Constructions and Loci · Study Notes · Edexcel IGCSE 4MA1 Higher

Compass-and-ruler constructions: bisectors, perpendiculars. Loci: equidistant patterns. Always show ALL construction arcs.

At a glance

Perpendicular bisector: same radius from each point, connect intersections.
Angle bisector: arc through both rays, then equal arcs from intersections.
Locus from a POINT: CIRCLE.
Locus from a LINE: parallel line.
Locus from TWO POINTS: perpendicular bisector.
Locus from a CORNER: angle bisector.
ALWAYS leave construction arcs visible.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.6 — Construct perpendicular bisectors and angle bisectors using compasses.
4.6 — Find the locus of points satisfying given conditions.

Constructions

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Compass and ruler. Show all arcs.

Perpendicular bisector of $AB$ .

Set compass radius MORE than half $|AB|$ .
From $A$ : draw arc above and below $AB$ .
From $B$ (SAME radius): draw arcs that intersect the first arcs.
Connect the two intersection points.
The line passes through the midpoint and is perpendicular.

Equal-radius arcs from A and B cross at two points; the line through them is the perpendicular bisector.

Angle bisector of $\angle BAC$ .

From $A$ : draw an arc cutting both rays $AB$ and $AC$ . Call the intersections $P$ and $Q$ .
From $P$ : draw an arc into the angle.
From $Q$ (same radius): draw an arc that meets the first.
Connect $A$ to that intersection.
The line bisects the angle.

Perpendicular from a point to a line.

For point $P$ NOT on line $AB$ :

From $P$ : draw arcs cutting line $AB$ at two points (call them $X, Y$ ).
From $X, Y$ : draw arcs meeting on the OTHER SIDE of the line.
Connect $P$ to the new intersection.

Worked qualitative. Why does same-radius from two points give the perpendicular bisector?

Each arc is the locus of points equidistant from the centre.
Their intersection is equidistant from BOTH points.
The line connecting two equidistant points is the perpendicular bisector.

Edexcel tip. Construction arcs MUST BE VISIBLE. Erasing them loses you the M1 mark.

Same radius for both arcs.
Show all construction arcs.
Use ruler ONLY to connect points.
Sharp pencil for accuracy.

Loci — equidistant patterns

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Standard loci based on what's equidistant.

Locus = path/region of points satisfying a condition.

Standard loci:

From	Locus
Single point	Circle (centre at point, radius = distance)
Single line	Pair of parallel lines (distance away each side)
Two points (equidistant)	Perpendicular bisector
Two lines (equidistant)	Angle bisector
Inside region	Intersection of multiple loci

A locus a fixed distance from a point is a circle; from a line it is a pair of parallel lines.

Worked examples:

'Within $5$ cm of $P$ .' Disc centred at $P$ , radius $5$ .

' $3$ cm from line $AB$ .' Two parallel lines, $3$ cm above and below $AB$ .

'Closer to $A$ than $B$ .' Region on $A$ 's side of the perpendicular bisector of $AB$ .

'Within $4$ cm of $A$ AND closer to $A$ than $B$ .' Intersection of disc and half-plane.

Combined loci. Often Edexcel asks for the region satisfying TWO conditions. Find each locus, shade where they overlap.

Worked qualitative. Why does 'closer to $A$ than $B$ ' give a half-plane?

Equal distance: perpendicular bisector.
Closer to $A$ : the side OF the bisector that contains $A$ .
That's a half-plane.

Edexcel tip. When asked for a region (not just a curve), SHADE the area. Mark schemes credit the shading.

From point: circle.
From line: parallel lines.
Equidistant from 2 points: perp bisector.
Equidistant from 2 lines: angle bisector.
Combined: intersect loci.

Quick recap

Perpendicular bisector: equal-radius arcs from both endpoints.
Angle bisector: arc through rays, then equal arcs from intersections.
Locus from point: circle.
Locus from two points: perpendicular bisector.
Always show construction arcs.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Perpendicular bisector: arcs from $A$ + $B$ .
Angle bisector: arcs from $A$ through both rays, then equal arcs.
Locus from point: CIRCLE.
Locus equidistant: perpendicular bisector / angle bisector.

How it’s examined

Constructions on Higher Tier (3-5 marks). Examiner reports flag (1) erasing arcs, (2) using different radii, (3) wrong locus shape.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on construction , written exactly the way a tutor would explain them at the board.

1Construct a perpendicular bisector
Foundation• construction, bisector
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Question
Describe how to construct the perpendicular bisector of line segment $AB$ .
Step-by-step solution
1. Step 1
  Set compass to MORE than half the length of $AB$ .
2. Step 2
  From $A$ , draw arcs above and below the line.
3. Step 3
  Without changing the compass, do the same from $B$ .
4. Step 4
  Two pairs of arcs intersect — one pair above, one below the line. Draw a line through both intersections. This is the perpendicular bisector.
Answer
Draw arcs from $A$ , then $B$ , with same compass setting. Connect the intersections.
Examiner tip
Mark schemes credit ARCS being clearly visible. Don't erase them.
2Construct an angle bisector
Higher• construction, angle bisector
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Question
Bisect angle $\angle BAC$ .
Step-by-step solution
1. Step 1
  From $A$ , draw an arc that crosses both $AB$ and $AC$ . Let the intersections be $P$ and $Q$ .
2. Step 2
  From $P$ and $Q$ , draw arcs (same radius) that intersect.
3. Step 3
  Draw a line from $A$ through this intersection.
Answer
Arc from $A$ through both rays. Equal arcs from intersections meet inside angle. Line from $A$ through that point bisects.
3Locus equidistant from two points
Higher• locus
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Question
Sketch the locus of points equidistant from $A(0, 0)$ and $B(8, 0)$ .
Step-by-step solution
1. Step 1
  Equidistant from two points = perpendicular bisector of the line segment joining them.
2. Step 2
  Midpoint of $A, B$ is $(4, 0)$ . Perpendicular bisector is the vertical line $x = 4$ .
Answer
$x = 4$ (vertical line through $(4, 0)$ ).
4Locus equidistant from a line
Higher• locus
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Question
Describe the locus of points $5$ cm from a fixed point $P$ .
Step-by-step solution
1. Step 1
  Equidistant from a single point = a CIRCLE centered at that point.
2. Step 2
  Centre at $P$ , radius $5$ cm.
Answer
Circle centre $P$ , radius $5$ cm.
Examiner tip
Loci patterns: equidistant from a POINT → CIRCLE. From a LINE → parallel line. From TWO POINTS → perpendicular bisector. From a CORNER → angle bisector.
5Choosing the right scale on a protractor [1 mark]
Getting started• Paper 1H/2H style — 1 mark• measuring angles, protractor
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Question
Kofi measures angle $ABC$ with a protractor. The centre of the protractor is on $B$ and the arm $BA$ lies along the zero line.

The arm $BC$ passes through the mark where the two scales on the protractor read $48$ and $132$ .

Angle $ABC$ is obtuse.

Write down the size of angle $ABC$ . [1]
Step-by-step solution
1. Step 1
  The two scales always add up to 180°, because they measure the angle from opposite ends of the zero line: $48 + 132 = 180$ .
2. Step 2
  B1 The angle is obtuse, so it is the reading greater than 90°: angle $ABC = 132$ °.
Answer
132°
Examiner tip
Examiner tip: Always decide whether the angle is acute or obtuse before reading a protractor. Better still, read the scale that has its zero on the arm you lined up — here the one that starts at $0$ on $BA$ .

Common mistake: Writing $48$ °, the reading on the other scale. It is measured from the opposite end of the zero line, so it is the angle between $BC$ and the extension of $AB$ , not angle $ABC$ .
6Constructing an equilateral triangle [2 marks]
Getting started• Paper 1H/2H style — 2 marks• construction, equilateral triangle, compasses
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Question
Use ruler and compasses only to construct an equilateral triangle with sides of length $5.8$ cm.

You must show all your construction lines. [2]
Step-by-step solution
1. Step 1
  M1 Draw a line $PQ = 5.8$ cm. Set the compasses to exactly $5.8$ cm against the ruler and draw an arc centred on $P$ above the line.
2. Step 2
  A1 Without changing the compasses, draw an arc centred on $Q$ crossing the first arc at $R$ . Join $PR$ and $QR$ , and leave both arcs on the page.
Answer
An equilateral triangle with each side $5.8$ cm (within $2$ mm) and both construction arcs visible.
Examiner tip
Examiner tip: One compass setting does everything here: the base and both arcs are all $5.8$ cm, so every angle of the triangle is 60°. This is also how a 60° angle is constructed without a protractor.

Common mistake: Measuring the 60° angles with a protractor instead of using arcs. The question says ruler and compasses only, so a protractor-drawn triangle with no arcs scores 0.
7Scale drawing to find an angle of elevation [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• scale drawing, angle of elevation
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Question
A point $X$ is on level ground, $20$ m from the foot $F$ of a vertical tower. The top $T$ of the tower is $13$ m above the ground.

(a) Using a scale of $1$ cm to $2$ m, make a scale drawing of triangle $XFT$ . [2]

(b) Use your drawing to find the angle of elevation of $T$ from $X$ . [1]
Step-by-step solution
1. Step 1
  B1 (a) Convert and draw the ground: $20 \div 2 = 10$ cm, so draw $XF = 10$ cm.
2. Step 2
  B1 The tower is vertical, so draw a line at 90° to $XF$ at $F$ , and mark $T$ at $13 \div 2 = 6.5$ cm up it. Join $XT$ .
3. Step 3
  B1 (b) Measure angle $FXT$ with the protractor centred on $X$ : about 33°, accepted from 32° to 35°.
Answer
33° (accept 32° to 35°)
Examiner tip
Examiner tip: The angle of elevation is measured at the observer, $X$ , upwards from the horizontal. Check with trigonometry: $\tan^{-1}\left(\dfrac{13}{20}\right) = 33.0$ °.

Common mistake: Measuring the angle at the top of the tower, angle $XTF$ , which is about 57° (that is, 90° minus 33°). It is at the top of the tower, not at the observer, so it is not the angle of elevation.
8A regular hexagon inside a circle [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• construction, regular hexagon, compasses
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Question
(a) Use ruler and compasses only to construct a regular hexagon $ABCDEF$ whose vertices lie on a circle of radius $4$ cm. You must show all your construction lines. [3]

(b) Measure the length of the diagonal $AC$ . [1]
Step-by-step solution
1. Step 1
  M1 (a) Draw a circle of radius $4$ cm with centre $O$ . Mark any point $A$ on the circle. Keeping the compasses at $4$ cm, put the point on $A$ and draw an arc cutting the circle at $B$ .
2. Step 2
  M1 Still at $4$ cm, step round the circle: arc from $B$ gives $C$ , from $C$ gives $D$ , then $E$ and $F$ . The sixth step should land back on $A$ .
3. Step 3
  A1 Join $A$ , $B$ , $C$ , $D$ , $E$ , $F$ in order. Each side should be $4$ cm (within $2$ mm).
4. Step 4
  B1 (b) Measure $AC$ : about $6.9$ cm, accepted from $6.8$ cm to $7.1$ cm.
  $AC = 2 \times 4 \sin 60^\circ = 4\sqrt{3} = 6.93\ldots \text{ cm}$
Answer
(b) $AC = 6.9$ cm (accept $6.8$ cm to $7.1$ cm)
Examiner tip
Examiner tip: The construction works because each triangle $OAB$ , $OBC$ , … has all three sides equal to the radius, so it is equilateral and the angle at $O$ is 60°; six of them make 360°. If the sixth step misses $A$ by more than a millimetre or two, your compasses slipped.

Common mistake: Changing the compass setting between steps, or stepping from the previous arc mark in the wrong direction. Keep the same $4$ cm radius throughout and always step from the newest point.
9Two perpendicular bisectors meeting inside a triangle [5 marks]
Stretch• Paper 1H/2H style — 5 marks• construction, perpendicular bisector, triangle
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Question
You must show all your construction lines.

(a) Use ruler and compasses to construct triangle $ABC$ with $AB = 8$ cm, $BC = 7$ cm and $AC = 6$ cm. [1]

(b) Use ruler and compasses only to construct the perpendicular bisectors of $AB$ and of $BC$ . Label the point where they meet $O$ . [2]

(c) Measure $OA$ , $OB$ and $OC$ , and give a reason why they are equal. [2]
Step-by-step solution
1. Step 1
  B1 (a) Draw $AB = 8$ cm; arcs of radius $6$ cm from $A$ and $7$ cm from $B$ cross at $C$ . Join up and keep the arcs.
2. Step 2
  M1 (b) For $AB$ : with a setting more than $4$ cm, draw equal arcs from $A$ and $B$ on both sides of $AB$ and rule through the two crossings. Repeat for $BC$ with a setting more than $3.5$ cm.
3. Step 3
  A1 Both lines accurate (each through the midpoint of its side at 90°), extended until they meet at $O$ .
4. Step 4
  B1 (c) Each distance measures about $4.1$ cm, accepted from $4.0$ cm to $4.3$ cm.
5. Step 5
  B1 Reason: $O$ is on the perpendicular bisector of $AB$ , so $OA = OB$ ; $O$ is also on the perpendicular bisector of $BC$ , so $OB = OC$ .
Answer
(c) $OA = OB = OC = 4.1$ cm (accept $4.0$ to $4.3$ cm); every point on the perpendicular bisector of a segment is the same distance from both ends of it
Examiner tip
Examiner tip: A third perpendicular bisector (of $AC$ ) would pass through the same point $O$ — a quick way to test the accuracy of your drawing. The accurate distance is $4.13$ cm.

Common mistake: Drawing a line from a vertex to the midpoint of the opposite side (a median) instead of a perpendicular bisector. A median usually does not meet the side at 90° and has no arcs, so it scores nothing in (b).
10Locating a boat from two angles on a scale drawing [5 marks]
Stretch• Paper 1H/2H style — 5 marks• scale drawing, two angles and a side
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Question
Two observers stand at points $A$ and $B$ on a straight shore. $AB = 400$ m.

A boat $S$ is out at sea. Angle $SAB = 58$ ° and angle $SBA = 47$ °.

(a) Using a scale of $1$ cm to $50$ m, make a scale drawing to show the position of the boat. [3]

(b) Use your drawing to find the real distance of the boat from $A$ . [2]
Step-by-step solution
1. Step 1
  B1 (a) $400 \div 50 = 8$ , so draw $AB = 8$ cm.
2. Step 2
  B1 At $A$ , measure 58° from $AB$ and draw a long line.
3. Step 3
  B1 At $B$ , on the same side of $AB$ , measure 47° from $BA$ and draw a long line. $S$ is where the lines cross.
4. Step 4
  M1 (b) Measure $AS$ (about $6.1$ cm) and convert with the scale.
  $6.1 \times 50 = 305 \text{ m}$
5. Step 5
  A1 Accept any answer from $295$ m to $310$ m (a measurement of $5.9$ cm to $6.2$ cm).
Answer
$305$ m from a $6.1$ cm measurement (accept $295$ m to $310$ m)
Examiner tip
Examiner tip: The accurate value from the sine rule is $\dfrac{400 \sin 47^\circ}{\sin 75^\circ} = 303$ m. On a $1$ cm to $50$ m scale, a $2$ mm drawing error is $10$ m, which is why the accepted range is so wide — but it is not wider than that, so draw carefully.

Common mistake: Putting 58° at $B$ and 47° at $A$ . The boat then ends up further from $A$ : the distance measures about $351$ m, well outside the range.

Question 1
Paper 1H/2H style — 1 mark1 mark
The dial of a speedometer is marked at $40$ km/h and at $60$ km/h. The gap between these marks is divided into $4$ equal intervals.

The needle points to the third mark after $40$ km/h.

Write down the speed shown. [1]
Model answer
One interval $= 20 \div 4 = 5$ km/h

$40 + 3 \times 5 = 55$

Answer: $55$ km/h
Why this scores
Mark scheme:
- B1 for $55$ cao
Why this scores full marks: The script finds the value of one interval before counting. Assuming each mark is $2$ km/h (giving $46$ ) or $1$ km/h (giving $43$ ) is the usual error.
Question 2
Paper 1H/2H style — 2 marks2 marks
Ravi is asked to draw a triangle $ABC$ with $AB = 5$ cm, angle $BAC = 110$ ° and angle $ABC = 75$ °.

(a) Explain why he cannot draw this triangle. [1]

(b) Angle $BAC$ stays at 110°. Write down the greatest whole number of degrees that angle $ABC$ could be. [1]
Model answer
(a) $110 + 75 = 185$ , which is more than $180$ . The angles in a triangle add up to 180°, so the two lines from $A$ and $B$ would not meet on that side of $AB$ .

(b) Angle $ABC < 180 - 110 = 70$ °

Answer: 69°
Why this scores
Mark scheme:
- B1 for a reason that uses the angle sum, e.g. $110 + 75 = 185 > 180$ oe
- B1 for $69$ cao
Why this scores full marks: The reason quotes the fact (angles in a triangle add up to 180°) and shows the numbers that break it. In (b) the candidate sees that 70° would leave nothing for the third angle, so the largest whole number is 69°.
Question 3
Paper 1H/2H style — 2 marks2 marks
On a scale drawing of a sports hall, a wall of real length $24$ m is drawn $6$ cm long.

(a) Work out the scale of the drawing, in the form $1$ cm represents $n$ m. [1]

(b) A door is $0.3$ cm wide on the drawing. Work out the real width of the door, in metres. [1]
Model answer
(a) $24 \div 6 = 4$

Answer: $1$ cm represents $4$ m

(b) $0.3 \times 4 = 1.2$

Answer: $1.2$ m
Why this scores
Mark scheme:
- B1 for $4$ (m) cao
- B1 for $1.2$ m, ft their scale from (a)
Why this scores full marks: Real length divided by drawing length gives the metres per centimetre. Dividing the other way ( $6 \div 24 = 0.25$ ) is the common slip, and would make the hall smaller than its own drawing.
Question 4
Paper 1H/2H style — 3 marks3 marks
(a) Use ruler and compasses to construct triangle $XYZ$ with $XY = 9$ cm, $YZ = 6$ cm and $XZ = 7.5$ cm. You must show all your construction lines. [2]

(b) Measure the size of the smallest angle in the triangle. [1]
Model answer
(a) $XY$ drawn $9$ cm. Arc radius $7.5$ cm centred on $X$ ; arc radius $6$ cm centred on $Y$ ; crossing at $Z$ . $XZ$ and $YZ$ joined; arcs left.

(b) The smallest angle is opposite the shortest side, $YZ$ , so it is angle $YXZ$ .

Answer: 41°
Why this scores
Mark scheme:
- M1 for $XY = 9$ cm and one arc of the correct radius from the correct end
- A1 for a complete triangle, sides within $2$ mm, both arcs shown
- B1 for an angle in the range 40° to 43°
Why this scores full marks: Before measuring, the candidate reasons that the smallest angle is opposite the shortest side, so only one angle needs measuring. The accurate value is 41.4°.
Question 5
Paper 1H/2H style — 3 marks3 marks
The line segment $AB$ has length $7$ cm.

(a) Use ruler and compasses only to construct the perpendicular bisector of $AB$ . You must show all your construction lines. [2]

(b) The perpendicular bisector cuts $AB$ at $M$ . $C$ is the point on the perpendicular bisector with $MC = 5$ cm. Measure the length of $AC$ . [1]
Model answer
(a) Compasses set to $5$ cm (more than $3.5$ cm). Arcs from $A$ and from $B$ , same setting, crossing above and below $AB$ . Line ruled through both crossings; $M$ marked on $AB$ .

(b) $C$ marked $5$ cm from $M$ along the bisector; $AC$ measured.

Answer: $AC = 6.1$ cm
Why this scores
Mark scheme:
- M1 for two pairs of intersecting arcs of equal radius, centred on $A$ and $B$
- A1 for the perpendicular bisector within $2$ mm of the midpoint and 2° of 90°
- B1 for $AC$ in the range $5.9$ to $6.3$ cm
Why this scores full marks: One compass setting is used for all four arcs, and the arcs stay visible. The candidate could check that $BC$ also measures $6.1$ cm, since every point on the perpendicular bisector is the same distance from $A$ and $B$ ; Pythagoras gives $\sqrt{3.5^{2} + 5^{2}} = 6.10$ cm.
Question 6
Paper 1H/2H style — 4 marks4 marks
(a) Use ruler and compasses only to construct a square $PQRS$ with sides of length $4.5$ cm. You must show all your construction lines. [3]

(b) Measure the length of the diagonal $PR$ . [1]
Model answer
(a) $PQ$ drawn $4.5$ cm and extended beyond $P$ .

Arc centred on $P$ cutting the line at two points, one each side of $P$ . Equal arcs from those two points crossing above $P$ ; line ruled from $P$ through the crossing (angle $QPS = 90$ °). $S$ marked $4.5$ cm up it.

Arcs of radius $4.5$ cm centred on $Q$ and on $S$ crossing at $R$ . $QR$ and $SR$ joined.

(b) Answer: $PR = 6.4$ cm
Why this scores
Mark scheme:
- M1 for a correct construction of a perpendicular at $P$ (arcs shown)
- A1 for a right angle at $P$ (±2°) with $PS = 4.5$ cm
- M1 for arcs of radius $4.5$ cm from $Q$ and $S$ to locate $R$
- B1 for $PR$ in the range $6.2$ to $6.5$ cm
Why this scores full marks: The right angle is made as the perpendicular bisector of a segment centred on $P$ , so no protractor is needed. The fourth vertex is located by arcs, not guessed. The accurate diagonal is $4.5\sqrt{2} = 6.36$ cm.
Question 7
Paper 1H/2H style — 4 marks4 marks
A triangular field $ABC$ has $AB = 120$ m, $BC = 90$ m and $AC = 100$ m.

(a) Using a scale of $1$ cm to $20$ m, use ruler and compasses to construct a scale drawing of the field. You must show all your construction lines. [3]

(b) Use your drawing to find the size of angle $ABC$ . [1]
Model answer
(a) $120 \div 20 = 6$ cm, $90 \div 20 = 4.5$ cm, $100 \div 20 = 5$ cm.

$AB$ drawn $6$ cm. Arc radius $5$ cm centred on $A$ ; arc radius $4.5$ cm centred on $B$ ; crossing at $C$ . Triangle completed; arcs left.

(b) Answer: angle $ABC = 55$ °
Why this scores
Mark scheme:
- B1 for all three drawing lengths $6$ , $4.5$ and $5$ cm
- M1 for $AB$ drawn and arcs of the correct radii from $A$ and $B$
- A1 for a complete triangle, sides within $2$ mm, arcs shown
- B1 for an angle in the range 53° to 56°
Why this scores full marks: The lengths are converted and written down first, then the triangle is constructed with compasses, not guessed. Angles are not changed by a scale drawing, so the measured 55° is the real angle — no conversion is needed (the accurate value is 54.6°).
Question 8
Paper 1H/2H style — 5 marks5 marks
You must show all your construction lines.

(a) Use ruler and compasses to construct triangle $ABC$ with $AB = 10$ cm, $BC = 8$ cm and $AC = 7$ cm. [1]

(b) Use ruler and compasses only to construct the bisector of angle $BAC$ and the bisector of angle $ABC$ . The bisectors meet at $I$ . [3]

(c) Measure the length of $AI$ . [1]
Model answer
(a) $AB$ drawn $10$ cm; arcs of $7$ cm from $A$ and $8$ cm from $B$ crossing at $C$ .

(b) At $A$ : arc cutting $AB$ and $AC$ ; equal arcs from those points; line from $A$ through the crossing.

At $B$ : arc cutting $BA$ and $BC$ ; equal arcs; line from $B$ through the crossing.

The lines meet at $I$ .

(c) Answer: $AI = 5.0$ cm
Why this scores
Mark scheme:
- B1 for a correct triangle with arcs shown
- M1 for a correct angle-bisector construction at $A$
- M1 for a correct angle-bisector construction at $B$
- A1 for both bisectors within 2° and $I$ marked
- B1 for $AI$ in the range $4.9$ to $5.2$ cm
Why this scores full marks: Each bisector has its own complete set of arcs, and both lines are ruled long enough to cross. The candidate can check the drawing: the bisector of angle $ACB$ must pass through the same point $I$ . The accurate length is $AI = 5.02$ cm.
Question 9
Paper 1H/2H style — 5 marks5 marks
You must show all your construction lines.

(a) $AB$ is a line segment of length $9$ cm. Use ruler and compasses only to construct a right angle at $B$ . Complete the right-angled triangle $ABC$ with $BC = 5$ cm. [2]

(b) Use ruler and compasses only to construct the perpendicular bisector of $AC$ . It meets $AB$ at $D$ . [2]

(c) Measure the length of $AD$ . [1]
Model answer
(a) $AB$ drawn $9$ cm and extended beyond $B$ . Arcs from $B$ cut the line at two points either side of $B$ ; equal arcs from those points cross; line ruled from $B$ through the crossing (90°). $C$ marked $5$ cm along it; $AC$ joined.

(b) Compasses set to more than half of $AC$ ; arcs from $A$ and $C$ , same setting, crossing on both sides of $AC$ ; line ruled through the crossings, meeting $AB$ at $D$ .

(c) Answer: $AD = 5.9$ cm
Why this scores
Mark scheme:
- M1 for a correct construction of the perpendicular at $B$ (arcs shown)
- A1 for a right angle (±2°) with $BC = 5$ cm and the triangle completed
- M1 for two pairs of intersecting arcs of equal radius from $A$ and $C$
- A1 for the perpendicular bisector of $AC$ drawn and $D$ marked
- B1 for $AD$ in the range $5.7$ to $6.0$ cm
Why this scores full marks: Both constructions are complete, with every arc shown. The candidate can check the result: $D$ is on the perpendicular bisector of $AC$ , so $DC$ should equal $AD$ . Algebraically, $(9 - x)^{2} = x^{2} + 25$ gives $BD = 3.11$ cm and $AD = 5.89$ cm.
Question 10
Paper 1H/2H style — 5 marks5 marks
$AB$ is a line segment of length $6$ cm.

(a) Use ruler and compasses only to construct an angle of 75° at $A$ , with $AB$ as one arm. You must show all your construction lines. [4]

(b) $C$ is the point on the other arm of the angle with $AC = 5$ cm. Measure the length of $BC$ . [1]
Model answer
(a) $BA$ extended beyond $A$ . Perpendicular at $A$ constructed (equal arcs either side of $A$ , then crossing arcs): angle of 90° to $AB$ .

Arc centred on $A$ cutting $AB$ at $X$ ; same setting, arc centred on $X$ cutting the first arc at $Y$ : angle $XAY = 60$ °.

Angle between $AY$ (60°) and the perpendicular (90°) bisected with equal arcs; line ruled from $A$ through the crossing. It makes $60 + 15 = 75$ ° with $AB$ .

(b) Answer: $BC = 6.7$ cm
Why this scores
Mark scheme:
- M1 for a correct construction of 90° at $A$
- M1 for a correct construction of 60° at $A$
- M1 for bisecting the angle between the 60° and 90° lines
- A1 for a 75° angle (±2°) with all arcs shown
- B1 for $BC$ in the range $6.6$ to $6.9$ cm
Why this scores full marks: The candidate sees $75 = \dfrac{60 + 90}{2}$ and builds it from two standard constructions and one bisection. Every set of arcs is left visible. By the cosine rule, $BC = \sqrt{6^{2} + 5^{2} - 60 \cos 75^\circ} = 6.74$ cm.

4MA1 Constructions and Loci

At a glance

What you’ll learn

Constructions

Loci — equidistant patterns

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Construct a perpendicular bisector

2Construct an angle bisector

3Locus equidistant from two points

4Locus equidistant from a line

5Choosing the right scale on a protractor [1 mark]

6Constructing an equilateral triangle [2 marks]

7Scale drawing to find an angle of elevation [3 marks]

8A regular hexagon inside a circle [4 marks]

9Two perpendicular bisectors meeting inside a triangle [5 marks]

10Locating a boat from two angles on a scale drawing [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10