Volume & Surface Area of Prisms - Standard Maths

What makes a prism

A prism is a solid with a constant cross-section: cut it anywhere across its length and you get the same shape. A triangular prism, a hexagonal nut and a plain cuboid are all prisms. The repeated shape is the cross-section, and the distance it runs is the length.

Cross-section: The flat shape you see when you slice straight through a solid. For a prism it stays identical along the whole length.

Volume of a prism

Think of the prism as many thin copies of the cross-section stacked along its length. That gives one clean rule:

\[ V = \text{area of cross-section} \times \text{length} \]

So the whole job splits in two: find the area of the end face, then multiply by how long the prism is. This is exactly why a cuboid works out as \(l \times w \times h\): its cross-section is a rectangle of area \(l \times w\).

Find the end face first

Always identify the cross-section before touching the length. The length is simply a multiplier once the area of that end shape is known, so getting the end face right is where the marks live.

Surface area from the net

The surface area is the total area of every face. The safest way to see them all is to imagine unfolding the prism into its net: a flat pattern that, when folded, wraps the solid exactly. For any prism the net is the two identical end faces plus a set of rectangles, one for each side of the cross-section.

A useful shortcut for the rectangles: together they form one long band whose height is the length of the prism and whose width is the perimeter of the cross-section. So:

\[ \text{Surface area} = 2 \times (\text{cross-section area}) + (\text{cross-section perimeter}) \times \text{length} \]

A full worked example

Worked example

A triangular prism has a right-angled triangle cross-section with base 6 cm and height 8 cm. The prism is 10 cm long. Find its volume and its surface area.

Area of the triangular end: \(\tfrac12 \times 6 \times 8 = 24\) cm\(^2\).

Volume: \(V = 24 \times 10 = 240\) cm\(^3\).

The sloping side of the triangle (hypotenuse): \(\sqrt{6^2 + 8^2} = \sqrt{100} = 10\) cm. So the three sides are 6, 8 and 10 cm.

Two triangular ends: \(2 \times 24 = 48\) cm\(^2\). Three rectangles: perimeter \(\times\) length \(= (6 + 8 + 10) \times 10 = 24 \times 10 = 240\) cm\(^2\).

Total surface area: \(48 + 240 = 288\) cm\(^2\).

Volume \(= 240\) cm\(^3\), surface area \(= 288\) cm\(^2\)

Where this is assessed

Choosing the cross-section and applying the rule is Criterion A (Knowing and understanding). Setting out the net and each face with correct units earns Criterion C (Communicating). Sizing a real object such as a ramp or a chocolate bar box is Criterion D (Applying maths in context). Keep volume in cm\(^3\) and surface area in cm\(^2\).

Check yourself

1. A prism has a cross-sectional area of 12 cm\(^2\) and a length of 7 cm. Find its volume. +

\(V = \text{area of cross-section} \times \text{length} = 12 \times 7 = \mathbf{84}\) cm\(^3\).

2. A triangular prism has a triangle cross-section of base 5 cm and height 4 cm, and is 9 cm long. Find its volume. +

Triangle area \(= \tfrac12 \times 5 \times 4 = 10\) cm\(^2\). Volume \(= 10 \times 9 = \mathbf{90}\) cm\(^3\).

3. Find the surface area of a cuboid measuring 3 cm by 4 cm by 5 cm. +

A cuboid has three pairs of faces: \(3\times4\), \(4\times5\) and \(3\times5\). Surface area \(= 2(3\times4 + 4\times5 + 3\times5) = 2(12 + 20 + 15) = 2 \times 47 = \mathbf{94}\) cm\(^2\).

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