Constructions, bisectors, scale drawings & maps – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

An architect makes a scale drawing of a footbridge.

The scale of the drawing is 1 cm represents 5 m.

(a) On the drawing, the length of the footbridge is 13.6 cm.
Work out the real length of the footbridge. [1]

(b) A support tower on the footbridge is 42 m tall.
Work out the height of the tower on the scale drawing. [1]

The architect makes a second drawing of the footbridge using a scale of 1 : 200

(a) $68$ (m)B1
(b) $8.4$ (cm)B1
(c) converts real length to cm, e.g. $68 \times 100 = 6800$ (cm), or uses 1 cm represents 2 m, e.g. $68 \div 2$ ft their (a)M1
(c) $34$ (cm) ft $\text{their (a)} \div 2$A1
Answer: (a) 68 m (b) 8.4 cm (c) 34 cm
Examiner tip: In (c), change the real length into centimetres before dividing by 200; dividing 68 by 200 gives 0.34, which is in metres, not centimetres.

2Grade 1–3FoundationCalculator3 marks

The map shows the positions of three villages, Ashby, Brenton and Cole Hill.

The scale bar on the map shows that 2 cm on the map represents 1 km.

(a) On the map, the distance from Ashby to Brenton is 7.4 cm.
Use the scale bar to work out the real distance from Ashby to Brenton. [1]

(b) Write the scale of the map as a ratio in the form $1 : n$ [2]

(a) $3.7$ (km)B1
(b) correct conversion of units, e.g. $1 \text{ km} = 100\,000 \text{ cm}$ or $1 \text{ cm}$ represents $0.5 \text{ km} = 50\,000 \text{ cm}$, or $2 : 100\,000$ seenM1
(b) $1 : 50\,000$A1
Answer: (a) 3.7 km (b) 1 : 50 000
Examiner tip: A ratio scale has no units, so both sides must be in the same unit: convert kilometres to centimetres (1 km = 100 000 cm) before simplifying.

3Grade 4–5Calculator3 marks

The line $AB$ is drawn below. $AB = 8$ cm.

(a) Use ruler and compasses only to construct triangle $ABC$ with $AC = 6$ cm and $BC = 5$ cm.
You must show all your construction lines. [2]

(b) Measure the size of angle $ACB$ in your triangle. [1]

(a) an arc of radius 6 cm centred on $A$ or an arc of radius 5 cm centred on $B$ (within 2 mm), or a correct triangle drawn with no arcsB1
(a) fully correct triangle with $C$ within 2 mm of the correct position and both construction arcs visibleB1
(b) angle in the range $91^\circ$ to $95^\circ$ ft their triangle within $\pm 2^\circ$B1
Answer: (a) correct construction with arcs (b) $93^\circ$ (accept $91^\circ$–$95^\circ$)
Examiner tip: Leave both compass arcs visible: a correct triangle with no arcs only gets 1 of the 2 construction marks.

Stuck on Constructions, bisectors, scale drawings & maps?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Grade 4–5Calculator4 marks

The line segment $AB$ is drawn below. $AB = 7$ cm.

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

The perpendicular bisector crosses $AB$ at the point $M$.

(b) Write down the length of $AM$. [1]

The point $P$ lies on the perpendicular bisector. $PA = 5.2$ cm.

(a) two pairs of intersecting arcs of equal radius (radius greater than 3.5 cm), centred on $A$ and $B$M1
(a) correct line drawn through the intersections of the arcs, within 2 mm of the true bisectorA1
(b) $3.5$ (cm)B1
(c) $5.2$ (cm) and reason, e.g. every point on the perpendicular bisector is the same distance from $A$ and $B$ accept triangle $PAB$ is isoscelesB1
Answer: (a) correct construction (b) 3.5 cm (c) 5.2 cm, as points on the perpendicular bisector are equidistant from A and B
Examiner tip: Set your compasses to more than half of AB and keep the same radius for both pairs of arcs, otherwise the arcs will not cross on the bisector.

5Grade 4–5FoundationCalculator5 marks

The diagram shows a plan of a bedroom, drawn on a grid of centimetre squares.

The scale of the plan is $1 : 50$

(a) Work out the real length and the real width of the bedroom. Give your answers in metres. [2]

Kofi is going to cover the whole floor of the bedroom with flooring.

The flooring is sold in whole square metres. It costs \$14.50 for each square metre.

(b) Work out the cost of the flooring Kofi needs to buy. [3]

(a) length $5$ (m)B1
(a) width $3.5$ (m)B1
(b) $5 \times 3.5$ or $17.5$ (m²) ft their (a)M1
(b) rounds their area up to a whole number of square metres and multiplies by 14.50, e.g. $18 \times 14.50$M1
(b) \$261 (accept \$261.00)A1
Answer: (a) 5 m by 3.5 m (b) \$261
Examiner tip: Flooring is sold in whole square metres, so 17.5 m² must be rounded up to 18 m²; 17.5 × 14.50 = \$253.75 loses the final mark.

6Grade 4–5Calculator4 marks

The diagram shows a straight line with a point $A$ marked on it.

(a) Use ruler and compasses only to construct an angle of $60^\circ$ at $A$, with one arm along the line $AB$. Then bisect your $60^\circ$ angle to make an angle of $30^\circ$ at $A$.
You must show all your construction lines. [3]

A point $T$ lies on the angle bisector you drew in part (a). The distance from $T$ to the line $AB$ is 2.5 cm.

(b) Write down the shortest distance from $T$ to the other arm of the $60^\circ$ angle. [1]

(a) constructs a $60^\circ$ angle at $A$: arc centred on $A$ crossing $AB$, and an arc of the same radius from that crossing point, with the arm drawn through the intersectionM1
(a) bisects the $60^\circ$ angle: two arcs of equal radius from the points where an arc centred on $A$ crosses the two armsM1
(a) bisector drawn, giving a $30^\circ$ angle at $A$ within $\pm 2^\circ$, with all arcs visibleA1
(b) $2.5$ (cm) (points on an angle bisector are equidistant from the two arms)B1
Answer: (a) 60° angle constructed and bisected to give 30° (b) 2.5 cm
Examiner tip: An angle drawn with a protractor scores nothing here: the 60° must come from equal arcs (an equilateral triangle) and the 30° from bisecting it.

7Grade 4–5Calculator4 marks

A drone flies from its launch point $L$ in a straight line for 6 km on a bearing of $070^\circ$ to a checkpoint $P$.

It then flies in a straight line for 4 km on a bearing of $200^\circ$ to a point $Q$.

(a) Make an accurate scale drawing of the drone's flight. Use a scale of 1 cm represents 1 km. [2]

(b) Use your scale drawing to find
(i) the real distance from $L$ to $Q$
(ii) the bearing of $Q$ from $L$. [2]

(a) $LP$ drawn 6 cm long on a bearing of $070^\circ$ (within 2 mm and $\pm 2^\circ$), with a north line at $L$B1
(a) $PQ$ drawn 4 cm long on a bearing of $200^\circ$ (within 2 mm and $\pm 2^\circ$), measured from a north line at $P$B1
(b)(i) answer in the range $4.4$ to $4.8$ kmB1
(b)(ii) answer in the range $109^\circ$ to $115^\circ$B1
Answer: (b)(i) 4.6 km (4.4–4.8) (ii) $112^\circ$ (109°–115°)
Examiner tip: Draw a new north line at P before measuring the 200° bearing; measuring 200° from the line LP instead is the most common error.

8Grade 6–7HigherCalculator4 marks

On a map, a straight path of length 1.8 km is shown by a line of length 4.5 cm.

(a) Find the scale of the map. Give your answer in the form $1 : n$ [2]

On the same map, a lake has an area of 6 cm².

(b) Work out the real area of the lake. Give your answer in km². [2]

(a) $1.8 \text{ km} = 180\,000$ cm and $180\,000 \div 4.5$ or 1 cm represents 0.4 kmM1
(a) $1 : 40\,000$A1
(b) uses square of length scale, e.g. $1 \text{ cm}^2$ represents $0.4^2 = 0.16$ km², or $6 \times 40\,000^2$ (cm²)M1
(b) $0.96$ (km²)A1
Answer: (a) 1 : 40 000 (b) 0.96 km²
Examiner tip: Areas scale by the square of the length scale: 6 × 0.4 = 2.4 km² is the most common wrong answer.

9Grade 6–7Calculator5 marks

The diagram shows a scale drawing of a rectangular park $ABCD$.

The scale is 1 cm represents 20 m. On the drawing, $AB = 6$ cm and $AD = 4$ cm.

A fountain $F$ is to be built in the park so that

$F$ is the same distance from $A$ as it is from $B$
$F$ is the same distance from the side $AB$ as it is from the side $AD$.

(a) Use ruler and compasses only to construct the position of $F$. Show all your construction lines. [3]

(b) Use your drawing to find the real distance of the fountain from corner $A$. Give your answer in metres. [2]

(a) correct construction of the perpendicular bisector of $AB$ with arcs visibleB1
(a) correct construction of the bisector of angle $DAB$ with arcs visibleB1
(a) $F$ marked at the intersection of the two bisectors, within 2 mm of the correct point (3 cm from $AB$, 3 cm from $AD$)B1
(b) $AF$ measured (in the range 4.1 to 4.4 cm) and multiplied by 20 ft their FM1
(b) answer in the range $82$ to $88$ mA1
Answer: (a) F at the intersection of the perpendicular bisector of AB and the bisector of angle A (b) 85 m (82–88 m)
Examiner tip: Equidistant from two points means a perpendicular bisector; equidistant from two lines means an angle bisector — mixing these up loses all three construction marks.

10Grade 8–9HigherCalculator5 marks

Maya is asked to draw a triangle $ABC$ in which $AB = 9$ cm, angle $BAC = 40^\circ$ and $BC = 7$ cm.

She draws $AB$ and the $40^\circ$ angle at $A$, then uses compasses to draw an arc of radius 7 cm centred on $B$.

(a) Explain why Maya can draw two different triangles that fit the information. [1]

(b) Calculate the two possible lengths of $AC$. Give your answers correct to 3 significant figures. [4]

(a) the arc (radius 7 cm, centre $B$) crosses the arm through $A$ at two points, so there are two possible positions for $C$ accept 7 cm is shorter than 9 cm but longer than the perpendicular distance $9\sin 40^\circ \approx 5.79$ cm from $B$ to the armB1
(b) $\dfrac{\sin C}{9} = \dfrac{\sin 40^\circ}{7}$ or cosine rule $7^2 = 9^2 + x^2 - 2 \times 9 \times x \cos 40^\circ$M1
(b) $C = 55.7^\circ$ (and $124.3^\circ$) or $x^2 - 13.79x + 32 = 0$A1
(b) complete method to find $AC$ for either triangle, e.g. $B = 84.3^\circ$ and $\dfrac{7 \sin 84.3^\circ}{\sin 40^\circ}$, or uses quadratic formula on their quadraticM1
(b) $10.8$ (cm) and $2.95$ (cm) accept $10.83$–$10.84$ and $2.95$–$2.96$A1
Answer: (b) $AC = 10.8$ cm or $2.95$ cm
Examiner tip: The sine rule gives only the acute angle 55.7°; you must also use 180° − 55.7° = 124.3° to find the second triangle.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session