Constructions & loci – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator2 marks

Describe fully the locus of points that are

(a) exactly 3 cm from a fixed point P, [1]

(b) the same distance from point A as from point B. [1]

(a) a circle, centre P, radius 3 cm (both centre and radius needed; allow 'a circle of radius 3 cm around P')B1
(b) the perpendicular bisector of AB (accept 'a line through the midpoint of AB at right angles to AB')B1
Answer: (a) circle, centre P, radius 3 cm (b) perpendicular bisector of AB
Examiner tip: 'A circle' alone is not a full description — say where its centre is and how big its radius is.

2Grade 4–5Calculator4 marks

AB is a line of length 8 cm.

(a) Describe how to construct the perpendicular bisector of AB using ruler and compasses only.
Say what must be left on your diagram. [2]

(b) Point P lies on the perpendicular bisector of AB. PA = 5 cm.
Work out the distance from P to the line AB. [2]

(a) two pairs of intersecting arcs drawn with compasses centred on A and on B, with the SAME radius (more than 4 cm), arcs left visibleB1
(a) straight line drawn through both intersection points of the arcs (it crosses AB at its midpoint, 4 cm from A, at $90°$)B1
(b) $\sqrt{5^2 - 4^2}$ (using half of AB = 4 cm)M1
(b) $3$ cmA1
Answer: (a) arcs of equal radius from A and B; line through their intersections (b) 3 cm
Examiner tip: Never rub out your construction arcs — without them the construction mark is lost even if the line is in the right place.

3Grade 4–5Non-calculator3 marks

The diagram shows angle XOY $= 70°$.

(a) Describe how to construct the bisector of angle XOY using ruler and compasses only. [2]

(b) Write down the size of each of the two angles formed by the bisector. [1]

(a) an arc centred on O cutting both OX and OY, then two arcs of equal radius centred on the two crossing points, intersecting inside the angleB1
(a) straight line drawn from O through the intersection of the last two arcs, all construction arcs left visibleB1
(b) $35°$B1
Answer: (a) arc from O cutting both arms; equal arcs from those points; line from O through their intersection (b) $35°$
Examiner tip: Keep the compass width the same for the two arcs drawn from the arms — otherwise the line will not bisect the angle.

4Grade 4–5Calculator5 marks

ABCD is a rectangular garden. AB = 10 m and BC = 6 m.
A scale drawing uses 1 cm to represent 1 m.

A tree is to be planted in the garden so that it is

nearer to AB than to AD, and
more than 4 m from C.

(a) Describe the construction you would draw and the region you would shade to show where the tree can be planted. [3]

(b) The tree is planted at T, which is 3 m from AB and 7 m from AD.
Show that T satisfies both conditions. [2]

(a) describes constructing the bisector of angle DAB (accept 'bisect the angle at A', 'line from A at $45°$ to AB'; allow a description of the compass method: arc from A cutting AB and AD, equal arcs from those points, line from A through the crossing)B1
(a) describes an arc (or circle) with centre C and radius 4 cm (accept 4 m, the real distance)B1
(a) correct region described: inside the rectangle, on the AB side of (below) the angle bisector and outside the arc centre CB1
(b) distance $TC = \sqrt{(10-7)^2 + (6-3)^2}$ $(= \sqrt{18} = 4.24...)$ (allow $TC^2 = 3^2 + 3^2 = 18$)M1
(b) $4.24... > 4$ (allow $18 > 16$ if comparing squares) and 3 m from AB is less than 7 m from AD, so T satisfies both conditionsA1
Answer: (a) bisect angle DAB; draw an arc centre C, radius 4 cm; shade inside the rectangle, below the bisector (AB side) and outside the arc (b) TC = 4.24 m > 4 m and 3 m < 7 m
Examiner tip: 'More than 4 m from C' means OUTSIDE the arc — shading inside the circle is the most common slip.

5Grade 4–5Calculator5 marks

ABCD is a rectangular field. AB = 12 m and AD = 8 m.
A goat is tied to a post at corner A with a rope of length 6 m.

(a) Describe the boundary of the region of the field the goat can reach. [1]

(b) Work out the area of the field the goat can reach. Give your answer correct to 1 decimal place. [2]

(a) a quarter circle (arc) of radius 6 m, centre A, inside the fieldB1
(b) $\pi \times 6^2 \div 4$M1
(b) $28.3$ m² (from $28.27...$)A1
(c) $\dfrac{28.27...}{12 \times 8} \times 100$M1
(c) $29.5\%$ (accept $29.4$ to $29.5$)A1
Answer: (a) quarter circle, radius 6 m, centre A (b) 28.3 m² (c) 29.5%
Examiner tip: The rope reaches a quarter of a circle because the corner is a right angle — using the full circle area gives an answer 4 times too big.

6Grade 8–9HigherCalculator4 marks

Two radio masts, A and B, are 10 km apart.
The signal from mast A reaches up to 7 km. The signal from mast B reaches up to 6 km.
The edges of the two signal regions meet at the points P and Q.

(a) Describe the region that receives a signal from both masts. [1]

(b) Work out the distance PQ. Give your answer correct to 3 significant figures. [3]

(a) the region inside both circles: inside the circle centre A radius 7 km and inside the circle centre B radius 6 kmB1
(b) cosine rule for angle PAB: $\cos A = \dfrac{7^2 + 10^2 - 6^2}{2 \times 7 \times 10}$ $(A = 36.18...°)$, or for angle PBA: $\cos B = \dfrac{6^2 + 10^2 - 7^2}{2 \times 6 \times 10}$ $(B = 43.53...°)$, or finds $x = 5.65$ where $7^2 - x^2 = 6^2 - (10-x)^2$M1
(b) half of PQ $= 7\sin(36.18...°)$ or $6\sin(43.53...°)$ or $\sqrt{7^2 - 5.65^2}$ $(= 4.132...)$, then doublesM1
(b) $8.26$ km (from $8.26498...$; accept answers in the range $8.26$ to $8.27$ from rounded working, e.g. using $A = 36.2°$)A1
Answer: (a) the overlap of the two circles (b) 8.26 km
Examiner tip: PQ is perpendicular to AB and bisected by it — find half of PQ in a right-angled triangle, then remember to double it.

7Grade 1–3FoundationNon-calculator4 marks

The diagram shows a quadrilateral PQRS.

(a) Which side of PQRS is parallel to PQ? [1]

(b) Which side of PQRS is perpendicular to PQ? [1]

(d) Write down the mathematical name of the quadrilateral PQRS. [1]

(a) SR (accept RS)B1
(b) PS (accept SP)B1
(c) angle PQR (accept RQP, $\angle PQR$, $P\hat{Q}R$; do not accept angle Q on its own)B1
(d) trapezium (accept right-angled trapezium; do not accept parallelogram or kite)B1
Answer: (a) SR (b) PS (c) angle PQR (d) trapezium
Examiner tip: When naming an angle with three letters, the vertex where the angle sits must be the middle letter, so the angle at Q is PQR, not QPR.

8Grade 4–5FoundationNon-calculator4 marks

Triangle ABC has

AB = 8 cm
angle BAC $= 60°$
AC = 5 cm

(a) Use ruler and compasses only to construct triangle ABC.
You must not use a protractor and you must show all your construction lines. [3]

(b) Measure the length of BC on your diagram. [1]

(a) line AB drawn of length 8 cm (accept 7.8 cm to 8.2 cm)B1
(a) correct construction of $60°$ at A: an arc centred on A cutting AB, and an arc of the same radius centred on that crossing point, with the second arm drawn from A through where the two arcs meet (arcs visible; angle within $\pm 2°$)B1
(a) point C marked on the $60°$ arm with AC = 5 cm (accept 4.8 cm to 5.2 cm) and BC joined to complete the triangleB1
(b) $7$ cm (accept 6.8 cm to 7.2 cm; ft their accurate diagram)B1
Answer: (a) accurate construction with arcs shown (b) BC = 7 cm
Examiner tip: The 60° angle comes from an equilateral triangle, so the compasses must NOT be changed between the two arcs, and both arcs must be left on the page.

9Grade 4–5Calculator5 marks

The diagram shows a house H and a straight road.
A path is to be laid from H to the road.

(a) Describe how to construct, using ruler and compasses only, the line from H that is perpendicular to the road. [2]

The perpendicular meets the road at F. HF = 40 m.

(b) Explain why the path should be laid along HF if it is to be as short as possible. [1]

A second path is laid in a straight line from H to a bus stop R on the road. HR = 58 m.

(a) arc drawn with centre H cutting the road at two points (e.g. X and Y)B1
(a) arcs of the same radius centred on X and on Y crossing on the other side of the road (from H), then a straight line from H through that crossing point to the road; all arcs left visibleB1
(b) the perpendicular distance from a point to a line is the shortest distance (accept 'any other path from H to the road would be the hypotenuse of a right-angled triangle, so longer than HF')C1
(c) $\sqrt{58^2 - 40^2}$ (accept $58^2 - 40^2 = 1764$ or $FR^2 + 40^2 = 58^2$)M1
(c) $42$ mA1
Answer: (a) arc from H cutting the road twice; equal arcs from those two points; line from H through their crossing (b) the perpendicular is the shortest distance from a point to a line (c) 42 m
Examiner tip: In (c) HR is the hypotenuse, so you subtract the squares; adding them (58² + 40²) is the most common lost mark.

10Grade 4–5Non-calculator4 marks

The diagram shows a straight line $\ell$ and a point P on the line.

(a) Describe how to construct, using ruler and compasses only, the line through P that is perpendicular to $\ell$. [2]

(b) Explain how Mia can use your construction to make an angle of $45°$ at P without a protractor. [1]

(c) Mia's $45°$ line and the line $\ell$ make two angles at P.
Write down the size of the larger of these angles. [1]

(a) compasses centred on P, two arcs cutting $\ell$ at equal distances either side of P (e.g. at A and B)B1
(a) compasses opened wider, arcs of the same radius from A and from B crossing above (or below) the line, then a straight line through P and that crossing point; arcs left visibleB1
(b) bisect one of the $90°$ angles at P (arc from P cutting both arms, equal arcs from those two points, line from P through their crossing) (accept 'bisect the right angle')B1
(c) $135°$B1
Answer: (a) equal arcs from P cutting the line at A and B; equal, wider arcs from A and B; line from P through the crossing (b) bisect the 90° angle (c) 135°
Examiner tip: In (a) the second pair of arcs must be drawn with a radius bigger than PA, otherwise they will not cross.

11Grade 4–5Calculator5 marks

The diagram shows a line segment AB of length 6 cm.

A region R contains all the points that are 2 cm or less from the line segment AB.

(a) Describe fully the shape of the boundary of region R. [2]

(b) Work out the area of region R.
Give your answer correct to 3 significant figures. [3]

(a) two straight lines, each 6 cm long, parallel to AB and 2 cm from it (one on each side)B1
(a) two semicircles of radius 2 cm, centred on A and on B, joining the ends of the parallel lines (do not accept a rectangle with square corners)B1
(b) area of the rectangle $6 \times 4\ (= 24)$M1
(b) area of the two semicircles $\pi \times 2^2\ (= 12.56...)$M1
(b) $36.6$ cm² (from $36.566...$; accept $36.56$ or $36.57$, e.g. from $\pi = 3.14$; $24 + 4\pi$ alone is A0 as 3 s.f. is asked for)A1
Answer: (a) two parallel lines 6 cm long, 2 cm either side of AB, joined by semicircles of radius 2 cm centred on A and B (b) 36.6 cm²
Examiner tip: Beyond the ends of AB the nearest point of the segment is A or B itself, so the corners are semicircles, not square corners.

12Grade 6–7HigherNon-calculator6 marks

A is the point $(1, 2)$ and B is the point $(7, 6)$.

(a) Find an equation of the locus of points that are the same distance from A as from B.
Give your answer in the form $ax + by = c$, where $a$, $b$ and $c$ are integers. [4]

(b) Show that the point $(2, 7)$ is the same distance from A as from B. [2]

(a) midpoint of AB $= (4, 4)$M1
(a) gradient of AB $= \dfrac{6-2}{7-1} = \dfrac{2}{3}$ and perpendicular gradient $= -\dfrac{3}{2}$M1
(a) substitutes $(4, 4)$ into $y = -\tfrac{3}{2}x + c$ or uses $y - 4 = -\tfrac{3}{2}(x - 4)$M1
(a) $3x + 2y = 20$ (oe with integers, e.g. $6x + 4y = 40$)A1
(b) finds both distances (or squared distances): $\sqrt{1^2 + 5^2}$ and $\sqrt{5^2 + 1^2}$, or substitutes into their line: $3(2) + 2(7)$M1
(b) both distances $= \sqrt{26}$ (or $3(2) + 2(7) = 20$ with (a) correct), with a conclusionA1
Answer: (a) $3x + 2y = 20$ (b) both distances are $\sqrt{26}$
Examiner tip: The locus is the perpendicular bisector, so it must pass through the midpoint AND use the negative reciprocal of the gradient of AB.

13Grade 6–7HigherCalculator4 marks

Two straight hedges meet at the corner O of a field.
The angle between the hedges is $50°$.

A water tap T is to be placed in the field so that it is the same distance from both hedges.
OT = 20 m.

(a) Write down the name of the locus of points that are the same distance from both hedges. [1]

(b) Work out the shortest distance from the tap to each hedge.
Give your answer correct to 3 significant figures. [3]

(a) the bisector of the angle between the hedges (accept 'angle bisector of angle O')B1
(b) a right-angled triangle formed by the perpendicular from T to a hedge, with hypotenuse OT = 20 m and angle at O of $25°$ (may be seen on a diagram or in working)M1
(b) $20 \times \sin 25°$ (accept $20 \cos 65°$)M1
(b) $8.45$ m (from $8.452...$)A1
Answer: (a) the angle bisector (b) 8.45 m
Examiner tip: The shortest distance to a line is the perpendicular distance, so OT is the hypotenuse; using 50° instead of 25° or 20 × cos 25° are the usual errors.

14Grade 1–3FoundationNon-calculator4 marks

The diagram shows a map of a park. P and Q are two gates. M is a phone mast.

The dashed line is the locus of points that are the same distance from P as from Q.
The circle is the locus of points that are exactly 300 m from M.

(a) Write down the letter of the region that contains the points that are nearer to P than to Q and less than 300 m from M. [1]

(b) Describe the points in region D. [2]

(a) AB1
(b) more than 300 m from M (accept 'outside the circle', 'further than 300 m from the mast')B1
(b) nearer to Q than to P (accept 'closer to gate Q')B1
(c) $3$ cmB1
Answer: (a) A (b) more than 300 m from M and nearer to Q than to P (c) 3 cm
Examiner tip: Describe a region using BOTH loci: say which side of the dashed line it is on and whether it is inside or outside the circle.

15Grade 8–9HigherCalculator5 marks

ABCD is a square courtyard of side 10 m.

A lamp at A lights every point in the courtyard that is 10 m or less from A.
A lamp at B lights every point in the courtyard that is 10 m or less from B.

Work out the area of the courtyard that is lit by both lamps.
Give your answer correct to 3 significant figures.
You must show your working. [5]

identifies the region as inside both quarter circles (arcs centre A and centre B, radius 10 m) and that the arcs meet at a point X with AX = BX = AB = 10 m, so triangle ABX is equilateral (angle $XAB = 60°$)P1
sector area with angle $60°$: $\dfrac{60}{360} \times \pi \times 10^2\ (= 52.35...)$P1
area of equilateral triangle ABX: $\tfrac{1}{2} \times 10 \times 10 \times \sin 60°\ (= 43.30...)$P1
complete method: $2 \times 52.35... - 43.30...$ (or triangle $+ 2 \times$ segment, where segment $= 52.35... - 43.30... = 9.058...$)P1
$61.4$ m² (from $61.418...$; accept $61.4$ to $61.5$ from rounded working)A1
Answer: 61.4 m²
Examiner tip: The overlap is two 60° sectors that share the equilateral triangle, so the triangle must be subtracted once; using quarter circles (90°) is the usual error.

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