To draw an angle of 65°: Draw a ray. Place protractor centre on the endpoint. Mark 65° on the correct scale. Join mark to endpoint with a straight line.
To measure an angle: Place centre on vertex, align base line with one arm, read where the other arm crosses the scale.
💡 If the angle looks obtuse but your reading is acute, you've used the wrong scale — swap to the other number!
Part 2: Perpendicular Bisector of a Line Segment
The perpendicular bisector cuts a line segment in half at 90°. Every point on it is equidistant from both endpoints.
Step 1: Draw line segment AB.
Step 2: Open compass to more than half of AB. Place point on A and draw arcs above and below the line.
Step 3: Keep the same radius. Place point on B and draw arcs crossing the first pair.
Step 4: Join the two intersection points. This line is the perpendicular bisector.
✅ Check: the bisector should cross AB at its midpoint and form 90° angles.
Part 3: Angle Bisector
The angle bisector splits an angle into two equal halves.
Step 1: Draw angle BAC at vertex A.
Step 2: Place compass on A, draw an arc that cuts both arms at points P and Q.
Step 3: Place compass on P, draw an arc in the interior. Same radius from Q, cross that arc at point R.
Step 4: Draw ray AR. This bisects angle BAC.
💡 Keep the compass radius the same for steps 3 & 4 — that's the key!
Part 4: Constructing Triangles
Given
Method
SSS (3 sides)
Draw base AB. Arc of length AC from A, arc of length BC from B — intersection is C.
SAS (2 sides + included angle)
Draw base AB. Use protractor to draw the included angle at A. Mark length AC along that ray. Join BC.
ASA (2 angles + included side)
Draw base AB. Draw angle at A, draw angle at B — where the two rays meet is C.
💡 Always draw the base line first, then build upwards!
Part 5: Perpendicular from a Point to a Line & Locus
Perpendicular from point P to line L:
Open compass at P to reach past the line. Draw an arc cutting the line at two points X and Y. Bisect segment XY — the bisector from P is the perpendicular.
Locus = the set of all points satisfying a condition:
📌 Equidistant from a fixed point → the locus is a circle centred at that point.
📌 Equidistant from a fixed straight line → the locus is two parallel lines on either side.
📌 Equidistant from two fixed points → the locus is the perpendicular bisector of the segment joining them.
📌 Equidistant from two intersecting lines → the locus is the pair of angle bisectors.
🗺️ Practical use: A mobile mast covers all phones within 2 km — the coverage boundary is a circle (locus).
💡 Worked Examples
Example 1: Drawing an Angle of 124°
Use a protractor and ruler to draw an angle of 124°.
Step 1: Draw a horizontal ray with a ruler. Mark the starting point O.
Step 2: Place the protractor centre exactly on O and align the base line with the ray. 124° is obtuse, so use the outer scale (the one reading 0° on the right).
Step 3: Make a small mark at 124° on the paper.
Step 4: Remove the protractor. Draw a ray from O through the mark.
Answer: The two rays form a 124° obtuse angle. Label it.
💡 124° > 90° so it is obtuse. If you read 56° by mistake, you used the wrong scale!
Example 2: Perpendicular Bisector of AB (AB = 8 cm)
Construct the perpendicular bisector of a line segment AB where AB = 8 cm.
Step 1: Draw AB = 8 cm.
Step 2: Set compass to 5 cm (> 4 cm = half of AB). Draw arcs above and below from A.
Step 3: Same radius from B. Arcs cross at points P (above) and Q (below).
Step 4: Draw line PQ. It crosses AB at M — the midpoint (4 cm from each end) at 90°.
✅ PQ is the perpendicular bisector. Every point on PQ is exactly the same distance from A as from B.
Example 3: Triangle SSS — sides 5 cm, 6 cm, 7 cm
Construct triangle ABC with AB = 7 cm, BC = 5 cm, AC = 6 cm.
Step 1: Draw base AB = 7 cm.
Step 2: Set compass to 6 cm (= AC). Draw a long arc from A.
Step 3: Set compass to 5 cm (= BC). Draw arc from B. Mark the intersection as C.
Step 4: Draw AC and BC. Triangle complete!
💡 If the two arcs don't cross, check your compass radius matches the side lengths exactly.
Example 4: Locus — Signal Range
A radio mast at point M can send a signal exactly 3 km in every direction. Describe and draw the locus of the maximum-range boundary.
Step 1: The locus is the set of all points exactly 3 km from M.
Step 2: This is a circle, centre M, radius 3 km.
Step 3: Any phone inside the circle gets signal. The boundary is the circle itself.
🗺️ Real-world application: emergency services use locus circles to plan coverage zones.
📐 Interactive Construction Studio
🧭 Virtual Compass Tool
Click on the canvas to set the compass centre. Then drag outward to set the radius and draw an arc. Draw multiple arcs to practise constructions!
Click = set centre · Drag = draw arc · Multiple arcs supported
📏 Animated Step-by-Step Construction
🔺 Triangle Builder
🎯 Locus Explorer
Click the canvas to place a point, then watch the locus animate!
✏️ Exercise 1: Angle Measurement
Choose the correct type or size of angle.
✏️ Exercise 2: Identify Construction Steps
Select the correct next step or missing instruction for each construction.
✏️ Exercise 3: Triangle from Given Info
Identify what type of construction (SSS/SAS/ASA) is needed, or answer a triangle question.
🗺️ Exercise 4: Locus Descriptions
Choose the correct shape of the locus for each scenario.
🔺 Exercise 5: Mixed Construction Problems
Apply all your construction knowledge.
📝 Practice Questions
What type of angle is 135°? Choose: acute / right / obtuse / reflex.
An angle measures 47°. Is it acute or obtuse?
Using a protractor, describe the steps to draw an angle of 78°.
Name the tool used alongside a ruler to make accurate geometric constructions without measuring.
To construct a perpendicular bisector of AB, you must open the compass to more than _______ the length of AB.
Where do the two pairs of arcs cross in the perpendicular bisector construction? What is that point called?
True or false: every point on the perpendicular bisector of AB is the same distance from A and B.
In an angle bisector construction, why must you keep the compass width the same when drawing arcs from P and Q?
You are given sides AB = 5 cm, AC = 7 cm, and angle BAC = 40°. What type of triangle construction is this?
You are given AB = 6 cm, BC = 8 cm, CA = 10 cm. What type of construction is this?
For ASA construction with AB = 5 cm and angles at A and B, where does point C appear?
Describe the locus of all points exactly 4 cm from a fixed point P.
A dog is tied to a post by a 3 m lead. Describe the area the dog can reach.
Describe the locus of points equidistant from a straight road.
What is the locus of points equidistant from two fixed points A and B?
A phone must be within 500 m of a mast. Draw a rough sketch of the boundary locus.
Two towns are 10 km apart. A new school must be equidistant from both. What is the locus?
A point moves so it is always 2 cm from a straight line segment. Describe the full locus shape (include the end sections).
In a triangle construction, you are given: angle A = 60°, angle B = 70°, AB = 6 cm. Which method is used?
A treasure is buried equidistant from two trees and also 3 m from a wall. How many possible burial spots are there?
Obtuse
Acute (47° < 90°)
Draw a ray, place protractor centre at endpoint, align baseline, mark 78° on inner scale, join mark to endpoint.
Compass
Half (more than ½ of AB)
The arcs cross at two points above and below. The line joining them is the perpendicular bisector. The crossing on AB is the midpoint.
True
To ensure the two arcs from P and Q are the same size so they cross at the correct point R on the angle bisector.
SAS (Side-Angle-Side)
SSS (Side-Side-Side)
C is where the two rays from A and B intersect.
A circle centre P, radius 4 cm.
A circle (and interior) with centre at the post, radius 3 m.
Two parallel lines, one on each side of the road, equidistant from it.
The perpendicular bisector of segment AB.
A circle of radius 500 m centred on the mast.
The perpendicular bisector of the 10 km segment joining the two towns.
Two parallel lines + two semicircles at the ends = a stadium (discorectangle) shape.
ASA (Angle-Side-Angle)
Two possible spots (intersection of the perpendicular bisector and the locus line parallel to the wall).
🏆 Challenge: Multi-Step & Reasoning Problems
A point P is equidistant from two parallel lines L₁ and L₂ which are 8 cm apart. Describe and sketch the locus of P.
A goat is tied by a 6 m rope to the corner of a 4 m × 4 m square barn. Describe the locus of the furthest point the goat can reach. (Hint: the barn blocks part of the circle.)
Construct triangle ABC where AB = 9 cm, angle A = 50°, and BC = 7 cm. Explain why two triangles might be possible and how you decide which to draw.
Two radio masts A and B are 12 km apart. Mast A covers a radius of 8 km and mast B covers 9 km. Describe and sketch the region covered by both masts (i.e. the intersection of their loci).
You have constructed the perpendicular bisector of AB and the angle bisector of angle PQR. Prove (using a construction argument) that a point on the perpendicular bisector of AB is always equidistant from A and B.
A nature reserve is bounded by two straight fences meeting at an angle of 80°. A watering hole must be equidistant from both fences. Describe and construct the locus of the watering hole.
Using only a compass and straight edge (no protractor), explain how you could construct a 90° angle at a point on a line.
A point Q moves so it is always 3 cm from a fixed point P and also 3 cm from a fixed straight line L. The line L passes 5 cm from P. How many positions can Q have? Explain using locus ideas.
A line parallel to both L₁ and L₂, exactly 4 cm from each (halfway between them).
The goat can sweep a 270° arc of radius 6 m around the corner. Around the adjacent corners it can wrap: 90° arcs of radius 2 m (6−4) on two sides. The full locus is a combination of these arcs.
With SSA (two sides and a non-included angle), two triangles may be possible — one acute, one obtuse. Both arcs from C may cross BC. The problem context or an added constraint (e.g. angle C is acute) resolves the ambiguity.
Both loci are circles: circle A (centre A, radius 8 km) and circle B (centre B, radius 9 km). Since 8+9=17>12, they overlap. The intersection region is a lens shape bounded by arcs of both circles.
Let M be any point on the perp. bisector of AB. By construction, the arcs from A and B cross at M with equal compass radii → MA = MB. This holds for every point on the bisector by the arc method. ∎
The locus is the angle bisector of the 80° angle, i.e. the ray that makes 40° with each fence. Construct: arc from vertex cuts both fences, equal arcs from those points, join vertex to intersection.
Draw a line and mark point P. Use the perpendicular bisector construction: place compass on P, draw arc cutting the line at X and Y. Bisect XY — the bisector through P is perpendicular (90°) to the line.
The locus of points 3 cm from P is a circle (radius 3). The locus of points 3 cm from L is two parallel lines, 3 cm each side of L. L is 5 cm from P, so one parallel line is 5−3=2 cm from P and one is 5+3=8 cm from P. The circle of radius 3 intersects the closer parallel line at 2 points. So Q has 2 positions.