Cambridge O Level Mathematics: Mensuration (4024)

This study guide is aligned to the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for examination in 2025, 2026 and 2027. It covers all of syllabus topic 5, Mensuration, sections 5.1 to 5.5. The syllabus is not tiered. Paper 1 does not allow a calculator and Paper 2 requires a scientific calculator, so practise leaving answers in terms of π as well as working to 3 significant figures.

Condensed recall is in the revision notes. Exam-style questions are in the mensuration practice set. Use the course hub and the printable checklist to track progress, and take a free 10-minute diagnostic if you are unsure where to start.

Section	What you must be able to do
5.1	Use metric units of mass, length, area, volume and capacity, and convert between them
5.2	Find perimeter and area of a rectangle, triangle, parallelogram and trapezium
5.3	Find circumference, area, arc length and sector area (minor and major sectors)
5.4	Find surface area and volume of a cuboid, prism, cylinder, sphere, pyramid and cone
5.5	Solve problems on compound shapes and solids, and on parts of shapes and solids

Which formulas are given

The syllabus says a list of formulas is included on page 2 of the examination papers. For this topic it covers the area of a triangle, the area and circumference of a circle, the curved surface area of a cylinder and of a cone, the surface area of a sphere, and the volume of a prism, pyramid, cylinder, cone and sphere. Except for the triangle, area formulas are not given, so you must know the rectangle, parallelogram and trapezium formulas. The syllabus does not list arc length or sector area as given, so learn them as fractions of the circle (section 5.3). Cuboid formulas, total surface areas and the hemisphere are not in the list either, so you build them from the pieces.

5.1 Units of measure

The syllabus names mm, cm, m, km; mm², cm², m², km²; mm³, cm³, m³; ml and l; g and kg. Area and volume factors are the length factor squared and cubed.

From	To	Multiply by
1 cm	mm	10
1 cm²	mm²	10² = 100
1 m²	cm²	100² = 10 000
1 km²	m²	1000² = 1 000 000
1 m³	cm³	100³ = 1 000 000
1 cm³	ml	1
1 l	cm³	1000
1 m³	l	1000
1 kg	g	1000

Worked example 1. A tank measures 150 cm by 80 cm by 60 cm. Find its capacity in litres.

Volume = 150 × 80 × 60 = 720 000 cm³. 1 litre = 1000 cm³, so capacity = 720 000 ÷ 1000 = 720 litres.

Worked example 2. A field has area 0.35 km². Give this in m². An engraved plate has area 4500 mm². Give this in cm².

0.35 × 1 000 000 = 350 000 m². 4500 ÷ 100 = 45 cm².

Going to a smaller unit multiplies. Going to a larger unit divides.

5.2 Area and perimeter

Perimeter is the distance round the outside. Area is the surface enclosed.

Shape	Area	Given in exam?
Rectangle, length l, width w	l × w	No
Triangle, base b, perpendicular height h	½bh	Yes
Parallelogram, base b, perpendicular height h	bh	No
Trapezium, parallel sides a and b, height h	½(a + b)h	No

The height is always measured at right angles to the base. The sloping side of a parallelogram or trapezium is used for the perimeter, never for the area.

Worked example 3. A parallelogram has base 14 cm, a sloping side of 8 cm and a perpendicular height of 6.5 cm. Find its area and perimeter.

Area = 14 × 6.5 = 91 cm². Opposite sides are equal, so perimeter = 2 × (14 + 8) = 44 cm.

Worked example 4. A trapezium has parallel sides 5 cm and 13 cm and area 54 cm². Find the distance between the parallel sides.

54 = ½ × (5 + 13) × h 54 = 9h h = 6 cm.

5.3 Circles, arcs and sectors

For a circle of radius r, the area is πr² and the circumference is 2πr. Both are in the list of formulas. Answers may be asked for in terms of π, which means leave π as a symbol, for example 20π cm.

An arc is part of the circumference. A sector is the region between two radii and an arc. If the angle at the centre is θ degrees, the sector is a fraction θ/360 of the whole circle:

• arc length = (θ/360) × 2πr
• sector area = (θ/360) × πr²

A sector with θ less than 180° is a minor sector. A sector with θ greater than 180° is a major sector. The same two formulas work for both. For a major sector, either use the large angle directly or subtract the minor sector from the whole circle.

Worked example 5. A sector has radius 12 cm and angle 75°. Find the arc length, the area and the perimeter, leaving each answer in terms of π first, then rounding to 3 significant figures.

Fraction = 75/360 = 5/24. Arc length = (5/24) × 2π × 12 = 5π cm = 15.7 cm. Area = (5/24) × π × 12² = (5/24) × 144π = 30π cm² = 94.2 cm². The perimeter of the sector includes both radii: 12 + 12 + 5π = 24 + 5π cm = 39.7 cm.

Worked example 6. A major sector has radius 6 cm and angle 210°. Find its area in terms of π.

Area = (210/360) × π × 6² = (7/12) × 36π = 21π cm².

Worked example 7. The arc of a sector of radius 14 cm has length 7π cm. Find the angle.

7π = (θ/360) × 2π × 14 7π = (θ/360) × 28π θ/360 = 1/4 θ = 90°.

5.4 Surface area and volume

A prism is any solid with a uniform cross-section. The syllabus gives a cylindrical sector as an example, so a cylinder, a triangular prism and a half-cylinder are all prisms.

Solid	Volume	Surface area
Cuboid l by w by h	lwh	2(lw + lh + wh)
Prism, cross-section area A, length l	Al	2A + (perimeter of cross-section) × l
Cylinder	πr²h	curved 2πrh, plus 2πr² for both ends
Cone	⅓πr²h	curved πrl, plus πr² for the base
Sphere	(4/3)πr³	4πr²
Pyramid, base area A, height h	⅓Ah	base plus the triangular faces

In the cone, h is the perpendicular height and l is the sloping edge. They are linked by l² = r² + h². In a pyramid, the slant height of a triangular face is different from the vertical height, and you may need Pythagoras to get it.

Worked example 8. A cone has base radius 7 cm and perpendicular height 24 cm. Find the sloping edge, the total surface area and the volume, in terms of π.

l² = 7² + 24² = 49 + 576 = 625, so l = 25 cm. Curved surface area = π × 7 × 25 = 175π. Base = π × 7² = 49π. Total surface area = 175π + 49π = 224π cm². Volume = ⅓ × π × 49 × 24 = 392π cm³.

Worked example 9. A sphere has radius 4.5 cm. Find its volume and surface area in terms of π.

Volume = (4/3) × π × 4.5³ = (4/3) × 91.125π = 121.5π cm³. Surface area = 4 × π × 4.5² = 4 × 20.25π = 81π cm².

Worked example 10. A triangular prism has a right-angled triangle cross-section with sides 6 cm, 8 cm and 10 cm. The prism is 15 cm long. Find its volume and total surface area.

Cross-section area = ½ × 6 × 8 = 24 cm². Volume = 24 × 15 = 360 cm³. Perimeter of cross-section = 6 + 8 + 10 = 24 cm. Total surface area = 2 × 24 + 24 × 15 = 48 + 360 = 408 cm².

Worked example 11. A square-based pyramid has base edge 10 cm and vertical height 12 cm. Find its volume and total surface area.

Volume = ⅓ × 10² × 12 = 400 cm³. Each triangular face has base 10 cm. Its height is the slant height, found from a right-angled triangle with sides 12 and half the base, 5: slant² = 12² + 5² = 169, so slant = 13 cm. Four faces = 4 × ½ × 10 × 13 = 260. Total surface area = 100 + 260 = 360 cm².

Worked example 12. An open cylindrical cup has radius 3.5 cm and height 9 cm. Find its capacity in ml and the area of material needed.

Volume = π × 3.5² × 9 = 110.25π = 346.36 cm³, so capacity = 346 ml (3 significant figures). Open means no top: area = base + curved surface = π × 3.5² + 2π × 3.5 × 9 = 12.25π + 63π = 75.25π = 236 cm² (3 significant figures).

5.5 Compound shapes and parts of shapes

A compound shape or solid is built from simpler pieces. Break it up, find each piece, then add. For a shape with a hole or a shaded region, subtract.

For perimeter and surface area only the outside counts. A join between two pieces is inside the shape.

Worked example 13. A shape is a rectangle 20 cm by 12 cm with a semicircle of diameter 12 cm attached to one 12 cm side. Find its area and perimeter.

Semicircle radius = 6 cm. Area = 20 × 12 + ½ × π × 6² = 240 + 18π = 297 cm² (3 significant figures). Perimeter = 20 + 20 + 12 + half the circumference of the circle = 52 + π × 6 = 52 + 6π = 70.8 cm (3 significant figures). The 12 cm side where the semicircle is attached is not part of the perimeter.

Worked example 14. A solid half-cylinder has radius 4 cm and length 10 cm. Find its volume and total surface area, in terms of π.

Cross-section is a semicircle of area ½π × 4² = 8π cm². Volume = 8π × 10 = 80π cm³. Surface area has four parts: two semicircular ends, the curved face and the flat rectangle. Ends = 2 × 8π = 16π. Curved face = ½ × 2π × 4 × 10 = 40π. Flat face = 8 × 10 = 80. Total = 56π + 80 cm².

Worked example 15. A cone of height 18 cm and base radius 9 cm has a smaller cone of height 6 cm and radius 3 cm cut off the top. Find the volume and total surface area of the frustum that remains.

Frustum volume = large cone − small cone. Large cone = ⅓ × π × 9² × 18 = 486π. Small cone = ⅓ × π × 3² × 6 = 18π. Volume = 468π cm³ = 1470 cm³ (3 significant figures).

For the curved surface, the sloping edges are l = √(18² + 9²) = √405 = 9√5 and, for the small cone, √(6² + 3²) = √45 = 3√5. Large curved surface = π × 9 × 9√5 = 81√5 π. Small curved surface = π × 3 × 3√5 = 9√5 π. Curved surface of frustum = 72√5 π = 506 cm². The frustum also has a circular base of radius 9 and a circular top of radius 3: 81π + 9π = 90π. Total surface area = 72√5 π + 90π = 789 cm² (3 significant figures).

The small cone is an enlargement of the large one with scale factor 1/3.

Parts of shapes and solids

“Part of” questions use a fraction. A quarter circle, a hemisphere and a wedge cut from a cylinder are fractions of a whole. Find the whole, take the fraction, then add any new flat faces made by the cut. A hemisphere of radius r has curved surface 2πr² and volume (2/3)πr³, and a solid hemisphere also has a flat circular face of area πr².

Rounding and accuracy

Give non-exact answers to 3 significant figures unless told otherwise, avoid rounding before the final answer, and use π from your calculator or 3.142.

Common errors

• Converting m² to cm² with a factor of 100 instead of 10 000.
• Using the sloping side of a parallelogram or trapezium as the height.
• Using the diameter in place of the radius.
• Using the perpendicular height of a cone in the curved surface formula, where the sloping edge is needed.
• Forgetting the two radii when asked for the perimeter of a sector.
• Counting a join between two parts of a compound shape in the perimeter or surface area.
• Including a lid on an open container, or leaving out the base of a pyramid.

Frequently asked questions

What does mensuration cover in Cambridge O Level Mathematics 4024? Sections 5.1 to 5.5 cover metric units and conversions, perimeter and area of 2D shapes, circles, arcs and sectors, surface area and volume of solids, and compound shapes and parts of shapes.

How do you find the area of a trapezium? Use area = ½(a + b)h, where a and b are the parallel sides and h is the perpendicular distance between them. Do not use the sloping side as the height.

How do you find the surface area of an open container? Leave out the lid. For an open cylindrical cup, the area is the base plus the curved surface, πr² + 2πrh.

How do you find the perimeter of a compound shape? Add only the outside edges. A join between two parts of the shape, such as the side where a semicircle meets a rectangle, is not part of the perimeter.

How do you convert cm³ to litres? Divide by 1000, because 1 litre = 1000 cm³. A tank of 720 000 cm³ holds 720 litres.

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for examination in 2025, 2026 and 2027, Cambridge International Education, topic 5 Mensuration.