Surface Area of Prisms
The surface area of a prism is the total area of all its faces: two matching ends plus the rectangles around the sides. Drawing a net helps to check that no face is missed. In NSW Year 10 Advanced maths the ends include triangles, trapeziums, rhombuses and hexagons, and Pythagoras' theorem is often needed to find a missing side first (NESA outcome MA5‑ARE‑C‑01).
You will find surface areas of rectangular and triangular prisms, open and closed, and of prisms with trapezium, rhombus, hexagon and cross-shaped ends. Practical problems include wrapping a gift, a Perspex display cover, a water trough, a loading ramp in plywood sheets and painting a room around a door and window. Harder questions compare two boxes with the same volume and form a quadratic equation for the size of a box. It is part of Year 10 Advanced Measurement.
Theory
The surface area of a prism is the total area of all its faces: two equal ends plus the rectangular sides. The sides together have area perimeter of the cross-section \(\times\) length. This Year 10 Advanced maths page covers triangular and trapezium prisms with Pythagoras' theorem, open containers, painting and costing problems, and equations for an unknown side.
A prism is a solid with two identical, parallel ends (the cross-section) joined by rectangular faces. A right prism's side faces are at right angles to its ends.
The surface area (SA) is the total area of all the faces, measured in square units such as \(\text{cm}^2\) or \(\text{m}^2\). A net, the solid unfolded flat, shows every face so none is missed.
An open prism, such as a trough or a box without a lid, leaves out the missing faces. In real problems, read carefully which faces are covered, painted or made.
Often one side of the cross-section is not given. Find it with Pythagoras' theorem, by drawing a height in an isosceles triangle or a trapezium.
Rectangular prism (\(\ell \times w \times h\)) and cube (edge \(a\)):
| Cross-section | Area |
|---|---|
| triangle | \(\dfrac{1}{2}bh\) |
| trapezium | \(\dfrac{1}{2}(a + b)h\) |
| rhombus or kite | \(\dfrac{1}{2}xy\) (diagonals) |
| equilateral triangle, side \(a\) | \(\dfrac{\sqrt{3}}{4}a^2\) |
| regular hexagon, side \(a\) | \(6 \times \dfrac{\sqrt{3}}{4}a^2\) |
How to find the surface area of a prism
- Identify the cross-section and the length. Sketch a net if the shape is unusual.
- Find any missing side of the cross-section with Pythagoras' theorem (Examples 1 and 2).
- Add the two ends and the side faces (perimeter \(\times\) length). Leave out any open or uncovered faces.
- State the answer with square units, exact or rounded as asked.
Problems built on surface area
- Open or partly covered: list the faces that count, e.g. a display cover with no base, a trough with no top, a ramp with no bottom.
- Composite cross-sections: for an L-shape or a plus sign, the perimeter is the total length of all outside edges; the area is the sum of the parts.
- Painting and covering: subtract doors and windows, multiply by the number of coats, divide by the coverage, then round up to whole tins or sheets (Example 3). Add a waste allowance first if one is given, e.g. \(\times 1.1\) for \(10\%\).
- Comparing designs: boxes with the same volume can need different amounts of material; work out both surface areas.
- Folding a net: an open box made by cutting squares from the corners of a sheet has surface area equal to the sheet minus the squares.
- Cutting a solid: cutting creates new faces, so the total surface area increases.
- Unknown dimension: write the surface area in terms of the pronumeral, set it equal to the given value and solve; reject negative lengths (Example 4).
The third rectangle needs the hypotenuse of the end, so find it first.
| \(h^2\) | \(=\) | \(9^2 + 12^2 = 81 + 144 = 225\) |
| \(h\) | \(=\) | \(15\text{ cm}\) |
Two triangular ends, plus the side faces as perimeter \(\times\) length:
| \(\text{Ends}\) | \(=\) | \(2 \times \dfrac{1}{2} \times 9 \times 12 = 108\) |
| \(\text{Sides}\) | \(=\) | \(20 \times (9 + 12 + 15) = 720\) |
| \(\text{SA}\) | \(=\) | \(108 + 720 = \mathbf{828}\text{ cm}^2\) |
Drop a perpendicular from a top corner. The extra \(12\) cm of the bottom is shared equally, \(6\) cm each side.
| \(s^2\) | \(=\) | \(6^2 + 8^2 = 100\) |
| \(s\) | \(=\) | \(10\text{ cm}\) |
| \(\text{Ends}\) | \(=\) | \(2 \times \dfrac{1}{2}(10 + 22) \times 8 = 256\) |
| \(\text{Perimeter}\) | \(=\) | \(10 + 22 + 10 + 10 = 52\) |
| \(\text{Sides}\) | \(=\) | \(52 \times 30 = 1560\) |
| \(\text{SA}\) | \(=\) | \(256 + 1560 = \mathbf{1816}\text{ cm}^2\) |
(i) the area to paint
(ii) the number of tins and the cost.
(i) The floor is not painted. The four walls together are perimeter \(\times\) height.
| \(\text{Walls}\) | \(=\) | \(2(3 + 2.4) \times 2.1 = 22.68\) |
| \(\text{Roof}\) | \(=\) | \(3 \times 2.4 = 7.2\) |
| \(\text{Door}\) | \(=\) | \(1.8 \times 0.8 = 1.44\) |
| \(A\) | \(=\) | \(22.68 + 7.2 - 1.44 = \mathbf{28.44}\text{ m}^2\) |
(ii) Two coats double the area.
| \(\text{Paint}\) | \(=\) | \(\dfrac{2 \times 28.44}{12} = 4.74\text{ L}\) |
| \(\text{Tins}\) | \(=\) | \(4.74 \div 2 = 2.370 \to \mathbf{3}\) |
| \(\text{Cost}\) | \(=\) | \(3 \times \$38.50 = \mathbf{\$115.50}\) |
Round the number of tins up: \(2\) tins would not be enough.
Two square faces and four rectangles, each \(x\) by \(6\).
| \(2x^2 + 4 \times 6x\) | \(=\) | \(128\) |
| \(2x^2 + 24x - 128\) | \(=\) | \(0\) |
| \(x^2 + 12x - 64\) | \(=\) | \(0\) |
| \((x + 16)(x - 4)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(4 \text{ or } x = -16\) |
A length must be positive, so \(\mathbf{x = 4}\): the box is \(4\) cm by \(4\) cm by \(6\) cm. Check: \(2(16) + 4(24) = 32 + 96 = 128\ \checkmark\)
Common pitfalls
Frequently asked questions
How do you find the surface area of a prism?
Add the areas of all its faces. For any prism, \(\text{SA} = 2 \times\) area of cross-section \(+\) perimeter of cross-section \(\times\) length.
How do you find the surface area of a triangular prism?
Find the area of the triangle and double it for the two ends. Then add the three rectangles, which together are the perimeter of the triangle times the length. Use Pythagoras' theorem for any missing side of the triangle.
What is the difference between surface area and volume?
Surface area is the total area of the outside faces, in square units, such as the amount of wrapping paper. Volume is the space inside, in cubic units, such as how much the box holds.
How do you find the surface area of an open box?
Add only the faces that are there. A box with no lid has a base and four sides, so leave out the top.
How much paint do I need for a room?
Find the area of the walls and ceiling, subtract doors and windows, multiply by the number of coats, and divide by the area one litre covers. Round up to whole tins.