Surface Area of Prisms
Surface Area of Prisms in NSW Year 10 Core maths finds the total area of all the faces of a prism (NESA outcome MA5‑ARE‑C‑01). Picturing the net helps: a rectangular prism has three pairs of matching rectangles, and a triangular prism has two triangular ends and three rectangular sides.
You will find the surface areas of cubes and of rectangular, triangular and trapezoidal prisms, use Pythagoras' theorem to find a missing side first, and work backwards from a surface area to a missing height. You will also solve practical problems: cardboard for a cereal box, paint for a garden shed, tiles for a swimming pool, and the surface area of a cube with a square hole cut through it. It is part of Year 10 Core Measurement.
Theory
The surface area of a prism is the total area of all its faces, the area of its net. This Year 10 maths page covers cubes, rectangular and triangular prisms, trapezium prisms, open boxes, practical painting and tiling problems, and working backwards to a missing length.
A right prism is a solid with two identical, parallel ends joined by rectangles. The shape of the end (the cross-section) names the prism: rectangular, triangular, trapezium and so on.
The surface area (SA) is the total area of all the faces. If you unfold the prism flat you get its net; the surface area is the area of the net.
A rectangular prism has three pairs of matching faces. A triangular prism has two triangular ends and three rectangles. The rectangles around any prism fit together into one long rectangle whose length is the perimeter of the end and whose width is the length of the prism.
In real problems some faces are missing or hidden: a pool or planter has no top, a shed floor is not painted, and where two solids are glued together the touching faces are no longer on the surface.
| Prism | Faces | Surface area |
|---|---|---|
| Cube, edge \(s\) | 6 equal squares | \(6s^{2}\) |
| Rectangular prism | 3 pairs of rectangles | \(2(lw + lh + wh)\) |
| Any right prism | 2 ends + rectangles | \(2 \times\) end area \(+\) perimeter of end \(\times\) length |
How to find the surface area of a prism
- Sketch the prism or its net and list every face. Check which faces are really there (a lid? a floor?).
- Find any missing side lengths, using Pythagoras' theorem if needed. Put all lengths in the same unit.
- Calculate each face area. For matching faces, find one and multiply.
- Add the face areas and give square units.
Problems
- Paint, tiles or material: find the area that is covered, then multiply by the price per square metre.
- Working backwards: write the surface area with a pronumeral for the unknown length, set it equal to the given area and solve.
- Hole through a solid: total \(=\) outside surface \(-\) the two openings \(+\) the inside walls.
- Stacked or glued solids: add both surface areas, then subtract the touching area twice (once from each solid).
There are three pairs of matching faces. Find one face of each pair, add, then double.
| \(\text{top}\) | \(=\) | \(9 \times 5 = 45\) |
| \(\text{front}\) | \(=\) | \(9 \times 3 = 27\) |
| \(\text{end}\) | \(=\) | \(5 \times 3 = 15\) |
| \(SA\) | \(=\) | \(2(45 + 27 + 15)\) |
| \(\) | \(=\) | \(2 \times 87 = \mathbf{174}\text{ cm}^2\) |
The rectangles need all three sides of the triangle, so find the sloping side \(x\) first.
| \(x^{2}\) | \(=\) | \(9^{2} + 12^{2} = 225\) |
| \(x\) | \(=\) | \(15\text{ cm}\) |
| \(\text{two ends}\) | \(=\) | \(2 \times \dfrac{1}{2} \times 12 \times 9 = 108\) |
| \(\text{rectangles}\) | \(=\) | \((9 + 12 + 15) \times 20 = 720\) |
| \(SA\) | \(=\) | \(108 + 720 = \mathbf{828}\text{ cm}^2\) |
Change every length to metres: \(50\text{ cm} = 0.5\text{ m}\) and \(40\text{ cm} = 0.4\text{ m}\). There is no top, so count the floor once.
| \(\text{floor}\) | \(=\) | \(1.2 \times 0.5 = 0.6\) |
| \(\text{long sides}\) | \(=\) | \(2 \times 1.2 \times 0.4 = 0.96\) |
| \(\text{short sides}\) | \(=\) | \(2 \times 0.5 \times 0.4 = 0.4\) |
| \(A\) | \(=\) | \(0.6 + 0.96 + 0.4\) |
| \(\) | \(=\) | \(\mathbf{1.96}\text{ m}^2\) |
To cost the timber, multiply by the price per square metre. At \(\$35\) per square metre: \(1.96 \times 35 = \$68.60\).
Write the surface area in terms of \(h\), then solve. There are two square faces and four equal rectangles.
| \(2 \times 6^{2} + 4 \times 6h\) | \(=\) | \(240\) |
| \(72 + 24h\) | \(=\) | \(240\) |
| \(24h\) | \(=\) | \(168\) |
| \(h\) | \(=\) | \(\mathbf{7}\text{ cm}\) |
Check: \(72 + 4 \times 6 \times 7 = 72 + 168 = 240\).
Common pitfalls
Frequently asked questions
What is the surface area of a prism?
It is the total area of all the faces of the prism, the same as the area of its net. It is measured in square units.
How do you find the surface area of a rectangular prism?
Find the area of the top, the front and the side, add them, then double: \(SA = 2(lw + lh + wh)\).
How do you find the surface area of a triangular prism?
Add the areas of the two triangular ends to the three rectangles. The rectangles together equal the perimeter of the triangle times the length of the prism.
What is the difference between surface area and volume?
Surface area measures the outside of a solid in square units, like the amount of wrapping paper. Volume measures the space inside in cubic units, like the amount of water it holds.
How do you find the surface area of an open box?
Add the areas of the base and the four sides, but leave out the top. That is the closed surface area minus the area of the missing lid.