4: Join AC and BC. Triangle ABC has all sides = 5 cm, all angles = 60°.

Example 2 – Perpendicular Bisector as Locus

P is a point equidistant from A(0,0) and B(6,0). Describe the locus.

Answer: The perpendicular bisector of AB — the vertical line x = 3

Any point (3, y) is: distance from A = √(9+y²), distance from B = √(9+y²) ✓ Equal!

Example 3 – Locus at Fixed Distance

Describe the locus of points 3 cm from a fixed point P.

Answer: A circle with centre P and radius 3 cm

Example 4 – Region Satisfying Two Conditions

Region closer to A than B, AND within 5 cm of A.

Closer to A than B: on A's side of the perpendicular bisector of AB

Within 5 cm of A: inside a circle radius 5 centred on A

Combined region: intersection of these two areas (draw both, shade overlap)

🎨 Visualizer

🔧 Construction Animator

Watch key constructions step by step.

📍 Locus Explorer

Select a locus type to visualize it.

📐 Triangle SSS Calculator

Enter 3 sides. Check if they form a valid triangle and see its angles.

a: b: c:

Exercise 1 – Triangle Inequality

A triangle is valid if the sum of any two sides is greater than the third. Enter 1 if valid, 0 if not.

Exercise 2 – Angles in Triangles (SSS)

Use the cosine rule: cos A = (b²+c²−a²)/(2bc). Give angles to nearest degree.