Surface Area and Nets

A net is a solid unfolded

Cut along enough edges of a cardboard box and flatten it. What you get is a net: every face lying flat, still joined, ready to fold back up.

That is the whole idea behind surface area. Once a solid is unfolded, every face is a flat shape whose area you already know how to find, so the total is an addition.

$$\text{surface area} = \text{sum of the areas of all the faces}$$

Nothing new is being learned here. It is rectangles, triangles and circles added together.

Surface area against volume

	Measures	Units
surface area	the outside	square units
volume	the inside	cubic units

Surface area is how much wrapping paper a box needs. Volume is how much fits inside it. Two solids can match on one and differ on the other.

A rectangular prism

A box unfolds into six rectangles in three matching pairs:

Pair	Each area	Both
top and bottom	$lw$	$2lw$
front and back	$lh$	$2lh$
the two ends	$wh$	$2wh$

$$SA = 2lw + 2lh + 2wh$$

A box $5$ by $3$ by $2$:

$$2(15) + 2(10) + 2(6) = 30 + 20 + 12 = 62 \text{ square units}$$

The pairing is worth using as a check. Opposite faces of a box are identical, so every area should appear twice.

A cube

A cube is the case where all three measurements match, so all six faces are the same square:

$$SA = 6s^2$$

A cube of side $4$: $6(16) = 96$ square units.

A cylinder

Unfold a tin can and three pieces fall out: two circular ends and one rectangle that was the curved side.

$$SA = \underbrace{2\pi r^2}_{\text{two circles}} + \underbrace{2\pi r h}_{\text{the side}}$$

The rectangle is the part worth thinking about. Its height is the cylinder’s height $h$. Its width is the circumference, $2\pi r$, because that edge used to wrap the whole way round the circular end.

$$\text{side area} = \underbrace{2\pi r}_{\text{width}} \times \underbrace{h}_{\text{height}}$$

So the label on a tin, laid flat, is a rectangle as wide as the tin’s circumference. A cylinder of radius $3$ and height $10$:

$$2\pi(9) + 2\pi(3)(10) = 18\pi + 60\pi = 78\pi \approx 245$$

A square pyramid

A square pyramid unfolds into a square base and four triangles:

$$SA = s^2 + 4\left(\tfrac{1}{2} s \ell\right) = s^2 + 2s\ell$$

The $\ell$ is the slant height — the distance up the sloping face, not the vertical height of the pyramid. Using the vertical height gives an answer that is too small, because the slope is longer than the drop.

The two are related by Pythagoras: the slant height is the hypotenuse of a right triangle whose legs are the vertical height and half the base.

Why nets are worth drawing

Sketching the net first removes the two errors that account for most lost marks.

A missing face. Adding areas from a picture of a solid means adding faces you cannot see. A net shows all of them at once, so a box gives six rectangles and nothing is quietly left out.

The wrong measurement. On a net it is visible which length belongs to which face — that the cylinder’s rectangle is as wide as the circumference, that a pyramid’s triangles use the slant height. In a three-dimensional drawing those lengths are foreshortened and readily mixed up.

It also settles questions a formula cannot. An open-topped box has five faces rather than six, and no memorized formula will tell you that. The net will.

Worked examples

Common mistakes

Practice problems