OxfordAQA International GCSE Mathematics: Units, area, volume and surface area (9260) – Revision Notes

These revision notes are for OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), section 3.3.2 Mensuration and calculation, references G14 to G16. Core content is also assessable at Extension. Content marked Extension only is for Extension students. For full explanations and worked examples, use the study guide.

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

G14: units

Type	Facts
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 is 1000 ml, or 1000 cm³; a cubic metre is 1000 litres
Time	1 h = 60 min; 1 min = 60 s
Area	Square the length factor: a square metre is 10 000 cm²; a square centimetre is 100 mm²
Volume	Cube the length factor: a cubic metre is 1 000 000 cm³; a cubic centimetre is 1000 mm³

Compound units

Quantity	Formula	Units seen
Speed	distance ÷ time	km/h or m/s
Density	mass ÷ volume	g/cm³ or kg/m³

Rearranged: distance = speed × time; mass = density × volume.

Change	Rule
km/h to m/s	÷ 3.6
m/s to km/h	× 3.6
g/cm³ to kg/m³	× 1000

Worked reminder: a cyclist rides 27 km in 1 h 30 min. Speed = 27 ÷ 1.5 = 18 km/h, which is 18 ÷ 3.6 = 5 m/s. A metal bar of volume 250 cm³ and mass 1.75 kg has density 1750 ÷ 250 = 7 g/cm³, which is 7000 kg/m³.

Time

• Convert minutes to hours by dividing by 60: 1 h 45 min = 1.75 h.
• 24 hour clock: add 12 to pm hours (7:15 pm = 19:15). 00:00 is midnight, 12:00 is noon.
• 12:40 am is 00:40. 12:40 pm is 12:40.
• To find an arrival time, add the hours and minutes separately and carry over 60 minutes.

Method: converting units

1. Note how many small lengths fit into the large one (for example, 100 cm in each metre).
2. Square that number for area, or cube it for volume.
3. Multiply when going to a smaller unit, divide when going to a larger unit.
4. Check the size: a smaller unit gives a bigger number.

G15: area and prisms

Shape	Area
Rectangle	length × width
Triangle	½ × base × perpendicular height
Parallelogram	base × perpendicular height
Trapezium	½ × (sum of parallel sides) × h

• Use the perpendicular height. Never use the sloping side.
• Prism: V = Ah, where A is the cross-sectional area and h is the length.
• Prism surface area: two cross-sections, plus the cross-section’s perimeter multiplied by the length.
• Reverse problems: h = V ÷ A, or A = V ÷ h.

Method: prism volume

1. Identify the cross-section (the face that repeats).
2. Find its area using the correct formula.
3. Multiply by the length.
4. State units cubed.

Worked reminder: a trapezium has parallel sides 5 and 11 and height 4. Area = ½ × 16 × 4 = 32. Used as the end of a prism 3.5 long, V = 32 × 3.5 = 112.

G16: circles and composite shapes

Quantity	Formula
Circumference	C = 2πr = πd
Area	A = πr²
Semicircle perimeter	πr + 2r (arc plus diameter)
Semicircle area	½πr²
Quarter circle area	¼πr²
Ring (annulus)	π(R² − r²), with R outside, r inside

• Diameter = 2 × radius. Halve before squaring.
• “In terms of π” means leave π as a symbol: 25π, not 78.5.
• Working back from C: r = C ÷ (2π). From A: r = √(A ÷ π).

Method: composite shape

1. Sketch and label every length, working out missing ones by adding or subtracting.
2. Split into simple shapes, or find a large shape minus a smaller one.
3. For area: add or subtract the areas.
4. For perimeter: add only the outside edges, including arcs. Leave out any join inside the shape.
5. Round only at the end.

Worked reminder: a 10 cm square has a quarter-circle of radius 4 cm removed from one corner. Area = 100 − 4π = 87.4 cm². Perimeter = 10 + 10 + 6 + 6 + (¼ × 2π × 4) = 32 + 2π = 38.3 cm.

Worked reminder: semicircle of radius 5. Perimeter = 5π + 10 = 25.7. Area = ½ × π × 25 = 12.5π = 39.3.

Extension only: 3D solids

Solid	Volume	Surface area
Sphere (radius r)	(4/3)πr³	4πr²
Hemisphere	(2/3)πr³	3πr² (including flat face)
Cone (radius r, height h, slant l)	(1/3)πr²h	πrl curved; plus πr² for base
Pyramid (base area A, height h)	(1/3)Ah	base plus triangular faces
Cylinder	πr²h	2πrh curved; plus 2πr² for two ends

• Slant height: l² = r² + h². For a square-based pyramid, the triangular faces have height s with s² = h² + (half base)².
• The cylinder row is included for use in composite solids. Cylinder volume is Core, as a prism (V = Ah).
• Composite solids: add the volumes. For surface area, count only the faces on the outside. A hemisphere on top of a cylinder hides the circle where they join.

Frustums

A frustum is a cone or pyramid with a smaller, similar cone or pyramid removed from the top.

1. Find the scale factor from the heights: small ÷ large.
2. Multiply the large solid’s lengths by the scale factor to get the small solid’s lengths.
3. Volume of frustum = large volume − small volume.
4. Curved surface of a cone frustum = large curved area − small curved area. Add the two circular faces for a total surface area.

Worked reminder: a cone of height 15 cm and radius 6 cm has a top cone of height 6 cm removed. Scale factor = 6 ÷ 15 = 0.4, so the top cone has radius 2.4 cm. Large volume = (1/3) × π × 36 × 15 = 180π. Small volume = (1/3) × π × 5.76 × 6 = 11.52π. Frustum = 168.48π = 529.3 cm³.

Worked reminder: cone, radius 3 and height 4, so l = 5. Volume = (1/3) × π × 9 × 4 = 12π. Curved area = π × 3 × 5 = 15π. Total area = 15π + 9π = 24π.

Must-know distinctions

• Radius versus diameter: area uses r, circumference can use either.
• Perpendicular height versus slant height: volume uses the height, curved area of a cone uses the slant height.
• Perimeter (a length, in cm) versus area (cm²) versus volume (cm³).
• Exact (in terms of π) versus rounded: give what the question asks for.
• Capacity (litres) and volume (cm³, m³) measure the same thing in different units.

Quick self-test

1. Convert 0.8 m³ to litres.
2. Convert 7.5 cm² to mm².
3. Convert 108 km/h to m/s.
4. A material has density 0.9 g/cm³. Find the mass, in kg, of 2000 cm³ of it.
5. Write 7:15 pm in 24 hour form, and 00:40 in 12 hour form.
6. Find the area of a triangle with base 9 cm and perpendicular height 6.4 cm.
7. Find the area of a trapezium with parallel sides 5 cm and 11 cm and height 4 cm.
8. A circle has diameter 10 cm. Give its circumference and area in terms of π.
9. A prism has cross-sectional area 18 cm² and length 7.5 cm. Find its volume.
10. A semicircle has diameter 10 cm. Find its perimeter to 3 significant figures.
11. (Extension only) A ball has radius 3 cm. Give its volume in terms of π.
12. (Extension only) A square-based pyramid has base side 6 cm and height 7 cm. Find its volume.

Answers

1. 0.8 × 1000 = 800 litres
2. 7.5 × 100 = 750 mm²
3. 108 ÷ 3.6 = 30 m/s
4. 0.9 × 2000 = 1800 g = 1.8 kg
5. 19:15 and 12:40 am
6. ½ × 9 × 6.4 = 28.8 cm²
7. ½ × (5 + 11) × 4 = 32 cm²
8. Circumference 10π cm; area π × 5² = 25π cm²
9. 18 × 7.5 = 135 cm³
10. 5π + 10 = 25.7 cm
11. (4/3) × π × 27 = 36π cm³
12. (1/3) × 36 × 7 = 84 cm³

Where marks are usually lost

• Using the plain length factor on an area or volume, instead of squaring or cubing it.
• Treating 1 h 45 min as 1.45 h.
• Using the slanted side of a parallelogram or trapezium as the height.
• Squaring the diameter instead of the radius.
• Writing a decimal such as 78.5 where the answer should keep π.
• Adding an internal join into a composite perimeter, or leaving out a curved edge.
• Rounding part-way through and then losing accuracy in the final answer.
• Leaving off units, or giving cm instead of cm² or cm³.
• Putting the perpendicular height into πrl, or forgetting the base circle in a total surface area.
• Subtracting the small solid’s volume from the wrong solid when finding a frustum.

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.