Skip to topic
    ← Back to course topics

    Mensuration and calculation — AQA GCSE Mathematics

    Test yourself on Mensuration and calculation with AQA GCSE practice questions.

    Start free

    7 days Premium · Then free forever · No card, no charge

    Mensuration and calculation explained

    Every measurement carries a unit, and most of the work here is changing between units safely.

    Read the full explanation

    Lengths step by 10, 100 and 1000: 1 cm = 10 mm, 1 m = 100 cm, 1 km = 1000 m. A square unit scales by the square of the length factor, so a square metre is 10 000 square centimetres, and a cubic unit scales by the cube, so a cubic metre is a million cubic centimetres. Capacity links to volume through 1 ml = 1 cm³ and 1 litre = 1000 cm³. Mass runs 1 kg = 1000 g and 1 tonne = 1000 kg. Time is not decimal: 90 minutes is 1.5 hours, not 1.9 hours. Money is written with two decimal places. A question often mixes units on purpose, for example a tank 40 cm by 50 cm by 1.2 m, so change everything into one unit first and then calculate.

    measure line segments and angles in geometric figures, including interpreting maps and scale drawings and use of bearings

    Lengths are read with a ruler to the nearest millimetre, and angles with a protractor placed so its centre sits on the vertex and its zero line lies along one arm. A scale of 1 : 50 000 means that 1 cm on the paper stands for 50 000 cm on the ground, which is 0.5 km. Going from the drawing to real life you multiply by the scale; going back the other way you divide. A bearing is measured from north, turning clockwise, and is always written with three figures, so due east is 090° and a direction 25° clockwise from north is written 025°. Example: if the bearing of B from A is 070°, the bearing of A from B is 070° + 180° = 250°. If the bearing is above 180°, subtract 180° instead: the back bearing of 250° is 250° − 180° = 070°. A typical question asks you to measure a line, convert it with the scale, then measure or work out a bearing.

    know and apply formulae to calculate: area of triangles, parallelograms, trapezia; volume of cuboids and other right prisms (including cylinders)

    Area of a triangle is ½ × base × perpendicular height. Area of a parallelogram is base × perpendicular height. Area of a trapezium is ½(a + b)h, where a and b are the parallel sides and h is the distance between them. In all three the height must be at right angles to the base, never the sloping edge. For any prism the volume is the area of the cross-section multiplied by the length, so a cylinder of radius r and height h has volume πr²h. Example: a trapezium with parallel sides 5 cm and 9 cm and height 4 cm has area ½ × (5 + 9) × 4 = 28 cm², and a prism with that cross-section and a length of 10 cm holds 280 cm³. Some questions give the volume and ask for a missing length, which means dividing rather than multiplying.

    know the formulae: circumference of a circle = 2πr = πd area of a circle = πr² calculate perimeters of 2D shapes, including circles areas of circles and composite shapes

    The distance round the edge is given by circumference = 2πr = πd, so pick the radius version or the diameter version to match what you are told. The space inside is πr², and if the question gives a diameter you halve it before squaring. A composite shape is built from parts, so work out each part on its own line, then add or subtract. Example: a semicircle of radius 6 cm has area ½ × π × 6² = 18π ≈ 56.5 cm², while its perimeter is half the circumference plus the straight diameter, π × 6 + 12 ≈ 30.8 cm to three significant figures. Leave the answer as a multiple of π when an exact value is wanted, and otherwise keep the full value on the calculator and round once at the end.

    surface area and volume of spheres, pyramids, cones and composite solids

    Calculate volumes and surface areas for various 3D solids using standard formulae. A sphere of radius r has volume V = ⁴⁄₃πr³ and surface area SA = 4πr². A cone with radius r, perpendicular height h, and slant height l has V = ⅓πr²h and curved surface area πrl. A pyramid has V = ⅓ × base area × perpendicular height. Surface area is the sum of the areas of all exposed faces. For composite solids, calculate each part separately and sum them, but exclude any faces hidden where components join. For example, the surface area of a cone placed on a cylinder of the same radius is the cone's curved area (πrl), the cylinder's curved area (2πrh), and the cylinder's base area (πr²). The circular face where they meet is not included. Pythagoras' theorem is often needed to find a slant height (l) from a perpendicular height (h) and radius (r), as they form a right-angled triangle where l² = h² + r².

    calculate arc lengths, angles and areas of sectors of circles

    A sector is a slice of a circle, and the size of the slice is the fraction θ/360, where θ is the angle at the centre. So the arc length is that fraction of the whole circumference, and the sector area is that fraction of the whole area. The perimeter of a sector is the arc plus the two straight radii, which is the part most often left out. Example: a sector of radius 10 cm with an angle of 72° is one fifth of the circle, so the arc is 4π ≈ 12.6 cm and the area is 20π ≈ 62.8 cm². Questions also run backwards: given an arc length and a radius, rearrange to find the angle. On Higher tier the same fraction is used for a segment, which is the sector with the triangle taken away.

    apply the concepts of congruence and similarity, including the relationships between lengths in similar figures

    Two shapes are congruent when they are identical in shape and size, so one can be turned, reflected or slid exactly onto the other. For triangles the conditions are side-side-side, side-angle-side, angle-side-angle and right-angle-hypotenuse-side. Two shapes are similar when one is an enlargement of the other: matching angles are equal and matching sides are all in the same ratio. The method is to pair up corresponding sides, divide one known pair to get the scale factor, then multiply or divide to reach the length you want. Example: if 6 cm on the small shape matches 9 cm on the large one, the scale factor is 9 ÷ 6 = 1.5, so a side of 4 cm on the small shape matches 6 cm on the large one.

    including the relationships between lengths, areas and volumes in similar figures (Higher tier only)

    This is Higher tier only. When one figure is an enlargement of another with length scale factor k, every area is multiplied by k² and every volume by k³. Double the lengths and the surface area becomes four times as big while the volume becomes eight times as big. The method is always to get k first: from a pair of matching lengths directly, from a ratio of areas by taking the square root, or from a ratio of volumes by taking the cube root. Then move between lengths, areas and volumes using k, k² and k³. Example: two similar jugs have heights 6 cm and 9 cm, so k = 1.5, their surface areas are in the ratio 1 : 2.25 and their capacities in the ratio 1 : 3.375, and a small jug holding 200 ml matches a large one holding 675 ml.

    know the formulae for: Pythagoras’ theorem, a² + b² = c² and the trigonometric ratios, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse and tan θ = opposite/adjacent apply them to find angles and lengths in right-angled triangles in two dimensional figures

    The hypotenuse is the side opposite the right angle, and the three sides are linked by a² + b² = c², with c as the hypotenuse. To reach the hypotenuse you add the two squares and square root; to reach a shorter side you subtract the smaller square from the larger one and square root. When an angle is involved, first label the sides opposite, adjacent and hypotenuse with respect to that angle, then choose the ratio that uses the two sides the question mentions. Example: with an opposite side of 5 cm and a hypotenuse of 13 cm, sin θ = 5/13, so θ = 22.6° to one decimal place. Set the calculator to degrees, and use the inverse trigonometric key whenever the unknown is the angle rather than a length.

    apply them to find angles and lengths in right-angled triangles and, where possible, general triangles in two and three dimensional figures (Higher tier only)

    To find lengths and angles inside a 3D solid, identify a relevant 2D triangle and redraw it separately. Common tasks include finding the long diagonal of a cuboid or the angle between a line and a plane (e.g., an edge and the base). This is often a two-step process: first, use Pythagoras' theorem on one face (e.g., the base) to find a diagonal length. Second, use that length in a new right-angled triangle that cuts through the solid to find the required length or angle using SOHCAHTOA. For a cuboid measuring 3 cm by 4 cm by 12 cm, the base diagonal is 5 cm, the longest diagonal is 13 cm, and the angle it makes with the base is tan⁻¹(12/5) = 67.4°. If the identified triangle is not right-angled, the sine rule or cosine rule must be used instead.

    know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90° know the exact value of tan θ for θ = 0°, 30°, 45°, 60°

    These values are wanted whenever a question says give an exact answer, or when no calculator is allowed. The sines climb: sin 0° = 0, sin 30° = ½, sin 45° = √2/2, sin 60° = √3/2, sin 90° = 1. The cosines run the other way: cos 0° = 1, cos 30° = √3/2, cos 45° = √2/2, cos 60° = ½, cos 90° = 0. For the tangent, tan 0° = 0, tan 30° = √3/3, tan 45° = 1, tan 60° = √3. They come from two special triangles: half an equilateral triangle of side 2 gives shorter sides of 1 and √3 with hypotenuse 2, and a right-angled isosceles triangle with two sides of 1 has hypotenuse √2. Example: a hypotenuse of 8 cm at 60° gives an exact height of 4√3 cm.

    know and apply the sine rule, a/sin A = b/sin B = c/sin C and cosine rule, a² = b² + c² − 2bc cos A to find unknown lengths and angles (Higher content only)

    This is Higher tier only. These two rules take over when a triangle has no right angle. Each lower-case letter is the side opposite the capital letter of the same name. The first rule, a/sin A = b/sin B = c/sin C, fits a question giving two angles and a side, or two sides and an angle opposite one of them; flip it as sin A/a = sin B/b = sin C/c when the unknown is an angle. The second, a² = b² + c² − 2bc cos A, fits two sides with the angle between them, or all three sides when an angle is wanted, and rearranges to cos A = (b² + c² − a²)/(2bc). Example: with b = 7 cm, c = 9 cm and A = 40°, the second rule gives a² = 33.478, so a = 5.79 cm to three significant figures.

    know and apply Area = ½ab sin C to calculate the area, sides or angles of any triangle (Higher content only)

    This is Higher tier only. When you know two sides of a triangle and the angle trapped between them, the area is ½ab sin C, where a and b are those two sides and C is the angle where they meet. No perpendicular height is needed, which is why it suits a triangle where no height is marked on the diagram. Example: sides of 8 cm and 5 cm meeting at 30° enclose an area of ½ × 8 × 5 × sin 30° = 10 cm². The formula also runs backwards: given the area and the two sides you can find the angle, and given the area, one side and the angle you can find the other side. It also gives the area of a segment, once this triangle is subtracted from the sector around it.

    Your focus

    1. Write down the conversions between mm, cm, m and km, and between millilitres, cubic centimetres and litres.
    2. Convert every measurement in a mixed-unit question into one unit before calculating, such as 1.2 m into 120 cm.
    3. Explain why one square metre is 10 000 square centimetres while one metre is only 100 centimetres.
    Show all 44 objectives
    1. Explain why 90 minutes is 1.5 hours and not 1.9 hours.
    2. Measure a line to the nearest millimetre and an angle with the protractor centred on the vertex and zero along an arm.
    3. Calculate a real distance from a scale drawing by applying a scale such as 1 : 50 000 to the measured length.
    4. Write down a bearing as three figures clockwise from north and work out the back bearing by adding 180° when the bearing is below 180°, or subtracting 180° when it is above 180°.
    5. Explain why a measured answer is accepted inside a tolerance either side of the true value.
    6. Calculate the volume of a prism as cross-sectional area times length and state that the answer is the space it fills in cubic units.
    7. Apply half of (a + b) times h to a trapezium, substituting the perpendicular height rather than a sloping edge.
    8. Work out a missing length when the volume of a prism is given, dividing rather than multiplying.
    9. Explain why the height used in an area formula has to be at right angles to the base.
    10. Calculate the area of a circle from pi r squared, halving a given diameter first, and give the answer in square units.
    11. Work out the perimeter of a composite shape such as a semicircle by adding the straight edges to the curved part.
    12. Justify leaving an answer as a multiple of pi when an exact value is asked for, and rounding only once otherwise.
    13. Calculate the volume of a cone as a third of pi r squared h and state that the answer measures the space inside in cubic units.
    14. Work out a cone's slant height from its perpendicular height before finding the curved surface area.
    15. Explain why a face hidden where two pieces of a composite solid meet is left out of the surface area.
    16. Calculate an arc length or a sector area as the fraction theta over 360 of the whole circumference or area, with its unit.
    17. Work out the angle at the centre by rearranging when the arc length and the radius are the values given.
    18. Explain why the perimeter of a sector includes the two straight radii as well as the arc.
    19. Compare congruence and similarity, one meaning identical in shape and size and the other meaning one is an enlargement.
    20. Work out an unknown length in similar figures by pairing corresponding sides and applying the scale factor.
    21. Prove two triangles congruent by naming the condition and attaching a reason to each matching fact.
    22. Describe how areas scale by k squared and volumes by k cubed when every length scales by k.
    23. Calculate a matching area or volume by finding k first, square rooting an area ratio or cube rooting a volume ratio.
    24. Justify which way round the scale factor is applied by checking whether the answer should come out larger or smaller.
    25. Calculate a missing side using a squared plus b squared equals c squared, adding for the hypotenuse and subtracting for a shorter side.
    26. Apply sin, cos or tan by first labelling the sides opposite, adjacent and hypotenuse for the angle concerned.
    27. Explain why the inverse trigonometric key is needed when the unknown is an angle rather than a length.
    28. Draw the right-angled triangle that lies inside a solid on its own, away from the picture of the solid.
    29. Calculate the longest diagonal of a cuboid in two steps, finding the diagonal of the base first.
    30. Justify choosing the sine rule or the cosine rule when the triangle picked out of the figure has no right angle.
    31. Write down the exact values of sin and cos for 0, 30, 45, 60 and 90 degrees and of tan for 0, 30, 45 and 60 degrees.
    32. Show that these values come from half an equilateral triangle of side 2 and a right-angled isosceles triangle with sides of 1.
    33. Apply an exact value in a non-calculator question and simplify the surd, giving 4 root 3 rather than a rounded decimal.
    34. Compare when the sine rule fits and when the cosine rule fits, judging by the sides and angles the triangle gives you.
    35. Calculate an unknown side from the cosine rule, remembering to square root at the end.
    36. Work out an unknown angle by flipping the sine rule so that sin A over a equals sin B over b.
    37. Justify the choice of rule for a given triangle before any values are substituted.
    38. Calculate the area of a triangle from half ab sin C, where C is the angle between the two sides, giving square units.
    39. Work out a missing side or the included angle by rearranging the formula when the area is known.
    40. Explain why this formula suits a triangle with no perpendicular height marked on the diagram.
    41. Apply the formula to find the area of a segment by subtracting the triangle from the sector around it.

    Mensuration and calculation exam tips

    Quick Revision Summary (Key Takeaway)

    Mensuration and calculation covers finding perimeters, areas and volumes of 2D and 3D shapes, including circles, prisms, cylinders, cones and spheres, plus compound shapes and correct unit conversion. In AQA GCSE Mathematics you must recall and apply formulae such as pi r squared, pi d, and volume equals cross-sectional area times length, showing clear method and giving answers with correct units.

    Topic Overview

    Mensuration and calculation is the branch of geometry that deals with measuring lengths, perimeters, areas, surface areas and volumes of 2D and 3D shapes. In AQA GCSE Mathematics you must confidently use formulae for rectangles, triangles, parallelograms, trapezia, circles, prisms, cylinders, pyramids, cones and spheres, and apply these to compound shapes and real-world problems.

    This topic matters because it connects algebra, ratio, rounding and unit conversion in practical contexts such as construction, engineering and design. It also forms the foundation for later work on similarity, trigonometry and calculus, so secure understanding of formulae and units is essential for success in both Foundation and Higher tier exams.

    Key Concepts
    • →Perimeter is the distance around a 2D shape; for a circle it is called the circumference and is calculated using C = pi x d or C = 2 x pi x r.
    • →Area is measured in square units: rectangle = length x width, triangle = half x base x perpendicular height, parallelogram = base x perpendicular height, trapezium = half x (a + b) x height, circle = pi x r squared.
    • →Volume is measured in cubic units: for a prism or cylinder, volume = cross-sectional area x length; for a pyramid or cone, volume = one third x base area x perpendicular height; for a sphere, volume = four thirds x pi x r cubed.
    • →Surface area is the total area of all faces of a 3D shape; for a cylinder it is 2 x pi x r squared + 2 x pi x r x h, and for a sphere it is 4 x pi x r squared.
    • →Compound shapes must be split into simpler shapes, and missing lengths found using the given dimensions before calculating total area or volume.
    Marking Points
    • converting every measurement into a single unit before any calculation, even if the arithmetic that follows is wrong
    • the correct conversion factor, for example multiplying by 1000 to change litres into millilitres
    • the numerical answer following through from the converted values
    • the answer written with the correct unit where the question asks for a measure
    • a measured length or angle that falls inside the tolerance a mark scheme allows either side of the true value
    • applying the scale correctly to turn a measured length into a real distance, even if the measurement itself was slightly off
    • a bearing given as three figures measured clockwise from north
    • the final distance with a sensible unit such as kilometres
    • a back bearing found by adding 180° when the bearing is below 180°, or subtracting 180° when it is above 180°
    • choosing the correct formula, for example writing ½(a + b)h for a trapezium
    • substituting the perpendicular height rather than a slant length, even if the arithmetic then goes wrong
    • a correct cross-sectional area of a prism before it is multiplied by the length
    • the final value with square units for an area and cubic units for a volume
    • a correct substitution into circumference = 2πr or into area = πr², even if the arithmetic afterwards is wrong
    • halving a given diameter to get the radius before it is used
    • splitting a composite shape into recognisable parts and finding one of them correctly
    • including the straight edges when a perimeter contains a diameter or a radius, not only the curved part
    • an exact answer left in terms of π where that is asked for, or a correctly rounded decimal where it is not
    • using the correct formula for the given solid, for example V = ⅓πr²h for a cone.
    • using Pythagoras' theorem to find a slant height from a perpendicular height, or vice versa.
    • calculating the volume or surface area of each separate component of a composite solid correctly.
    • for surface area of a composite solid, summing only the areas of the external faces.
    • for a pyramid, calculating the surface area as the sum of the base area and the areas of all triangular faces.
    • stating the final answer with the correct units, e.g., cm³ for volume and cm² for area.
    • writing or using the fraction θ/360 correctly
    • multiplying that fraction by the full circumference for an arc, or by the full area for a sector
    • adding the two radii when the perimeter of the sector is asked for
    • a correct rearrangement when the angle is the unknown, even if the final value is wrong
    • the answer to the accuracy asked for, with the correct unit
    • identifying which sides correspond, for example by naming the vertices of the two triangles in matching order
    • a correct scale factor from a known pair of sides, even if it is then applied to the wrong side
    • multiplying or dividing by that scale factor to reach the unknown length
    • naming the congruence condition, such as side-angle-side, when a pair of triangles is to be proved congruent
    • a reason attached to each fact in a proof, such as angles being equal because they are alternate
    • finding the length scale factor, including by square rooting a ratio of areas or cube rooting a ratio of volumes
    • squaring it for an area or cubing it for a volume, even if the multiplication that follows is wrong
    • applying the factor the right way round for the direction of the enlargement
    • the final answer with square or cubic units as appropriate
    • a correct use of a² + b² = c² with the sides placed correctly
    • subtracting rather than adding when the unknown is a shorter side
    • a trigonometric statement that matches the labelled sides, for example tan 40° = x/7
    • a correct rearrangement, such as x = 7 tan 40°, even if the value is then worked out wrongly
    • the final length or angle rounded to the accuracy the question asks for
    • identifying and drawing the correct 2D triangle from within the 3D solid.
    • correctly calculating an intermediate length needed for a second calculation, e.g., the diagonal on the base of a cuboid.
    • selecting the correct trigonometric ratio (SOHCAHTOA) or rule (sine/cosine) for the identified triangle.
    • writing a correct trigonometric statement with values substituted correctly.
    • giving the final answer to the required degree of accuracy, with degrees for angles.
    • quoting the exact value that is needed, for example √3/2 rather than a rounded decimal
    • substituting that exact value into the calculation set up by the question
    • an answer left in exact form and simplified, such as 4√3 in place of an unsimplified surd
    • a correct simplification where the exact form is a fraction containing a surd
    • choosing the rule that fits the information the triangle gives you
    • a correct substitution into either rule, even if the arithmetic that follows is wrong
    • rearranging correctly to make the unknown side or angle the subject
    • taking the square root at the end when a length has been found from the cosine rule
    • the final answer to the accuracy the question asks for
    • using the formula with an angle that lies between the two sides substituted
    • a correct substitution of both sides and the included angle
    • finding the included angle first, by the cosine rule or by the angle sum of a triangle, where it is not given
    • a correct rearrangement when the area is known and a side or angle is the unknown
    • the area given with square units
    Examiner Tips
    • 💡Underline every unit in the question before you start, then decide the single unit you will work in.
    • 💡Square the length factor for an area conversion and cube it for a volume conversion, rather than reusing the length factor.
    • 💡Check the size of your answer: the same amount written in smaller units must come out as a bigger number.
    • 💡Draw the north line at the point you are measuring from before the protractor goes anywhere near the paper.
    • 💡Sense check every protractor reading against whether the angle looks acute or obtuse.
    • 💡Turn the scale into a sentence such as 1 cm represents 2 km before converting anything.
    • 💡For a back bearing, check whether the original is below or above 180° before deciding whether to add or subtract.
    • 💡Mark the perpendicular height on the diagram before you write a single number down.
    • 💡For any prism, think cross-sectional area first and length second; that covers cuboids and cylinders together.
    • 💡If the shape is awkward, cut it into a rectangle and a triangle, work out both, then add.
    • 💡Write the radius onto the diagram the moment you read a diameter, so you cannot substitute the wrong one.
    • 💡If an exact answer is asked for, stop at the multiple of π instead of pressing the equals key.
    • 💡Draw pencil lines to split a composite shape and label each part before you calculate anything.
    • 💡Label the radius (r), perpendicular height (h) and slant height (l) on the diagram before you start.
    • 💡For composite solids, list the separate shapes and the faces you need to calculate for each one before you begin.
    • 💡Keep π in the calculator throughout the calculation and only round the final answer to the required degree of accuracy.
    • 💡Write the fraction of the circle first, for example 45/360, and simplify it before touching the calculator.
    • 💡Ask yourself whether the answer should be a length or an area, then check the unit you have written matches.
    • 💡If the angle is the unknown, substitute everything you know first and rearrange afterwards; the substitution earns method credit on its own.
    • 💡Redraw the two triangles separately and the same way round before writing any ratio.
    • 💡Write the scale factor down and keep it visible; every later line of working uses it.
    • 💡In a congruence proof, list three matching facts with a reason each, then name the condition at the end.
    • 💡Write k, k² and k³ in a small table at the side and tick the one the question actually needs.
    • 💡Check the direction at the end: an answer for the larger figure must be bigger than the value you started from.
    • 💡If you are given a ratio of areas or volumes, convert it to the length factor before doing anything else.
    • 💡Label the three sides on the diagram before you decide between sine, cosine and tangent.
    • 💡If no angle is given and none is asked for, the question is Pythagoras rather than trigonometry.
    • 💡Keep the unrounded value on the calculator when the answer feeds into a second stage of the question.
    • 💡Always redraw the working triangle flat on the page and label it with the lengths and angles you know.
    • 💡If you cannot mark a right angle on your 2D sketch, you must use the sine rule or the cosine rule.
    • 💡Store an intermediate length in the calculator's memory rather than retyping a rounded version of it to maintain accuracy.
    • 💡Sketch the two special triangles in the margin at the start of a non-calculator paper and read the values off them.
    • 💡If a question says exact, leave the answer as a surd, a fraction, a whole number or a multiple of π, and never as a rounded decimal.
    • 💡Sense check a half-remembered value: sine grows from 0° to 90° while cosine shrinks.
    • 💡Label the triangle so every lower-case letter sits opposite its capital before you write a rule down.
    • 💡Two sides with the angle between them, or three sides, points to the cosine rule; a matching side and angle pair points to the sine rule.
    • 💡Write the rule out in full first and substitute afterwards; a clear statement of the rule shows the examiner your method, though credit depends on the mark scheme for that question.
    • 💡Mark the angle on the diagram and check that both named sides run out of it before substituting.
    • 💡If the included angle is missing, look first for the cosine rule or the angle sum of a triangle.
    • 💡For a segment, work out the sector and the triangle on separate lines, then subtract, labelling both clearly.
    • 💡Always write down the formula you are using before substituting numbers. This earns method marks even if you make an arithmetic error later.
    • 💡Show every stage of your working, including any unit conversions and the final rounding. AQA awards marks for correct method, not just the final answer.
    • 💡Check that your answer is sensible: for example, a volume in cm cubed should be much larger than the corresponding area in cm squared, and a circumference should be roughly three times the diameter.
    Common Mistakes
    • converting an area by multiplying by 100 instead of by 100 twice, so treating a square metre as 100 square centimetres
    • writing 2 hours 30 minutes as 2.3 hours instead of 2.5 hours
    • multiplying a length in centimetres by a length in metres without converting either one
    • giving a money answer as 4.5 rather than £4.50
    • reading the wrong ring of numbers on the protractor, so writing 130° for an angle that is clearly 50°
    • writing a bearing as 70° instead of 070°
    • measuring a bearing anticlockwise, or from the wrong point, so answering with the bearing of A from B when B from A was asked for
    • multiplying by the scale when converting a real distance back onto the drawing, where you should divide
    • adding 180° to a bearing above 180° and giving an answer over 360°, instead of subtracting 180°
    • using the sloping side of a triangle or parallelogram as the height
    • leaving out the ½ in the triangle or trapezium formula, so doubling the area
    • adding the parallel sides of a trapezium and then forgetting to halve
    • multiplying three dimensions together for a prism that is not a cuboid, instead of finding the cross-section first
    • substituting the diameter where the formula wants the radius, which makes an area four times too big
    • working out the area when the question asks for the perimeter, or the other way round
    • calling the curved arc of a semicircle its perimeter and leaving out the straight diameter
    • rounding partway through instead of keeping the full value until the final step
    • confusing slant height and perpendicular height, e.g. using slant height in a volume formula; correct by using perpendicular height for volume and slant height for curved surface area.
    • for a solid hemisphere's surface area, forgetting to add the area of the flat circular base (πr²); correct by visualising all external faces of the solid.
    • including a face that is not on the surface of a composite solid, such as where a cone is joined to a cylinder; correct by only including faces that are exposed to the air.
    • substituting the diameter instead of the radius into a formula, which is especially costly in cubed terms; correct by always halving the diameter to find the radius before substitution.
    • for a pyramid, omitting the base area or miscounting the triangular faces when calculating surface area; correct by listing all faces before starting calculations.
    • reaching for the area formula when an arc is wanted, or the circumference formula when an area is wanted
    • giving the arc length as the perimeter of the sector and ignoring the two straight edges
    • dividing by 360 once inside the fraction and again at the end
    • rounding the fraction of the circle to one or two decimal places before multiplying
    • matching sides by their position on the page rather than by the angles they sit between
    • using the scale factor upside down, so shrinking a length that ought to grow
    • assuming two shapes are similar because they look alike, without checking angles or ratios
    • quoting side-side-angle as a congruence condition, which does not fix a triangle
    • using the length scale factor for an area or a volume, so multiplying by k instead of by k² or k³
    • taking the square root of a ratio of volumes, or the cube root of a ratio of areas
    • finding k from a ratio of areas and then forgetting to cube it when a volume is wanted
    • reading a ratio the wrong way round, so dividing where you should multiply
    • adding the squares when finding a shorter side, which gives an answer longer than the hypotenuse
    • labelling opposite and adjacent from the right angle instead of from the angle named in the question
    • stopping at the squared value and forgetting to take the square root
    • leaving the calculator in radians, so an angle comes out as a small decimal
    • using the sine ratio with the adjacent side because the sides were never labelled
    • taking a length from the wrong face, such as an edge where the diagonal of the base is needed; correct by carefully tracing the required triangle on the 3D diagram before extracting it.
    • finding the wrong angle, for example the angle with a vertical line when the angle with the horizontal base was asked for; correct by marking the required angle clearly on your 2D sketch of the triangle.
    • rounding an intermediate length and using this rounded value in the next step; correct by using the calculator's memory function (ANS or STO) to store the exact value.
    • incorrectly assuming a triangle inside a solid is right-angled; correct by checking if the triangle's vertices correspond to perpendicular lines or planes in the solid; if not, use the sine or cosine rule.
    • swapping the sine and cosine values at 30° and 60°, since the same numbers appear in the other order
    • quoting a value for tan 90°, which is undefined
    • writing a rounded decimal when the question asks for an exact answer
    • reading ½ as the value of sin 45° rather than of sin 30°
    • pairing a side with an angle that is not opposite it
    • doing the subtraction before the multiplication, so subtracting 2bc first and then multiplying by the cosine
    • leaving the answer as the squared value instead of square rooting it
    • missing the second, obtuse angle when the sine rule is used to find an angle and the diagram allows it
    • using the cosine rule with an angle that does not lie between the two given sides
    • using an angle that is not between the two sides that have been named
    • leaving out the ½, which doubles the area
    • bringing a perpendicular height into this formula as well, so applying two area methods at once
    • forgetting that two different angles can share the same sine when the area is used to find an angle
    • Students often confuse radius and diameter, using the diameter in formulae that require the radius. Always halve the diameter first and label your diagram clearly.
    • Students sometimes use the slant height instead of the perpendicular height in area and volume formulae for triangles, parallelograms, trapezia, cones and pyramids. The perpendicular height is the vertical distance between the base and the opposite side or vertex.
    • Students frequently forget to convert units before calculating, for example mixing cm and m. Convert all lengths to the same unit at the start, and remember that 1 m squared = 10,000 cm squared and 1 m cubed = 1,000,000 cm cubed.
    Revision Plan
    1. 1Day 1-2: Revise all 2D area and perimeter formulae, including circles. Create a formula sheet from memory and check it against your textbook.
    2. 2Day 3-4: Practise compound 2D shapes, splitting them into rectangles, triangles and circles. Focus on finding missing lengths and using correct units.
    3. 3Day 5-6: Learn 3D volume and surface area formulae for prisms, cylinders, pyramids, cones and spheres. Complete at least ten mixed practice questions.
    4. 4Day 7-8: Work through past AQA GCSE exam questions on mensuration, timing yourself and marking your answers using the official mark scheme.
    5. 5Day 9-10: Review your mistakes, redo any incorrect questions, and create flashcards for formulae and unit conversions to test yourself regularly.
    Exam Question Types
    • 📋Calculate the area or perimeter of a compound 2D shape: split the shape into simpler parts, find missing lengths, and show all working. Remember to give units.
    • 📋Calculate the volume or surface area of a 3D shape such as a cylinder, cone or sphere: write the formula, substitute the correct radius or height, and round as instructed.
    • 📋Solve a real-world problem involving unit conversion, for example finding the cost of painting a wall or the capacity of a tank: convert units first, then apply the correct formula.
    • 📋Higher tier only: work with similar shapes to find missing lengths, areas or volumes using scale factors, or solve problems involving frustums and composite solids.
    Command Word Expectations (AQA)
    Calculate

    You must work out a numerical answer using the given information. Show clear method, include units, and round only if the question specifies a degree of accuracy.

    Work out

    Similar to calculate, but often used when the method is not immediately obvious. You must show all steps of your reasoning to gain full marks.

    Give your answer to 3 significant figures

    You must round your final answer to 3 significant figures. Do not round intermediate steps, or you may lose accuracy marks. State the rounded answer clearly.

    Show that

    You must demonstrate that a given result is true by showing every step of your working. The final answer is given, so all marks are for correct method and clear communication.

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Using the diameter instead of the radius in area and volume formulae, or forgetting to halve the diameter before substituting into pi r squared or four thirds pi r cubed.
    ❌ Weak Answer (Loses Marks):Area of circle with diameter 10 cm = pi x 10 squared = 314 cm squared.
    Example improved answer:Radius = 10 divided by 2 = 5 cm. Area = pi x 5 squared = 25 pi = 78.5 cm squared (to 3 significant figures).
    Examiner Tip: Always write down the radius first when the question gives a diameter. Underline the word diameter and immediately convert it to radius before using any formula.
    Pitfall: Mixing units within a calculation, for example using centimetres and metres together, or giving a volume in square units instead of cubic units.
    ❌ Weak Answer (Loses Marks):Volume = 3 x 4 x 5 = 60 cm squared.
    Example improved answer:All lengths are in cm, so volume = 3 cm x 4 cm x 5 cm = 60 cm cubed. Units are cubic because three lengths are multiplied.
    Examiner Tip: Convert all measurements to the same unit before calculating. Check that area answers use squared units and volume answers use cubed units.
    Step-by-Step Worked Solutions

    Question: A cylindrical water tank has a radius of 1.5 m and a height of 4 m. Calculate the volume of the tank in cubic metres, giving your answer to 3 significant figures. Use pi = 3.142.

    1. 1.Step 1: Identify the shape as a cylinder and write down the formula: volume = pi x r squared x h.
    2. 2.Step 2: Substitute r = 1.5 and h = 4: volume = 3.142 x 1.5 squared x 4.
    3. 3.Step 3: Calculate 1.5 squared = 2.25, then 3.142 x 2.25 = 7.0695, then 7.0695 x 4 = 28.278.
    4. 4.Step 4: Round to 3 significant figures and state units: 28.3 m cubed.
    Final Answer: The volume of the cylindrical tank is 28.3 m cubed (to 3 significant figures).

    Question: A compound shape is made from a rectangle of length 12 cm and width 5 cm with a semicircle of diameter 5 cm attached to one of the shorter sides. Calculate the total area of the shape, giving your answer to 1 decimal place. Use pi = 3.142.

    1. 1.Step 1: Split the compound shape into a rectangle and a semicircle.
    2. 2.Step 2: Area of rectangle = 12 x 5 = 60 cm squared.
    3. 3.Step 3: Radius of semicircle = 5 divided by 2 = 2.5 cm. Area of full circle = pi x 2.5 squared = 3.142 x 6.25 = 19.6375 cm squared. Area of semicircle = 19.6375 divided by 2 = 9.81875 cm squared.
    4. 4.Step 4: Total area = 60 + 9.81875 = 69.81875 cm squared.
    5. 5.Step 5: Round to 1 decimal place: 69.8 cm squared.
    Final Answer: The total area of the compound shape is 69.8 cm squared (to 1 decimal place).
    Active Recall Memory Test
    State the formula for the area of a circle and explain what each letter represents.
    Key Fact: Area = pi x r squared, where r is the radius of the circle. Pi is approximately 3.142.
    How do you calculate the volume of a prism or cylinder?
    Key Fact: Volume = cross-sectional area x length. For a cylinder, the cross-section is a circle, so volume = pi x r squared x h.
    What is the formula for the volume of a sphere?
    Key Fact: Volume = four thirds x pi x r cubed, where r is the radius.
    How many square centimetres are there in 1 square metre?
    Key Fact: There are 10,000 square centimetres in 1 square metre, because 1 m = 100 cm and 100 x 100 = 10,000.
    Frequently Asked Questions
    What is mensuration in GCSE maths?
    Mensuration is the branch of mathematics that deals with measuring lengths, areas, surface areas and volumes of shapes. In AQA GCSE Mathematics it includes using formulae for 2D shapes like rectangles, triangles and circles, and 3D shapes like prisms, cylinders, cones and spheres. You will also solve problems involving compound shapes and unit conversions.
    How do I find the area of a compound shape?
    Split the compound shape into simpler shapes such as rectangles, triangles and semicircles. Calculate the area of each part separately using the correct formula, then add or subtract the areas as appropriate. Make sure you find any missing lengths by using the dimensions given and label your diagram clearly.
    What is the difference between surface area and volume?
    Surface area is the total area of all the faces of a 3D shape and is measured in square units, such as cm squared. Volume is the amount of space inside the shape and is measured in cubic units, such as cm cubed. For example, the surface area of a cube with side 2 cm is 6 x 2 squared = 24 cm squared, while its volume is 2 cubed = 8 cm cubed.
    Do I need to memorise all the mensuration formulae for AQA GCSE?
    Some formulae are given in the exam, but many must be memorised. You should definitely memorise the area of a circle, circumference of a circle, volume of a prism, volume of a cylinder, volume of a cone, volume of a sphere and surface area of a sphere. Check the AQA formulae sheet for the full list of given formulae, and practise recalling the rest from memory.
    How do I avoid losing marks on unit conversions in mensuration questions?
    Always convert all measurements to the same unit before you start calculating. Remember that when converting areas you square the conversion factor, and when converting volumes you cube it. For example, 1 m = 100 cm, so 1 m squared = 10,000 cm squared and 1 m cubed = 1,000,000 cm cubed. Write down the conversion clearly as part of your working.
    What is the formula for the volume of a cone and how is it different from a cylinder?
    The volume of a cone is one third x pi x r squared x h, where r is the radius of the base and h is the perpendicular height. A cylinder has volume pi x r squared x h, so a cone is exactly one third of a cylinder with the same base and height. This relationship is useful to remember and can help you check your answers.