HookWhy an 18-inch pizza beats two 12-inch ones
In January 2019 the account Fermat's Library posted a fact that quietly annoyed millions of people: one 18-inch pizza contains more food than two 12-inch pizzas. It sounds wrong — 18 is only half as much again as 12 — until you remember that pizza is sold by diameter but eaten by area, and area does not grow at the same rate as width. A 12-inch pizza has area π × 6² ≈ 113 square inches; an 18-inch has π × 9² ≈ 254 square inches. Two 12-inch pizzas give about 226 square inches — still less than the single 18-inch. The width went up by half, but the area went up by (9 ÷ 6)² = 2.25 times, because area depends on the radius squared.
That gap between how a shape looks and how much it actually holds is the entire subject of mensuration — measuring lengths, areas and volumes and never being fooled by them. This section is where the formulae live: perimeter and area of every flat shape, circumference and area of circles, surface area and volume of solids, arc lengths and sectors, and then the two great tools for right-angled triangles — Pythagoras' theorem and trigonometry. These are the most calculator-heavy topics on the Foundation papers, and the marks turn on choosing the right formula, keeping full accuracy until the final line, and never confusing a length with an area.
ModelUnits — and the conversion that squares and cubes
The standard units are the metric backbone: millimetres, centimetres, metres and kilometres for length; grams, kilograms and tonnes for mass; millilitres and litres for capacity; seconds, minutes and hours for time; and pounds and pence for money. Compound units combine two of these — speed in metres per second or miles per hour, density in grams per cubic centimetre, a price in pounds per kilogram.
The trap that costs the most marks is not the linear conversions (10 mm = 1 cm, 1000 g = 1 kg) but the area and volume ones. Because 1 m = 100 cm, a square metre is 100 cm by 100 cm, which is 100 × 100 = 10,000 cm², not 100 cm². A cubic metre is 100 × 100 × 100 = 1,000,000 cm³. Whenever you convert an area you square the conversion factor; whenever you convert a volume you cube it. This is the pizza principle in disguise — the same reason a length ratio of 2 becomes an area ratio of 4.
Time is the other silent trap because it is not decimal: 1.5 hours is 90 minutes, not 150, and 2 hours 45 minutes is 2.75 hours. Converting a time to hours before dividing by it is where speed and rate questions are won or lost.
A rectangular garden is 3 m by 4 m. Find its area in cm². Method one, convert first: 3 m = 300 cm and 4 m = 400 cm, so area = 300 × 400 = 120,000 cm². Method two, area then convert: 3 × 4 = 12 m², and 1 m² = 10,000 cm², so 12 × 10,000 = 120,000 cm². The two agree — but a student who multiplies 12 by 100 gets 1,200 cm² and is out by a factor of 100. Always square the factor for areas.
MechanismArea formulae — and why each one works
A rectangle is base × height because it is a grid of unit squares, base squares across and height rows down. A parallelogram has the same area as a rectangle, base × height — imagine slicing the slanted triangle off one end and sliding it to the other to straighten it into a rectangle. Note the height is the perpendicular height, straight up, not the slanted side.
A triangle is ½ × base × height, because any triangle is exactly half of a rectangle (or parallelogram) that shares its base and height — pair a triangle with a copy of itself rotated 180° and they lock together into a parallelogram. Again the height is perpendicular to the base.
A trapezium is ½ × (a + b) × h, where a and b are the two parallel sides and h is the perpendicular gap between them. It works because you are averaging the two parallel sides — ½(a + b) is the width halfway up — and multiplying by the height. AQA prints the trapezium formula on the exam formula sheet, but you must know the triangle and parallelogram formulae from memory.
A trapezium has parallel sides of 6 cm and 10 cm, with a perpendicular height of 4 cm. Area = ½ × (6 + 10) × 4 = ½ × 16 × 4 = 32 cm². Sanity check by splitting it into a 6-by-4 rectangle (area 24) plus a triangle with base 4 and height 4 (area 8): 24 + 8 = 32 cm². The two methods matching is your proof the answer is right — and it is worth doing when the formula feels shaky.
ModelCircles — π ties the diameter to everything
Every circle obeys one magic ratio: the circumference divided by the diameter is always π ≈ 3.142, whatever the size of the circle. Rearranged, that gives the circumference C = πd = 2πr (since the diameter is twice the radius). The area is A = πr². You must memorise both — they are not on the formula sheet — and you must not swap them: circumference uses the radius once (it is a length), area uses it squared (it is a region).
The reason area needs r² is exactly the pizza fact from the introduction. Double the radius and the circumference merely doubles, but the area quadruples, because you have scaled the shape in two directions at once. That single-squared-versus-not distinction is the most common circle error in the country.
When a question says 'give your answer in terms of π', stop before you reach for the π button — leave the answer as, say, 25π cm², which is exact. When it wants a decimal, use the calculator's π key rather than typing 3.14, and write the full display down before rounding to the accuracy asked for, usually one or two decimal places or three significant figures.
A circle has radius 7 cm. Circumference = 2 × π × 7 = 14π ≈ 43.98 cm. Area = π × 7² = π × 49 = 49π ≈ 153.94 cm². Notice the units: the circumference is in cm (a length), the area in cm² (a region). A frequent slip is to compute π × 14 = 43.98 for the area by using the diameter — but area always uses the radius, squared, so it is π × 49, not π × 14.
MechanismComposite shapes, arcs and sectors — fractions of a whole
A composite shape is built from simpler ones. The method never changes: break it into rectangles, triangles and parts of circles, find each area, then add — or, for a shape with a hole, subtract. For the perimeter, trace all the way round the outside and add only the edges you actually walk along; a common error is to include an internal line that is not part of the boundary.
A sector is a slice of a circle, and an arc is the curved edge of that slice. Both are just a fraction of the whole circle, and the fraction is the angle at the centre over 360°. So the arc length is (θ ÷ 360) × πd, and the sector area is (θ ÷ 360) × πr². A quarter circle uses 90 ÷ 360 = ¼; a semicircle uses 180 ÷ 360 = ½. There is nothing new to learn here beyond the circle formulae — you are simply taking a share of them.
A sector has radius 10 cm and an angle of 72° at the centre. The fraction of a full circle is 72 ÷ 360 = 0.2, i.e. one fifth. Arc length = 0.2 × 2 × π × 10 = 0.2 × 20π = 4π ≈ 12.57 cm. Sector area = 0.2 × π × 10² = 0.2 × 100π = 20π ≈ 62.83 cm². Notice both start by finding the same fraction, 0.2 — find that first, then decide whether you are taking a share of the circumference (for the arc) or of the area (for the sector).
DataVolume and surface area — cross-sections and the formula sheet
A prism is any solid with the same cross-section all the way along, and its volume is beautifully simple: area of the cross-section × length. A cuboid is a prism with a rectangular cross-section, so its volume is length × width × height. A cylinder is a prism with a circular cross-section, so its volume is πr² × height. The prism formula is on the AQA formula sheet, but the thinking — find the cross-section area first, then multiply by the length — is what earns the marks.
Surface area is the total area of every face, so treat the solid as a flat net, find each face's area and add them up. A cylinder's surface area is two circles (2 × πr²) plus the curved part, which unrolls into a rectangle of width equal to the circumference: 2πr × h. For a sphere, cone or pyramid, AQA gives you the formulae on the sheet — sphere volume (4/3)πr³ and surface area 4πr², cone volume (1/3)πr²h and curved surface πrl — so your job is to substitute correctly and keep the units straight (volume in cubed units, surface area in squared units).
A cylinder has radius 5 cm and height 10 cm. Volume = πr²h = π × 5² × 10 = π × 25 × 10 = 250π ≈ 785.4 cm³. Surface area = two ends plus the curved side = 2 × π × 5² + 2 × π × 5 × 10 = 50π + 100π = 150π ≈ 471.2 cm². Keep the answer as 250π and 150π if the question says 'in terms of π' — those are exact, and reaching for the calculator would throw the exactness (and the final mark) away.
ModelPythagoras — the right-angled triangle's hidden rule
In any right-angled triangle, the square on the longest side equals the sum of the squares on the other two: a² + b² = c², where c is the hypotenuse, always the side opposite the right angle and always the longest. This is on the formula sheet, but the technique is what matters. To find the hypotenuse you add the squares of the two shorter sides and take the square root. To find a shorter side you subtract: rearrange to a² = c² − b². The direction — add for the long side, subtract for a short side — is the single most important decision in the whole topic, and getting it backwards is the classic Foundation error.
Pythagoras only works in right-angled triangles, so your first move is always to spot or mark the right angle. Real uses are everywhere: the diagonal of a TV screen, whether a ladder reaches a window, the straight-line distance between two points on a coordinate grid (which is just a right-angled triangle drawn on the axes).
A ladder 5 m long rests against a wall with its foot 1.5 m from the base. How high up the wall does it reach? The ladder is the hypotenuse (5 m), the ground distance is one shorter side (1.5 m), and the height is the other shorter side — so subtract: height² = 5² − 1.5² = 25 − 2.25 = 22.75, giving height = √22.75 ≈ 4.77 m. If you had added instead, you would have got √(25 + 2.25) ≈ 5.22 m — taller than the ladder, which is impossible, and that impossibility is your built-in check that you needed to subtract.
MechanismTrigonometry, exact values and similarity
Trigonometry links an angle to a ratio of sides in a right-angled triangle, through sin, cos and tan. Label the sides relative to the angle: the hypotenuse (opposite the right angle), the opposite (facing your angle) and the adjacent (beside it). Then SOH-CAH-TOA: sin = opposite ÷ hypotenuse, cos = adjacent ÷ hypotenuse, tan = opposite ÷ adjacent. To find a missing side, pick the ratio that uses the two sides you care about and rearrange. To find a missing angle, use the inverse buttons — sin⁻¹, cos⁻¹, tan⁻¹.
AQA expects you to know a few exact values without a calculator: sin 30° = cos 60° = ½, sin 60° = cos 30° = √3⁄2, sin 45° = cos 45° = √2⁄2, tan 45° = 1, sin 0° = 0 and cos 0° = 1. These turn up on the non-calculator paper precisely because a calculator is not allowed there.
Similar figures extend this: when two shapes are the same shape at different sizes, matching sides share one scale factor. Find that factor from a known pair, then multiply to get any missing length. (At Foundation you only work with the length relationships; how areas and volumes scale is Higher-tier.)
A right-angled triangle has an angle of 35° and a hypotenuse of 10 cm; find the opposite side. The ratio linking opposite and hypotenuse is sine: sin 35° = opposite ÷ 10, so opposite = 10 × sin 35° ≈ 10 × 0.5736 = 5.74 cm. Now the reverse — find an angle when the opposite is 7 and the adjacent is 10: tan θ = 7 ÷ 10 = 0.7, so θ = tan⁻¹(0.7) ≈ 35.0°. Choose the ratio by which two sides you have, then rearrange or invert.
CaseMeasuring, maps, scale drawings and bearings
The practical end of this section is reading real measurements off diagrams and maps. Line segments are measured with a ruler, angles with a protractor — line the protractor's centre on the vertex and its zero line along one arm, and read from the correct scale (there are two, going opposite ways). A scale such as 1 : 50,000 means 1 cm on the map is 50,000 cm — that is 500 m — in real life, so a map distance of 4 cm represents 2 km.
Bearings are how directions are given in navigation and always follow three rules: measured from north, turning clockwise, and written with three figures — so due east is 090°, due south is 180°, and a direction of 45° is written 045° with its leading zero. To find a bearing, draw the north line at your starting point and measure the clockwise angle to the target. The most common exam task combines this with a scale drawing: plot a journey from bearings and distances, then measure the final position off your diagram.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Papers 2 and 3 are calculator papers, and this is their heartland — expect a circle, a volume, a Pythagoras or trigonometry problem worth several marks each. AQA now supplies a formula sheet in the exam: it gives you the trapezium area, the prism volume, and the sphere, cone, Pythagoras and trigonometry formulae — but not the circumference or area of a circle, which you must still recall from memory. Know exactly what is and is not on that sheet before you walk in.
Mark schemes here are method-first. Write the formula, substitute the numbers, then evaluate — a correct substitution with a slipped final digit usually keeps most of the marks, while a bare wrong answer keeps none. Keep full calculator accuracy all the way through and only round on the final line, to the accuracy the question asks for; rounding partway through (using 3.14 for π, or 0.57 for sin 35°) can lose the accuracy mark. When a question says 'in terms of π' or 'give an exact answer', leave π and fractions in — a decimal there loses the mark even on a calculator paper.
Always write units, and make them match the dimension: cm for length, cm² for area, cm³ for volume. On Paper 1, the non-calculator paper, expect the exact trig values (sin 30° = ½, tan 45° = 1, sin 60° = √3⁄2) and Pythagoras with numbers that come out whole. For bearings and scale drawings, bring a protractor and ruler, draw the north line, and give bearings as three figures measured clockwise from north.