FG14, HG16 - Volumes of prisms and cylinders | Eduqas GCSE Mathematics

Calculate how much space a solid occupies from its cross-section and length. FG14 and HG16 cover cuboids, other right prisms and cylinders. You will also reverse volume calculations to find a length and connect cubic units to litres and millilitres.

Constant cross-sections

Two solid bars have identical triangular ends, but one bar is twice as long. Does its volume double? Think of the bar as a stack of identical thin slices.

Volume measures three-dimensional space in cubic units. A cuboid $6\,\mathrm{cm}$ wide, $4\,\mathrm{cm}$ high and $9\,\mathrm{cm}$ long has nine layers, each containing $6\times4=24$ unit cubes. The volume is $24\times9=216\,\mathrm{cm}^3$ .

Constant cross-section

A cross-section is the flat shape made by a slice through a solid. In a right prism, slices parallel to its identical end faces have the same polygonal shape and area throughout. Its length is the perpendicular distance between those ends. A right circular cylinder has circular cross-sections and follows the same volume method.

[DIAGRAM: asset_name: Constant cross-sections of prisms; asset_slug: prism-cross-sections; caption: Find the end-face area, then multiply by the perpendicular length between identical ends.; description: Oblique projections of a 6 cm by 4 cm rectangular cross-section and a right-triangular cross-section of base 6 cm and perpendicular height 4 cm. Both right prisms have length 9 cm; end areas 24 square cm and 12 square cm are labelled.]
Find the end-face area, then multiply by the perpendicular length between identical ends.

The slice argument works beyond whole centimetres: area times a small slice thickness gives a small volume; adding the slices gives area times the full perpendicular length.

Volume with a constant cross-section

V=A\ell

Here $A$ is the cross-sectional area and $\ell$ is the perpendicular length between the end faces. The prism rule is supplied on the additional formulae sheet. If $A$ is in square centimetres and $\ell$ in centimetres, $V$ is in cubic centimetres.

The height used to calculate a triangular end's area is different from the prism's length.

Worked example 1: a rectangular cross-section

Question. A right prism has a rectangular end $6\,\mathrm{cm}$ by $4\,\mathrm{cm}$ and length $9\,\mathrm{cm}$ . Find its volume.

Plan. Calculate the area of one end, then multiply by the length.

\begin{aligned} A&=6\times4=24\,\mathrm{cm}^2\\ V&=24\times9=216\,\mathrm{cm}^3 \end{aligned}

Answer. $216\,\mathrm{cm}^3$ .

Check. Group the dimensions differently: $6\times(4\times9)=6\times36=216$ . These are the same unit cubes grouped into different layers.

Keep the base, height and length, but replace the rectangular end by a triangle.

Worked example 2: a triangular cross-section

Question. A right triangular prism has end-face base $6\,\mathrm{cm}$ , perpendicular end-face height $4\,\mathrm{cm}$ and prism length $9\,\mathrm{cm}$ . Find its volume.

Plan. Halve the end's base-height product before multiplying by the prism length.

\begin{aligned} A&=\tfrac12\times6\times4=12\,\mathrm{cm}^2\\ V&=12\times9=108\,\mathrm{cm}^3 \end{aligned}

Answer. $108\,\mathrm{cm}^3$ .

Check. Two such prisms fit together to make the preceding cuboid: $2\times108=216\,\mathrm{cm}^3$ .

A pyramid narrows towards an apex, so its cross-section does not stay constant. Do not use $A\ell$ directly for the whole pyramid.

Watch out

Multiplying a triangular end's three side lengths is not its area. Use $\tfrac12 bh$ with a perpendicular height, then multiply by the prism length.

Use a supplied cross-sectional area.

Prisms and cylinders

Choose the area formula for the cross-section before using the prism rule. A trapezium needs its parallel sides and perpendicular separation; a cylinder needs its circular radius.

For a right cylinder, every circular slice has area $\pi r^2$ . Multiplying by its perpendicular height $h$ gives $V=\pi r^2h$ . This follows from the supplied circle-area and prism-volume rules. It is not the circumference times height: that product measures a surface area.

Worked example 3: a trapezium-ended prism

Question. A right prism has a trapezium cross-section with parallel sides $5\,\mathrm{cm}$ and $9\,\mathrm{cm}$ , separated by $3\,\mathrm{cm}$ . Its length is $12\,\mathrm{cm}$ . Find its volume.

Plan. Use the trapezium formula for one end, then multiply by $12$ .

\begin{aligned} A&=\tfrac12(5+9)\times3=21\,\mathrm{cm}^2\\ V&=21\times12=252\,\mathrm{cm}^3 \end{aligned}

Answer. $252\,\mathrm{cm}^3$ .

Check. Two identical end trapezia make a parallelogram of area $14\times3=42\,\mathrm{cm}^2$ . Twice the volume is $42\times12=504\,\mathrm{cm}^3$ , so one volume is $252\,\mathrm{cm}^3$ .

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FG14, HG16 - Volumes of prisms and cylinders

Constant cross-sections

Prisms and cylinders

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