Edexcel IGCSE Maths A (4MA1) · Foundation tier · Questions

IGCSE Maths 4MA1 Foundation Geometry and Trigonometry Questions

33 original exam-style questions on geometry and trigonometry for the Foundation tier, three for each sub-topic, from easier to harder. Try each one, then open the mark scheme and the worked solution. The marks are Edexcel style: M for method, A for accuracy, B for an independent result.

33 questions
Mark schemes and worked solutions
Grades 1 to 5
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More geometry and trigonometry practice →Geometry and Trigonometry notes

4.1 Angles, lines and triangles questions

Revise it first: Angles, lines and triangles notes and worked example.

Question 1 · 2 marks · grade 1

Write down the type of each angle: acute, right, obtuse or reflex.

(a) 35° (b) 90° (c) 150° (d) 260°

Show the answer, mark scheme and worked solution

B2 (a) acute (b) right (c) obtuse (d) reflex (B1 for 2 or 3 correct)

Acute: less than 90°. Right: exactly 90°. Obtuse: between 90° and 180°. Reflex: between 180° and 360°.

Answer: (a) acute (b) right (c) obtuse (d) reflex

Question 2 · 2 marks · grade 4

The angles of a triangle are in the ratio \(2 : 3 : 4\).

Work out the size of the largest angle.

Show the answer, mark scheme and worked solution

M1 for 180 ÷ 9 (= 20)
A1 80°

2 + 3 + 4 = 9 parts make 180°, so 1 part = 20°.
The largest angle is 4 × 20 = 80°.

Answer: \(80°\)

Question 3 · 4 marks · grade 5

In triangle \(ABC\), \(AB = AC\). Angle \(ABC = 2x°\) and angle \(BAC = (x + 20)°\).

Work out the size of each angle of the triangle.

Show the answer, mark scheme and worked solution

M1 for angle ACB = 2x (base angles of an isosceles triangle)
M1 for 2x + 2x + x + 20 = 180
A1 x = 32
A1 64°, 64° and 52°

AB = AC, so the base angles ABC and ACB are equal: both 2x.
2x + 2x + x + 20 = 180, so 5x = 160 and x = 32.
Angles: 64°, 64° and 52°.

Answer: \(\angle ABC = \angle ACB = 64°\), \(\angle BAC = 52°\)

4.2 Polygons questions

Revise it first: Polygons notes and worked example.

Question 4 · 2 marks · grade 1

Name a quadrilateral that has

(a) four equal sides and four right angles

(b) exactly one pair of parallel sides.

Show the answer, mark scheme and worked solution

B1 (a) square
B1 (b) trapezium

(a) A square has four equal sides and four right angles.
(b) A trapezium has exactly one pair of parallel sides.

Answer: (a) square (b) trapezium

Question 5 · 2 marks · grade 4

Each exterior angle of a regular polygon is 24°.

How many sides does the polygon have?

Show the answer, mark scheme and worked solution

M1 for 360 ÷ 24
A1 15

The exterior angles add up to 360°.
360 ÷ 24 = 15 sides.

Answer: \(15\)

Question 6 · 3 marks · grade 5

A hexagon has angles of 100°, 130°, 140°, 95°, \(x°\) and \(x°\).

Work out the value of \(x\).

Show the answer, mark scheme and worked solution

M1 for (6 − 2) × 180 (= 720)
M1 for 720 − 100 − 130 − 140 − 95 (= 255)
A1 127.5

The interior angles of a hexagon add up to (6 − 2) × 180 = 720°.
2x = 720 − 100 − 130 − 140 − 95 = 255.
x = 127.5.

Answer: \(x = 127.5\)

4.3 Symmetry questions

Revise it first: Symmetry notes and worked example.

Question 7 · 2 marks · grade 1

For a square, write down

(a) the number of lines of symmetry

(b) the order of rotational symmetry.

Show the answer, mark scheme and worked solution

B1 (a) 4
B1 (b) 4

(a) Two diagonals and two lines through the midpoints of opposite sides: 4 lines.
(b) The square looks the same 4 times in a full turn: order 4.

Answer: (a) \(4\) (b) \(4\)

Question 8 · 3 marks · grade 4

A regular polygon has 9 lines of symmetry.

(a) How many sides does it have?

(b) What is its order of rotational symmetry?

Show the answer, mark scheme and worked solution

B1 (a) 9
B1 (b) 9
B1 (c) 40°

A regular polygon with n sides has n lines of symmetry and order n.
(c) 360 ÷ 9 = 40°.

Answer: (a) \(9\) (b) \(9\) (c) \(40°\)

Question 9 · 3 marks · grade 5

The points \(A(2, 1)\), \(B(6, 1)\) and \(C(4, 5)\) are the vertices of an isosceles triangle.

(a) Write down the equation of the line of symmetry of triangle \(ABC\).

(b) \(D\) is the reflection of \(C\) in the line \(AB\). Write down the coordinates of \(D\).

Show the answer, mark scheme and worked solution

B1 (a) x = 4
B1 (b) (4, −3)
B1 (c) rhombus

(a) C is above the midpoint of AB, (4, 1), so the line of symmetry is x = 4.
(b) AB is the line y = 1. C is 4 units above it, so D is 4 units below: (4, −3).
(c) All four sides AC, CB, BD and DA have the same length (√20), and the shape is not a square, so ACBD is a rhombus.

Answer: (a) \(x = 4\) (b) \(D(4, -3)\) (c) rhombus

4.4 Measures questions

Revise it first: Measures notes and worked example.

Question 10 · 2 marks · grade 1

A car travels 150 km in 2 hours.

Work out its average speed.

Show the answer, mark scheme and worked solution

M1 for 150 ÷ 2
A1 75 km/h

Average speed = distance ÷ time = 150 ÷ 2 = 75 km/h.

Answer: \(75\) km/h

Question 11 · 2 marks · grade 4

Pressure = \(\frac{\text{force}}{\text{area}}\)

A force of 600 newtons acts on an area of 0.15 m\(^2\).

Work out the pressure in newtons per m\(^2\).

Show the answer, mark scheme and worked solution

M1 for 600 ÷ 0.15
A1 4000 N/m²

Pressure = 600 ÷ 0.15 = 4000 N/m².

Answer: \(4000\) N/m\(^2\)

Question 12 · 3 marks · grade 5

Ali drives 120 km at an average speed of 60 km/h. He then drives a further 90 km at an average speed of 45 km/h.

Work out his average speed for the whole journey.

Show the answer, mark scheme and worked solution

M1 for 120 ÷ 60 (= 2) or 90 ÷ 45 (= 2)
M1 for (120 + 90) ÷ (2 + 2)
A1 52.5 km/h

Time for the first part: 120 ÷ 60 = 2 hours. Second part: 90 ÷ 45 = 2 hours.
Total distance 210 km in 4 hours.
Average speed = 210 ÷ 4 = 52.5 km/h.

Answer: \(52.5\) km/h

4.5 Construction questions

Revise it first: Construction notes and worked example.

Question 13 · 1 mark · grade 1

Kim draws a line \(AB\) of length 6.4 cm and marks its midpoint \(M\).

How far is \(M\) from \(A\)?

Show the answer, mark scheme and worked solution

B1 3.2 cm

The midpoint is halfway along the line: 6.4 ÷ 2 = 3.2 cm.

Answer: \(3.2\) cm

Question 14 · 1 mark · grade 4

The bisector of angle \(ABC\) is constructed. Angle \(ABC = 74°\).

What is the angle between the bisector and the line \(BA\)?

Show the answer, mark scheme and worked solution

B1 37°

An angle bisector cuts the angle into two equal parts: 74 ÷ 2 = 37°.

Answer: \(37°\)

Question 15 · 3 marks · grade 5

A ship sails 12 km due north from a harbour \(H\) to a point \(A\), then 9 km due east to a point \(B\). A scale drawing is made using 1 cm to represent 2 km.

(a) Work out the lengths of \(HA\) and \(AB\) on the drawing.

(b) On the drawing, \(HB\) measures 7.5 cm. Work out the real distance \(HB\).

Show the answer, mark scheme and worked solution

B1 (a) HA = 6 cm
B1 (a) AB = 4.5 cm
B1 (b) 15 km

(a) 12 ÷ 2 = 6 cm and 9 ÷ 2 = 4.5 cm.
(b) 7.5 × 2 = 15 km. (Check with Pythagoras: √(12² + 9²) = √225 = 15.)

Answer: (a) \(HA = 6\) cm, \(AB = 4.5\) cm (b) \(15\) km

4.6 Circle properties questions

Revise it first: Circle properties notes and worked example.

Question 16 · 2 marks · grade 1

Complete each sentence with the correct word.

(a) A line from the centre of a circle to its circumference is a ______

(b) A straight line that touches a circle at exactly one point is a ______

Show the answer, mark scheme and worked solution

B1 (a) radius
B1 (b) tangent

(a) The radius joins the centre to the circumference.
(b) A tangent touches the circle at one point.

Answer: (a) radius (b) tangent

Question 17 · 3 marks · grade 4

\(PA\) and \(PB\) are tangents to a circle with centre \(O\), touching it at \(A\) and \(B\). Angle \(APB = 40°\).

Work out the size of the angle \(AOB\).

Show the answer, mark scheme and worked solution

B1 angles OAP and OBP are both 90°
M1 for 360 − 90 − 90 − 40
A1 140°

Each tangent is perpendicular to its radius, so angles OAP and OBP are 90°.
OAPB is a quadrilateral: angle AOB = 360 − 90 − 90 − 40 = 140°.

Answer: \(140°\)

Question 18 · 3 marks · grade 5

A circle has centre \(O\) and radius 17 cm. A chord \(CD\) has length 30 cm.

Work out the perpendicular distance from \(O\) to the chord \(CD\).

Show the answer, mark scheme and worked solution

B1 for half the chord = 15 cm
M1 for 17² − 15²
A1 8 cm

The perpendicular from the centre bisects the chord, so each half is 15 cm.
Distance = √(17² − 15²) = √64 = 8 cm.

Answer: \(8\) cm

4.7 Geometrical reasoning questions

Revise it first: Geometrical reasoning notes and worked example.

Question 19 · 2 marks · grade 2

Angle \(x\) and an angle of 115° lie next to each other on a straight line.

Work out the size of angle \(x\). Give a reason for your answer.

Show the answer, mark scheme and worked solution

B1 65°
B1 angles on a straight line add up to 180°

180 − 115 = 65°, because angles on a straight line add up to 180°.

Answer: \(x = 65°\) (angles on a straight line add up to \(180°\))

Question 20 · 2 marks · grade 4

In triangle \(PQR\), \(PQ = PR\) and angle \(QPR = 36°\). The side \(QR\) is extended to \(S\).

Work out the size of angle \(PRS\). Give reasons.

Show the answer, mark scheme and worked solution

B1 angle PQR = angle PRQ = 72° (base angles of an isosceles triangle are equal; angles in a triangle add up to 180°)
B1 108° (angles on a straight line add up to 180°)

PQ = PR, so the base angles are equal: (180 − 36) ÷ 2 = 72°.
Angle PRS = 180 − 72 = 108°, because angles on a straight line add up to 180°.

Answer: \(108°\)

Question 21 · 3 marks · grade 5

In triangle \(ABC\), \(AB = AC\) and angle \(BAC = 40°\). \(D\) is the point on \(BC\) such that \(AD\) bisects angle \(BAC\).

Work out the size of angle \(ADB\). Give reasons.

Show the answer, mark scheme and worked solution

B1 angle ABC = 70° (isosceles triangle, angles in a triangle add up to 180°)
B1 angle BAD = 20°
B1 90° (angles in triangle ABD)

AB = AC, so angle ABC = angle ACB = (180 − 40) ÷ 2 = 70°.
AD bisects angle BAC, so angle BAD = 20°.
In triangle ABD: angle ADB = 180 − 20 − 70 = 90°.

Answer: \(90°\)

4.8 Trigonometry and Pythagoras' theorem questions

Revise it first: Trigonometry and Pythagoras' theorem notes and worked example.

Question 22 · 2 marks · grade 2

A right-angled triangle has shorter sides of 6 cm and 8 cm.

Work out the length of the hypotenuse.

Show the answer, mark scheme and worked solution

M1 for 6² + 8²
A1 10 cm

Pythagoras: c² = 6² + 8² = 36 + 64 = 100.
c = 10 cm.

Answer: \(10\) cm

Question 23 · 2 marks · grade 4

In a right-angled triangle, the side opposite angle \(x\) is 5 cm and the side adjacent to \(x\) is 8 cm.

Work out the size of angle \(x\). Give your answer correct to 1 decimal place.

Show the answer, mark scheme and worked solution

M1 for tan x = 5/8
A1 32.0°

tan x = opposite ÷ adjacent = 5 ÷ 8 = 0.625.
x = tan⁻¹(0.625) = 32.005…° = 32.0°.

Answer: \(32.0°\)

Question 24 · 3 marks · grade 5

A ship sails 20 km from a port on a bearing of 060°.

Work out how far (a) east (b) north the ship is from the port. Give your answers correct to 3 significant figures.

Show the answer, mark scheme and worked solution

M1 for 20 × sin 60° or 20 × cos 30°
A1 (a) 17.3 km
A1 (b) 10 km (or 10.0 km)

The bearing 060° makes an angle of 60° with north.
(a) East: 20 × sin 60° = 17.32… = 17.3 km.
(b) North: 20 × cos 60° = 10 km.

Answer: (a) \(17.3\) km (b) \(10.0\) km

4.9 Mensuration of 2D shapes questions

Revise it first: Mensuration of 2D shapes notes and worked example.

Question 25 · 2 marks · grade 1

A rectangle is 9 cm long and 4 cm wide.

Work out (a) its area (b) its perimeter.

Show the answer, mark scheme and worked solution

B1 (a) 36 cm²
B1 (b) 26 cm

(a) 9 × 4 = 36 cm².
(b) 9 + 4 + 9 + 4 = 26 cm.

Answer: (a) \(36\) cm\(^2\) (b) \(26\) cm

Question 26 · 4 marks · grade 4

A semicircle has a diameter of 10 cm.

Work out (a) its perimeter (b) its area. Give your answers correct to 3 significant figures.

Show the answer, mark scheme and worked solution

M1 (a) for ½ × π × 10 + 10
A1 (a) 25.7 cm
M1 (b) for ½ × π × 5²
A1 (b) 39.3 cm²

(a) Curved edge: half of π × 10 = 15.707…; add the straight edge 10: 25.707… = 25.7 cm.
(b) Half of π × 5² = 12.5π = 39.26… = 39.3 cm².

Answer: (a) \(25.7\) cm (b) \(39.3\) cm\(^2\)

Question 27 · 3 marks · grade 5

A running track is made of a rectangle 100 m long and 60 m wide, with a semicircle on each of the 60 m ends. The track runs around the outside edge of the shape.

Work out the length of the track. Give your answer correct to the nearest metre.

Show the answer, mark scheme and worked solution

M1 for π × 60 (two semicircles make a full circle of diameter 60)
M1 for 2 × 100 + their 188.4…
A1 388 m

The two semicircular ends make one circle of diameter 60 m: π × 60 = 188.49… m.
The two straights: 2 × 100 = 200 m.
Total: 388.49… = 388 m.

Answer: \(388\) m

4.10 3D shapes and volume questions

Revise it first: 3D shapes and volume notes and worked example.

Question 28 · 1 mark · grade 1

A cuboid is 5 cm long, 4 cm wide and 3 cm high.

Work out its volume.

Show the answer, mark scheme and worked solution

B1 60 cm³

Volume = length × width × height = 5 × 4 × 3 = 60 cm³.

Answer: \(60\) cm\(^3\)

Question 29 · 3 marks · grade 4

A closed cylinder has a radius of 3 cm and a height of 8 cm.

Work out its total surface area. Give your answer correct to 3 significant figures.

Show the answer, mark scheme and worked solution

M1 for 2 × π × 3² (two circles)
M1 for 2 × π × 3 × 8 (curved surface)
A1 207 cm²

Two circular ends: 2 × π × 3² = 18π.
Curved surface: 2πrh = 2 × π × 3 × 8 = 48π.
Total = 66π = 207.3… = 207 cm².

Answer: \(207\) cm\(^2\)

Question 30 · 3 marks · grade 5

A cylindrical jug has a radius of 6 cm and holds 2 litres when full.

Work out the height of the jug. Give your answer correct to 3 significant figures.

Show the answer, mark scheme and worked solution

B1 2 litres = 2000 cm³
M1 for π × 6² × h = 2000
A1 17.7 cm

2 litres = 2000 cm³.
π × 6² × h = 2000, so h = 2000 ÷ 36π = 17.68… = 17.7 cm.

Answer: \(17.7\) cm

4.11 Similarity questions

Revise it first: Similarity notes and worked example.

Question 31 · 2 marks · grade 2

Two rectangles are similar. The smaller one is 4 cm by 6 cm. The shorter side of the larger one is 10 cm.

Work out the length of the longer side of the larger rectangle.

Show the answer, mark scheme and worked solution

M1 for a scale factor of 10 ÷ 4 (= 2.5)
A1 15 cm

Scale factor = 10 ÷ 4 = 2.5.
Longer side = 6 × 2.5 = 15 cm.

Answer: \(15\) cm

Question 32 · 2 marks · grade 4

At the same time of day, a post 1.5 m tall casts a shadow 2 m long and a tree casts a shadow 12 m long.

Work out the height of the tree.

Show the answer, mark scheme and worked solution

M1 for a scale factor of 12 ÷ 2 (= 6) or 1.5 ÷ 2 (= 0.75)
A1 9 m

The two triangles (object, shadow, sun's ray) are similar.
Scale factor = 12 ÷ 2 = 6, so the tree is 1.5 × 6 = 9 m tall.

Answer: \(9\) m

Question 33 · 3 marks · grade 5

Triangles \(ABC\) and \(PQR\) are similar, with \(A\) matching \(P\), \(B\) matching \(Q\) and \(C\) matching \(R\). \(AB = 8\) cm, \(AC = 6\) cm, \(PQ = 12\) cm and \(QR = 15\) cm.

Work out the lengths of (a) \(PR\) (b) \(BC\).

Show the answer, mark scheme and worked solution

M1 for a scale factor of 12 ÷ 8 (= 1.5)
A1 (a) 9 cm
A1 (b) 10 cm

Scale factor from ABC to PQR = 12 ÷ 8 = 1.5.
(a) PR = 6 × 1.5 = 9 cm.
(b) BC = 15 ÷ 1.5 = 10 cm.

Answer: (a) \(PR = 9\) cm (b) \(BC = 10\) cm

Every question here is original, written for IGCSE Math Revision rather than copied from Pearson papers, and each answer was re-solved independently before publishing.

Topic list for Edexcel IGCSE Maths 4MA1 Foundation geometry and trigonometry: angles, lines and triangles, polygons, symmetry, measures, construction, circle properties, geometrical reasoning, trigonometry and pythagoras' theorem, mensuration of 2d shapes, 3d shapes and volume, similarity — The 11 Foundation sub-topics of geometry and trigonometry.

Practise geometry and trigonometry

Unit 3 practice, Unit 5 practice with the Foundation filter on: questions written for Foundation and 4MA1 questions at grades 4 and 5, each with a mark scheme (IGCSE plan, with a free preview)
Edexcel IGCSE Maths 4MA1 past papers, to practise under exam conditions
IGCSE Maths formula sheet and the 4MA1 Higher revision notes when you are ready to go further