Mathematics · IGCSE 0580 · §5.1–5.5

Mensuration

Length, area and volume are one idea seen at three scales — and a single factor connects them all.

Mathematics · 0580 Extended Topic 8 of 12
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Congruence and similarity

a b × k ka kb length × k area × k² volume × k³
FIG 5.0 Enlarge a shape by k, area grows by k², volume by k³.

Two shapes are congruent when they match exactly, and similar when one is a scaled copy of the other. Similarity is the more powerful idea, because a single length ratio controls every measurement of the shape.

Definition
Congruent, similar and scale factor
Congruent shapes are identical in both shape and size — one is an exact copy of the other, allowing for rotation or reflection. Similar shapes have the same shape but not necessarily the same size: equal corresponding angles, and corresponding sides in a constant ratio. The scale factor k is the ratio of a length on the larger shape to the matching length on the smaller.

Congruent triangles

Two triangles are congruent if they match under one of four conditions: SSS (three sides), SAS (two sides and the included angle), ASA (two angles and a corresponding side) or RHS (right angle, hypotenuse and one side). Corresponding parts of congruent triangles are equal.

Similar shapes and scale factor

In similar shapes, corresponding angles are equal and corresponding sides share one scale factor k. Enlarging a length by k enlarges an area by k² and a volume by k³, because area is a product of two lengths and volume of three: A₂ = k²A₁, V₂ = k³V₁.

Worked example: two mathematically similar bottles have heights 12 cm and 18 cm. The smaller holds 500 ml. k = 18 ÷ 12 = 1.5. Capacity is a volume, so multiply by k³ = 1.5³ = 3.375: 500 × 3.375 = 1687.5 ml. The cube factor makes the taller bottle hold over three times as much.

Examiner note
To justify similarity, state the reason — e.g. "equal angles" or "sides in the same ratio". A bare answer with no reason forfeits the explanation mark.
Why this matters
Map scales, model figures and enlarged photographs all rely on the fact that area grows far faster than length.

Perimeter and area

Every area formula in this chapter is a variation on "base times height". Learn the four standard shapes, then treat any awkward figure as those shapes added together or cut away.

Definition
Perimeter
The total distance around the boundary of a shape, measured in length units.

The four standard areas

For a rectangle, area = length × width. For a parallelogram, area = base × perpendicular height. For a triangle, area = ½ × base × height. For a trapezium, average the two parallel sides and multiply by the distance between them: area = ½(a + b)h, where a and b are the parallel sides and h is the perpendicular height between them.

a b h
FIG 5.1 The height h is the perpendicular gap between the parallel sides a and b.

Worked example: a garden bed is a trapezium with parallel sides 8 m and 12 m, 5 m apart. Area = ½ × (8 + 12) × 5 = ½ × 20 × 5 = 50 m².

Examiner note
Only the area of a triangle (½bh) is given in the exam. The parallelogram and trapezium formulae must be recalled from memory. Area units square: 1 cm² = 100 mm² and 1 m² = 10 000 cm² — never convert an area with the length factor alone.
Why this matters
Flooring, paint coverage and land measurement are all compound-area problems dressed in context.

Circles, arcs and sectors

A sector is simply a fraction of a whole circle, and the fraction is set by its angle. Once you can find that fraction, arc length and sector area follow from the circumference and area of the full circle.

Definition
Arc and sector
An arc is part of the circumference between two radii — the larger piece is the major arc, the smaller the minor arc. A sector is the "pie slice" region bounded by two radii and the arc between them.

Fraction of the circle

A sector of angle θ occupies θ/360 of the circle. Multiply that fraction by the circumference 2πr for the arc length, and by the area πr² for the sector area: arc = θ/360 × 2πr, sector = θ/360 × πr².

r θ arc
FIG 5.2 The sector angle θ fixes what fraction of the circle the slice is.

Worked example: a sector has radius 10 cm and angle 72°. Fraction of circle = 72 ÷ 360 = 1/5. Arc = 1/5 × 2π × 10 = 4π cm. Sector area = 1/5 × π × 10² = 20π cm².

Examiner note
The perimeter of a sector is the arc plus two radii — a frequently dropped mark; do not confuse it with the arc length alone. If a question says "in terms of π", leave the π symbol in your answer — converting to a decimal can lose the accuracy mark.

Surface area and volume

The exam gives you every solid-volume formula and the curved surface areas — so the marks are won by choosing the right one, keeping the units cubic, and knowing when to add a flat face back on.

Definition
Prism and slant height
A prism is any solid with a uniform cross-section along its length; its volume is cross-section area × length. The slant height (l) of a cone is the distance up its sloping face — not the vertical height h. They are linked by Pythagoras: l² = r² + h².

Volumes you are given

Volume of a prism = Aℓ (cross-section × length); of a cylinder = πr²h; of a pyramid = ⅓Ah; of a cone = ⅓πr²h; of a sphere = 4/3 πr³. The cone and pyramid are exactly one third of the prism that boxes them in.

h r l
FIG 5.3 In a cone, r, h and the slant l form a right-angled triangle.

Worked example: a cone has base radius 6 cm and vertical height 8 cm. V = ⅓ × π × 6² × 8 = ⅓ × 288π = 96π cm³. The slant l = √(6² + 8²) = 10 cm would only be needed for surface area.

Examiner note
Cone and cylinder formulae give the curved surface only. For a total surface area you must add the circular base(s) yourself.
Why this matters
Packaging, tank capacity and the amount of metal in a can are volume and surface-area problems in disguise.

Compound solids and parts

A compound solid is only ever a set of familiar solids joined together. Split it into parts you know, work each one out, then add the volumes — taking care that surface areas leave out any face buried inside.

Definition
Hemisphere and frustum
A hemisphere is half a sphere: volume 2/3 πr³ and curved surface 2πr², each exactly half the sphere’s. A frustum is the solid left when the top of a cone or pyramid is sliced off parallel to the base.

Adding and subtracting parts

To find a volume, add the parts. A frustum is a large cone with a small cone removed, so its volume is the difference of the two. For a hemisphere, halve the sphere’s volume and curved surface, but remember its flat circular face πr² when a total surface is asked for.

h r hemisphere
FIG 5.4 A capsule tank — a cylinder capped by a hemisphere.

Worked example: a tank is a cylinder of radius 5 cm and height 12 cm topped by a hemisphere of radius 5 cm. Cylinder: πr²h = π × 5² × 12 = 300π. Hemisphere: 2/3 πr³ = 2/3 × π × 125 = 250/3 π. Total = 300π + 250/3 π = 1150/3 π ≈ 1204 cm³. The join circle is hidden, so it never enters a surface-area version.

Examiner note
For surface area of a composite solid, never count the hidden join face. Where two parts meet, that circle is inside the solid, not on its surface.
Why this matters
Silos, capsules and lampshades are all one basic solid with a part added or removed.

Exam advice

Common mistakes

Scaling area or volume by k instead of k² or k³
Multiplying a volume by the length factor loses both the method and the answer mark.
Using vertical height where slant height is needed
Curved surface area of a cone needs l, not h; substituting h gives a wrong figure and no marks.
Giving only the curved surface for a "total surface area"
Forgetting the base circle(s) drops the final accuracy mark on cones and cylinders.
Sector perimeter given as the arc alone
The two bounding radii must be added; leaving them out is a routine dropped mark.

Model answer

Two mathematically similar jugs have heights 10 cm and 15 cm. The smaller jug holds 240 ml. Calculate the capacity of the larger jug.
[3 marks]
M1
Find the length scale factor
k = 15 ÷ 10 = 1.5
M1
Cube it for the volume factor
1.5³ = 3.375
A1
Multiply the smaller capacity by the factor
240 × 3.375 = 810 ml

Recall checklist

  • State the four congruence conditions (SSS, SAS, ASA, RHS).
  • Explain how area and volume factors follow from k.
  • Calculate an arc length and a sector area.
  • Write the volume formulae for cone, pyramid and sphere.
  • Apply Pythagoras to find a cone's slant height.
  • Distinguish curved from total surface area.
  • Calculate the volume of a compound solid.
  • Convert between cm² and m², and cm³ and litres.

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