Inside and outside
Volume measures how much fits inside a solid, in cubic units. Surface area measures the outside: it is the total area of all the faces of the solid, in square units. It is the amount of paper needed to wrap the solid with no overlaps.
A cuboid has 6 rectangular faces. Take one whose edges are 4 cm, 3 cm and 2 cm long.
A cuboid whose edges are 4 cm, 3 cm and 2 cm long.
Unfold it flat
Cut along some of the edges and unfold the cuboid, so that all 6 faces lie flat in one piece. That flat shape is the cuboid’s net. Unfolding moves each face without stretching or shrinking it, so each face has the same area in the net as on the solid. The surface area of the cuboid is the area of its net.
In the net every face is a rectangle whose sides are edges of the cuboid, so every face can be measured.
The net of the 4 by 3 by 2 cuboid: six rectangles in one piece.
Three pairs of faces
Opposite faces of a cuboid are identical, so the 6 faces come in 3 matching pairs. Each pair uses two of the three edge lengths: one pair is 4 by 3, one is 4 by 2 and one is 3 by 2.
So only three areas need working out: 4 × 3 = 12 cm², 4 × 2 = 8 cm² and 3 × 2 = 6 cm².
The gold face is 4 by 3, and its partner is the other 4 by 3 rectangle, at the far right. The two 4 by 2 faces are above and below the gold one, and the two 3 by 2 faces are either side of it.
Add them up, twice
One face of each pair gives 12 + 8 + 6 = 26 cm². Every face has a partner of the same size on the opposite side, so the surface area is 2 × 26 = 52 cm².
For a cuboid of length l, width w and height h, the three areas are lw, lh and wh, so its surface area is 2(lw + lh + wh). A cube with edges of length s has 6 faces of , so its surface area is .
Count the faces as a check: the working used 3 areas, each twice, which is 6 faces.
An open box
A box with no lid has only 5 faces: leave the lid out of the total. An open box with the same edges, whose missing lid would be a 4 by 3 face, has a surface area of 52 − 12 = 40 cm².
Two slips
Stopping at 12 + 8 + 6 = 26 cm². That covers only three faces, one from each pair. Double it for the other three.
Multiplying the three edges, 4 × 3 × 2 = 24. That is the volume, in cm³, the number of 1 cm cubes that fit inside. The surface area adds the areas of the faces.
Worked example: The Net of an Open Box on Grid Paper
Question An open box has no lid. Its external length is 14 cm, its breadth 10 cm and its height 8 cm, and its walls are so thin that their thickness can be ignored. A student draws its net on a large sheet of grid paper whose squares are 2 cm by 2 cm, with every edge of the net along the grid lines. (a) How many grid squares does the net cover? (b) The box is then filled with cubes of edge 2 cm. How many cubes fit inside it?
1.The net has five faces: the base, two long walls and two short walls. There is no lid.
An open box: a base and four walls, no lid. 2.On the grid, 14 cm is 14 ÷ 2 = 7 squares, 10 cm is 10 ÷ 2 = 5 squares and 8 cm is 8 ÷ 2 = 4 squares.
14 cm is 7 squares, 10 cm is 5 and 8 cm is 4. 3.The base is 7 × 5 = 35 squares. Each long wall is 7 × 4 = 28 squares and each short wall 5 × 4 = 20 squares.
Base 7 × 5 = 35; long walls 28 each; short walls 20 each. 4.(a) The net covers 35 + 28 + 28 + 20 + 20 = 131 squares. Check by area: 140 + 2 × 112 + 2 × 80 = 524 cm2, and one square is 2 × 2 = 4 cm2, so 524 ÷ 4 = 131.
(a) 35 + 28 + 28 + 20 + 20 = 131 squares, and 524 ÷ 4 = 131. 5.(b) The walls are too thin to take up room, so the cubes fill the box in rows: 7 along the length, 5 across the breadth and 4 layers up the height. That is 7 × 5 × 4 = 140 cubes.
(b) 7 × 5 × 4 = 140 cubes fill the box.
Answer: (a) 131 (b) 140
Common mistakes
- Counting six faces, as for a closed box: the lid would add another 35 squares and give 166.
- Dividing the area by 2, the side of a grid square, instead of by 4, its area: 524 ÷ 2 = 262 counts each square twice.
The surface of a stack of cubes
When cubes are stacked, a face pressed against another cube, or against the table, is not part of the surface. Take ten cubes with 2 cm edges, standing on a table in two rows of three columns: the columns in the back row are 2, 3 and 2 cubes high, and those in the front row are 1 cube high each.
To count its exposed faces, look at the stack from each of the six directions: above, below, front, back, left and right. Every exposed face points in one of those directions. The columns stand side by side with no gaps between them, so each exposed face is seen, once, from the direction it faces.
Worked example: Exposed Faces of a Stack
Question The same stack, with columns 2, 3, 2 at the back and 1, 1, 1 at the front, stands on a table. How many faces of the small cubes are exposed to the air? What is the exposed area, the cubes being 2 cm?
1.From above: one top face per column, 6.
From above: one top per column, 6. 2.From the front and from the back: the tallest column at each of the three positions, 2 + 3 + 2 = 7 each way.
Front and back each see the tallest of each file: 2 + 3 + 2 = 7. 3.From the left and from the right: the tallest column in each row, 1 + 3 = 4 each way.
Left and right each see the tallest of each row: 1 + 3 = 4. 4.The bottom faces rest on the table and are not exposed.
The bottoms rest on the table. 5.Exposed faces = 6 + 7 + 7 + 4 + 4 = 28; each face is 2 × 2 = 4 cm2, so 112 cm2.
6 + 7 + 7 + 4 + 4 = 28 faces; 28 × 4 = 112 cm².
Answer: 28 faces; 112 cm2
Common mistakes
- Multiplying 10 cubes by 6 faces: most faces are pressed against another cube or the table.
- Counting the front view once and forgetting the back sees the same silhouette from behind.