OxfordAQA International GCSE Mathematics: Units, perimeter, area, volume and surface area (9260)

This study guide is for OxfordAQA International GCSE Mathematics (9260), written to the specification Version 5.1 (for exams May/June 2018 onwards). It covers section 3.3.2 Mensuration and calculation, references G14, G15 and G16. Core content is also assessable at Extension, so Core and Extension students both need everything except the statements marked Extension only.

Related pages: revision notes, practice questions, the course hub, the printable checklist and the free 10-minute diagnostics.

What this unit covers

Reference	What you must be able to do	Tier
G14	Use standard units (length, area, volume/capacity, mass, time, money); change between related units, including compound units such as speed and density; use the 12 and 24 hour clock	Core
G15	Find the area of triangles, parallelograms and trapezia; find the volume of a prism using V = Ah	Core
G16	Use circumference = 2πr = πd and area = πr²; find perimeters and areas of 2D shapes, including composite shapes; give answers in terms of π	Core
G16	Surface area and volume of spheres, pyramids, cones and composite solids, including frustums of pyramids and cones	Extension only

G14: units and conversions

Length, mass, capacity and time

Quantity	Facts to know
Length	A kilometre is 1000 metres; a metre is 100 centimetres; a centimetre is 10 millimetres
Mass	1 t = 1000 kg, 1 kg = 1000 g
Capacity	A litre holds 1000 ml, which is also 1000 cm³; a cubic metre holds 1000 litres
Time	1 h = 60 min, 1 min = 60 s

Area and volume conversions

The length conversion is squared for area and cubed for volume. Since 1 m = 100 cm:

• A square metre is 100 cm by 100 cm, so it holds 10 000 cm².
• A cubic metre is 100 cm by 100 cm by 100 cm, so it holds 1 000 000 cm³.

Example 1. Convert 3.6 m² to cm², and 0.45 litres to cm³.

3.6 m² = 3.6 × 10 000 = 36 000 cm²
0.45 litres = 0.45 × 1000 = 450 cm³

Compound units

A compound unit combines two quantities. Speed = distance ÷ time, with units such as km/h or m/s. Density = mass ÷ volume, with units such as g/cm³ or kg/m³. Rearrange as needed: distance = speed × time, mass = density × volume.

For km/h to m/s, multiply by 1000 to turn km into m and divide by 3600 to turn hours into seconds. Overall, divide by 3.6.

Example 2. Change 54 km/h to m/s.

54 km/h = 54 × 1000 m ÷ 3600 s = 15 m/s

Example 3. A block has mass 1.35 kg and volume 500 cm³. Work out its density in g/cm³, then in kg/m³.

Mass = 1.35 × 1000 = 1350 g
Density = 1350 ÷ 500 = 2.7 g/cm³
Multiply by 1000 to change g/cm³ into kg/m³: 2.7 × 1000 = 2700 kg/m³

The 12 and 24 hour clock

In 24 hour time, 00:00 is midnight and 12:00 is noon. Subtract 12 from the hours for pm times.

Example 4. A train leaves at 21:50 and the journey takes 5 h 35 min. Give the arrival time in 24 hour and 12 hour form.

21:50 + 2 h 10 min = 24:00 (midnight)
Remaining time: 5 h 35 min - 2 h 10 min = 3 h 25 min
Arrival: 03:25, which is 3:25 am the next day

Example 5. A van covers 187 km in 2 h 12 min. Find its average speed.

2 h 12 min = 2 + 12/60 = 2.2 h
Speed = 187 ÷ 2.2 = 85 km/h

G15: triangles, parallelograms, trapezia and prisms

Area formulae

Shape	Area
Triangle	½ × base × perpendicular height
Parallelogram	base × perpendicular height
Trapezium	half the sum of the parallel sides, multiplied by the height: ½ × (a + b) × h

Measure the height at right angles to the base, not along a sloping side.

Example 6. Find the areas of (a) a triangle with base 14.4 cm and perpendicular height 9.5 cm, (b) a parallelogram with base 8.6 cm, perpendicular height 5.5 cm and sloping side 6.8 cm, (c) a trapezium with parallel sides 7.5 cm and 12.5 cm that are 6.4 cm apart.

(a) ½ × 14.4 × 9.5 = 68.4 cm²
(b) 8.6 × 5.5 = 47.3 cm²   (the 6.8 cm side is not used)
(c) ½ × (7.5 + 12.5) × 6.4 = ½ × 20 × 6.4 = 64 cm²

Prisms: V = Ah

A prism has the same cross-section all the way along. Its volume is V = Ah, where A is the area of the cross-section and h is the length (or height) of the prism.

Example 7. A wedge-shaped block is 8 cm long. Its end face is a triangle with sides 9 cm, 12 cm and 15 cm, and the 9 cm and 12 cm sides meet at a right angle. Find its volume and total surface area.

End face area = ½ × 9 × 12 = 54 cm²
Volume = 54 × 8 = 432 cm³
Surface area = two end faces + the three rectangles
Two end faces = 2 × 54 = 108 cm²
Distance round the end face = 9 + 12 + 15 = 36 cm
Rectangles = 36 × 8 = 288 cm²
Total = 108 + 288 = 396 cm²

G16: circles and composite shapes

Circle formulae

Quantity	Formula
Circumference	C = 2πr or C = πd
Area	A = πr²

The radius r is half the diameter d. Square only the radius, not π.

Example 8. A circle has radius 7.5 cm. Find its circumference and area, in terms of π and to 1 decimal place.

C = 2 × π × 7.5 = 15π = 47.1 cm
A = π × 7.5² = 56.25π = 176.7 cm²

Example 9. A circle has circumference 30 cm. Find its area to 3 significant figures.

2πr = 30, so r = 15/π
A = π × (15/π)² = 225/π = 71.6 cm²

Answers in terms of π

If you are asked to answer in terms of π, keep the symbol π in the answer. Combine terms such as 49π + 12π = 61π. Give a decimal only if asked.

Composite shapes

Split the shape into rectangles, triangles and parts of circles, or take one shape away from another. For a perimeter, add only the outside edges, not the line where two parts join.

Example 10. A shape is a rectangle 12 cm by 7 cm with a semicircle of diameter 7 cm attached to one of the 7 cm sides. Find the perimeter and area, giving exact answers and then decimals.

Radius of semicircle = 3.5 cm
Semicircle arc = ½ × π × 7 = 3.5π
Perimeter = 12 + 12 + 7 + 3.5π = 31 + 3.5π = 42.0 cm (3 s.f.)
(the 7 cm joining side is inside, so only the far 7 cm side counts)
Area = 12 × 7 + ½ × π × 3.5² = 84 + 6.125π = 103.2 cm² (1 d.p.)

Example 11. A square of side 16 cm has a quarter-circle of radius 9 cm cut from one corner. Find the area and perimeter of what is left.

Area = 16² - ¼ × π × 9² = 256 - 20.25π = 192.4 cm² (1 d.p.)
Perimeter: the two removed 9 cm pieces of side are replaced by an arc
Arc = ¼ × 2 × π × 9 = 4.5π
Straight edges = 16 + 16 + (16 - 9) + (16 - 9) = 46
Perimeter = 46 + 4.5π = 60.1 cm (1 d.p.)

Example 12. A circular pond has radius 7 m and is surrounded by a path out to a radius of 11 m. Find the area of the path.

Area = π × 11² - π × 7² = 121π - 49π = 72π = 226.2 m² (1 d.p.)

Extension only: spheres, cones, pyramids and composite solids

Formulae

Solid	Volume	Surface area
Sphere, radius r	(4/3)πr³	4πr²
Cone, base radius r, height h, slant height l	(1/3)πr²h	Curved: πrl. Total: πrl + πr²
Pyramid, base area A, height h	(1/3)Ah	Base plus the triangular faces

The slant height l is the length down the sloping surface. Where it is not given, find it from the right-angled triangle formed by the height, the radius (or half the base) and the slant: l² = r² + h².

Example 13. A ball has diameter 18 cm. Find its surface area and volume, leaving π in your answers.

Radius = 9 cm
S = 4 × π × 9² = 324π = 1017.9 cm²
V = (4/3) × π × 9³ = (4/3) × 729π = 972π = 3053.6 cm³

Example 14. A cone has base radius 8 cm and height 15 cm. Find its volume, curved surface area and total surface area.

l² = 8² + 15² = 289, so l = 17 cm
V = (1/3) × π × 64 × 15 = 320π = 1005.3 cm³
Curved surface area = π × 8 × 17 = 136π = 427.3 cm²
Total = 136π + π × 8² = 200π = 628.3 cm²

Example 15. A pyramid has a square base of side 9 cm and perpendicular height 10 cm. Find its volume and total surface area.

V = (1/3) × 81 × 10 = 270 cm³
Each triangular face has base 9 and slant height s, where
s² = 10² + 4.5² (the 4.5 is half the base), so s = 10.966 cm
Four faces = 4 × ½ × 9 × 10.966 = 197.4 cm²
Total = 81 + 197.4 = 278.4 cm²

Composite solids

Example 16. A capsule-shaped store is a cylinder of radius 6 m and height 5 m with a hemisphere of the same radius fixed on top. Find its volume and the outside surface area including the base, in terms of π.

Cylinder volume = π × 6² × 5 = 180π
Hemisphere volume = ½ × (4/3) × π × 6³ = 144π
Total volume = 324π = 1017.9 m³

Curved side of cylinder = 2π × 6 × 5 = 60π
Circular base = π × 6² = 36π
Curved hemisphere = ½ × 4π × 6² = 72π
Total surface = 168π = 527.8 m²

Where the hemisphere sits on the cylinder, that circular join is hidden, so leave it out of the total.

Frustums

A frustum is what remains when a small cone (or pyramid) is cut off the top of a large one, made by a cut that runs parallel to the base. Find its volume or area by subtracting the small solid from the large one. The small solid is an enlargement of the large one, so every length is in the same ratio.

Example 17. A cone has radius 12 cm and height 20 cm. A smaller cone of height 8 cm is cut off the top. Find the volume and curved surface area of the frustum that remains.

Scale factor = 8/20 = 0.4, so the small cone has radius 12 × 0.4 = 4.8 cm
Large cone volume = (1/3) × π × 144 × 20 = 960π
Small cone volume = (1/3) × π × 4.8² × 8 = 61.44π
Frustum volume = 898.56π = 2822.9 cm³

Large slant = √(20² + 12²) = 23.32 cm
Small slant = 0.4 × 23.32 = 9.33 cm
Curved area = π × 12 × 23.32 - π × 4.8 × 9.33 = 738.6 cm²

Example 18. A square-based pyramid has base 12 cm and height 18 cm. The top 9 cm is cut off, leaving a frustum.

Small pyramid has height 9, so scale factor ½ and base 6 cm
Large volume = (1/3) × 144 × 18 = 864
Small volume = (1/3) × 36 × 9 = 108
Frustum = 864 - 108 = 756 cm³

Common errors

• Putting the diameter into πr² where the radius belongs.
• Using the sloping side as the height of a parallelogram, trapezium or triangle.
• Converting areas with the plain length factor, when the factor must be squared.
• Writing 1 h 30 min as 1.30 h rather than 1.5 h.
• Including an internal join in a composite perimeter.
• Putting the perpendicular height into πrl, which needs the slant height.
• Writing a decimal when the answer should keep π.

Next, condense this with the revision notes and test yourself with the practice questions.

Official syllabus

OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA. Section 3.3.2 Mensuration and calculation.