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Math 6 · 6.G.A.4 · 7.G.B.6

Surface area, and how it compares with volume

How much paper covers a present exactly once, and why can two boxes that hold the same amount need different amounts of paper?

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Surface area, and how it compares with volume
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When you reach for it

Wrapping a present, painting a jewelry box, cutting card for a box: each asks how much flat material covers the outside of a solid exactly once. That amount is the solid’s surface area. Filling the same box with sand asks a different question, about its volume.

The lesson takes two days. Part 1 (Examples 1 to 4) finds surface area by laying a solid flat as a net and adding the areas of the pieces. Part 2 (Examples 5 to 7) sets surface area beside volume: what each measures, why one can change while the other stays the same, and how much surface a solid has for each unit of space inside it.

The procedure

Part 1: adding up the faces

What surface area is. The surface area of a solid with flat faces is the total area of all its faces. It counts the unit squares needed to cover the outside once, with no gap and no overlap, so it is measured in square units: \text{cm}^2, \text{m}^2, \text{in}^2.

Why a net helps. Unfolding a solid into a net bends the sheet at the folds but does not stretch or shrink any face, so each piece has the same area as the face it came from. The area of the net is therefore the surface area, and every net of the same solid gives the same answer. The net also shows every face at once, so none is missed and none is counted twice.

The area of each piece. A rectangle is length times width, a triangle is half of base times height (Area of triangles), and a trapezoid or parallelogram is found as in Area of parallelograms and trapezoids. A length that is not given can be read from the edge it meets on folding, as in Nets of solids.

The faces of the usual solids.

  • A box (a rectangular prism) l long, w wide and h tall has six rectangles in three matching pairs, because opposite faces are the same size: two of l \times w, two of l \times h and two of w \times h. A cube has six equal squares.
  • A triangular prism has two triangles and three rectangles, one for each side of the triangle, each as long as the prism.
  • A pyramid has its base and one triangle for each side of the base. Each triangle stands on a side of the base, and its height runs up the middle of the sloping face to the top point. That length is the slant height. It lies along the face, so it is longer than the height of the pyramid, which runs straight down inside it. Here the slant height is always given.

Which faces count: the object decides. A box with no lid has five faces, not six. A glass case set over an object has no bottom. List the faces really covered before adding. Every length must be in the same unit before multiplying; centimeters give square centimeters.

Part 2: covering and filling

Two questions about one solid. Surface area counts unit squares covering the outside. Volume counts unit cubes filling the inside; for a box it is length times width times height, as in Volume of a box, with fraction edges as in Volume of a box with fraction edge lengths. Wrapping, painting, lining and the material a box is made of are surface area questions; how much a box holds is a volume question. The answers are in different units, \text{cm}^2 and \text{cm}^3, so they are never added or compared as numbers, just as a perimeter and an area are not (Perimeter versus area).

One can change while the other does not. The same unit cubes packed into a different box keep their volume, but the surface usually changes: a long thin box has more outside than a compact one. And two boxes made from the same amount of card can hold different amounts. Examples 5 and 6 show both.

Surface for each unit of volume. To compare how much outside solids have for their size, divide surface area by volume. The result is a rate (Rate and unit rate): square units of surface for each cubic unit of space. For cubes:

Edge (cm)Surface (\text{cm}^2)Volume (\text{cm}^3)Per \text{cm}^3
1616
22483
354272
496641.5
51501251.2
edge 1 edge 2 edge 3
Figure 1Cubes of edge 1, 2 and 3. Each face of the cube with edge s holds s \times s unit squares.

The rate falls as the cube grows. A cube with edge s has six faces of s \times s squares and holds s \times s \times s cubes, so

\frac{6 \times s \times s}{s \times s \times s} = \frac{6}{s}

since the two factors of s shared by numerator and denominator divide out, as in simplifying a fraction. Doubling the edge gives four times the surface but eight times the volume, so each cubic centimeter gets half as much surface. Big cubes have little surface for each unit of volume; small pieces have a lot. So anything that happens through a surface, such as ice melting into a drink, goes faster for many small pieces than for one big piece of the same total size.

Worked examples

Example 1

Wrapping a box-shaped present

Jorun wraps a present in a box 24 cm long, 15 cm wide and 9 cm tall. What is the least area of gift wrap that covers it?

front end back end top bottom 24 cm 15 cm 9 cm
Figure 2The net of the present box in Example 1.
Solution

The paper must cover all six faces, so the least area is the surface area. The net shows the faces in three matching pairs, because opposite faces of a box are the same size:

  • top and bottom: 24 \times 15 = 360\ \text{cm}^2 each;
  • front and back: 24 \times 9 = 216\ \text{cm}^2 each;
  • the two ends: 15 \times 9 = 135\ \text{cm}^2 each.

Adding both faces of each pair,

\begin{aligned} &2 \times 360 + 2 \times 216 + 2 \times 135 \\ &= 720 + 432 + 270 = 1{,}422\ \text{cm}^2 \end{aligned}

Real wrapping overlaps at a seam, so Jorun cuts a larger sheet; 1{,}422\ \text{cm}^2 is the least that could do it.

Example 2

A present with triangular ends

Kazimir’s present is in a long box whose ends are triangles with sides 10 cm, 10 cm and 12 cm. Each triangle is 8 cm high, measured from its 12 cm side, and the box is 30 cm long. How much gift wrap covers it exactly once?

10 cm 10 cm 12 cm 30 cm
Figure 3The net of the box in Example 2. Each triangle is 8 cm high, measured from its 12 cm side.
Solution

The net has two triangles and three rectangles 30 cm long, one rectangle for each side of the triangle.

A triangle’s area needs a base and the height that meets it at a right angle, so use the 12 cm side with the 8 cm height: \displaystyle \frac{1}{2} \times 12 \times 8 = 48\ \text{cm}^2. The two ends make 96\ \text{cm}^2. The 10 cm sides give the widths of two rectangles:

\begin{aligned} &10 \times 30 + 12 \times 30 + 10 \times 30 \\ &= 300 + 360 + 300 = 960\ \text{cm}^2 \end{aligned}

The surface area is 96 + 960 = 1{,}056\ \text{cm}^2. A check: in the net the three rectangles make one strip 10 + 12 + 10 = 32 cm wide and 30 cm long, and 32 \times 30 = 960 again.

Example 3

A pyramid-shaped present box

Liesl makes a present box shaped like a square pyramid. Its base is a square with sides of 9 cm, and each triangular face has slant height 12 cm. How much card does she need, with nothing extra for glue?

base 9 cm 12 cm
Figure 4The net of the pyramid box in Example 3. The dashed line is the slant height.
Solution

The net is the square with a triangle on each side. The square has area 9 \times 9 = 81\ \text{cm}^2. Each triangle stands on a 9 cm side, and its height is the slant height, 12 cm, measured on the face itself:

\frac{1}{2} \times 9 \times 12 = 54\ \text{cm}^2

Four triangles make 4 \times 54 = 216\ \text{cm}^2, and the card is 81 + 216 = 297\ \text{cm}^2. The finished box stands less than 12 cm tall, since its faces lean inward, but the height of the solid plays no part here.

Example 4

Lining a tray with no lid

Mikael lines the inside of the lift-out tray of a jewelry box with velvet, the bottom and the four walls. The tray is 20 cm long, 12 cm wide and 3.5 cm deep. Ignoring the thickness of the wood, how much velvet does he need?

bottom 20 cm 12 cm 3.5 cm
Figure 5The tray in Example 4 has five pieces, because it has no lid.
Solution

The tray has no lid, so its net has five pieces: the bottom with a wall on each side.

  • bottom: 20 \times 12 = 240\ \text{cm}^2;
  • the two long walls: 20 \times 3.5 = 70\ \text{cm}^2 each, 140\ \text{cm}^2 together;
  • the two short walls: 12 \times 3.5 = 42\ \text{cm}^2 each, 84\ \text{cm}^2 together.

The velvet is 240 + 140 + 84 = 464\ \text{cm}^2. A closed box of the same size would add a lid of another 240\ \text{cm}^2; only the object says whether the lid is there.

Example 5

Same stock cubes, different boxes

Ottar packs 12 stock cubes, each a cube with 2 cm edges, into a box that they fill exactly. Which boxes are possible, and which takes the least card?

Solution

The numbers of cubes along the three edges must multiply to 12: 1 \times 1 \times 12, 1 \times 2 \times 6, 1 \times 3 \times 4 or 2 \times 2 \times 3. Turning a box over gives no new box. Each cube is 2 cm across, so every length doubles. The table lists one face from each matching pair, and the surface area is twice their sum:

Box (cm)VolumeOne of each pairSurface
2 \times 2 \times 24964, 48, 48200
2 \times 4 \times 12968, 24, 48160
2 \times 6 \times 89612, 16, 48152
4 \times 4 \times 69616, 24, 24128

Volumes are in \text{cm}^3 and areas in \text{cm}^2. For the last row, 2 \times (16 + 24 + 24) = 2 \times 64 = 128. Every box holds the same twelve cubes, 96\ \text{cm}^3, yet the card runs from 200 down to 128\ \text{cm}^2. The 4 \times 4 \times 6 box, the closest to a cube, takes the least: packed tightly, more faces of the cubes press against other cubes, and those faces need no card.

Example 6

Same card, different boxes

Liesl makes two gift boxes. One is 5 cm by 5 cm by 25 cm, the other 5 cm by 10 cm by 15 cm. Show that both use 550\ \text{cm}^2 of card, and find which holds more.

Solution

The first box has faces of 25, 125 and 125\ \text{cm}^2, one of each pair, so its surface area is 2 \times (25 + 125 + 125) = 2 \times 275 = 550\ \text{cm}^2. The second has 5 \times 10 = 50, 5 \times 15 = 75 and 10 \times 15 = 150, so 2 \times (50 + 75 + 150) = 2 \times 275 = 550\ \text{cm}^2 as well.

The volumes are 5 \times 5 \times 25 = 625\ \text{cm}^3 and 5 \times 10 \times 15 = 750\ \text{cm}^3. The second box holds 750 - 625 = 125\ \text{cm}^3 more from exactly the same card. How much covers a box does not tell you how much it holds.

Example 7

Crushed ice

Mikael has an ice cube with edges of 6 cm. He cuts it into smaller cubes with edges of 3 cm, or 2 cm, or 1 cm. In each case find the total surface of the ice and the surface for each \text{cm}^3. Why does crushed ice cool a drink faster than a whole cube?

Solution

The whole cube holds 6 \times 6 \times 6 = 216\ \text{cm}^3 and has six faces of 36\ \text{cm}^2, a surface of 216\ \text{cm}^2: 1\ \text{cm}^2 for each \text{cm}^3. The two numbers match only by chance; one counts squares, the other cubes.

Cutting changes the surface, never the amount of ice, so the volume stays 216\ \text{cm}^3. With 3 cm cubes there are 6 \div 3 = 2 along each edge, so 2 \times 2 \times 2 = 8 pieces. In the table, areas are in \text{cm}^2:

Small edgePiecesOne pieceTotalPer \text{cm}^3
612162161
38544322
227246483
121661{,}2966

For example, 27 \times 24 = 648 and 648 \div 216 = 3. Each cut opens two new faces where there was ice inside before, and the last column is 6 divided by the small edge, as the rate \displaystyle \frac{6}{s} predicts.

Ice melts where it touches the warmer drink, at its surface. In 1 cm pieces the same ice has six times as much surface touching the drink, so warmth passes from the drink into the ice much faster.

Boundaries

Adding faces works for solids whose faces are flat polygons. A solid with a curved side, such as a tin, needs the distance round a circle, and comes later.

Each triangle of a pyramid needs the slant height. The height of the pyramid itself is a different length, and turning one into the other needs a later tool.

Surface area has no thickness. Card is thick, wrapping overlaps, and the inside of a thick-walled box is smaller than its outside, so real material needs a little more. Where two solids are joined, the faces pressed together are no longer on the surface.

Surface for each unit of volume depends on the unit (Exercise 25), so compare solids with this rate only in one unit.

Where this goes next

In Surface area of prisms and cylinders the strip of rectangles in a prism’s net becomes one formula for any prism, and the curved side of a cylinder unrolls into one more rectangle. Surface area of pyramids and cones does the same for pyramids and finds the slant height when only the height is known. The cube table returns in Scale factor, area and volume, which explains why making every length k times longer makes area k \times k times and volume k \times k \times k times as large.

Exercises

Part 1: surface area from a net

  1. A box’s net has a strip of four rectangles, each 6 cm tall, with widths 11, 5, 11 and 5 cm, and a lid 11 cm by 5 cm above and below the first rectangle. Find the surface area of the box.
  2. Find the surface area of a cube with edges of 7 cm.
  3. Find the surface area of a box \displaystyle 2\frac{1}{2} cm by 4 cm by 6 cm.
  4. A prism is 14 cm long. Its ends are right triangles with sides 9 cm, 12 cm and 15 cm, the right angle between the 9 cm and 12 cm sides. Find its surface area.
  5. A square pyramid has a base with sides of 7 cm, and each triangular face has slant height 11 cm. Find its surface area.
  6. A prism is 20 cm long. Each end is a trapezoid with parallel sides of 6 cm and 12 cm, 4 cm apart, and two slanted sides of 5 cm. Find its surface area.

Part 1: covering real things

  1. Ottar wants to wrap a present in a cube-shaped box with 12 cm edges. His sheet of gift wrap is 50 cm by 20 cm. Does the sheet have at least as much area as the box’s surface?
  2. Liesl paints a jewelry box 18 cm long, 10 cm wide and 7 cm tall, every outside face except the bottom, with two coats. How many square centimeters does she paint in all?
  3. A card box with no lid has a square bottom with 15 cm sides and is 6 cm tall. How much card is it made of?
  4. Jorun wraps six presents, each in a box 20 cm by 10 cm by 5 cm. A roll of gift wrap covers 3{,}000\ \text{cm}^2. Ignoring overlaps, how many rolls must she buy?
  5. A glass display case is a square pyramid with no base, set down over a vase. The square it covers has sides of 30 cm, and each glass face has slant height 25 cm. How much glass is used?
  6. Kazimir’s present box is a cube with 10 cm edges, with a square pyramid sitting on its top face as a lid. The pyramid’s base matches the top face and its slant height is 8 cm. How much paper covers the outside exactly once?

Part 1: reasoning

  1. Explain why the area of any net of a solid equals the solid’s surface area, and why two different nets of the same box have the same area.
  2. Explain why each triangle of a pyramid uses the slant height for its area, not the height of the pyramid.

Part 2: covering or filling

  1. Say whether each asks for surface area or volume: (a) the gift wrap for a present; (b) the sand that fills a sandpit; (c) the varnish on a jewelry box; (d) the number of stock cubes a box holds.
  2. Find the volume and the surface area of a box 4 cm by 5 cm by 6 cm, and say why the two numbers cannot be compared.
  3. Eighteen cubes with 1 cm edges are packed into a box they fill exactly. List the possible boxes and the surface area of each. Which takes the least card?
  4. Show that a box 1 cm by 2 cm by 8 cm and a box 2 cm by 3 cm by 4 cm have the same surface area. Which holds more?
  5. Two jewelry boxes are a cube with 4 cm edges and a box 2 cm by 4 cm by 8 cm. Show that they hold the same amount, and find which takes less wood, counting all six faces.

Part 2: surface for each unit of volume

  1. Find the surface for each \text{cm}^3 of a cube with 10 cm edges.
  2. Find the surface for each \text{cm}^3 of a box 2 cm by 3 cm by 5 cm. Does it have more or less surface for its size than a cube with 3 cm edges?
  3. An ice block 4 cm by 4 cm by 2 cm is cut into cubes with 1 cm edges. Find the surface for each \text{cm}^3 before and after cutting, and how many times as much surface the ice has after.
  4. Which has more surface for each \text{cm}^3: one cube with 8 cm edges, or eight cubes with 4 cm edges? Find the total surface of each.

Part 2: reasoning

  1. Explain why cutting a solid into pieces never changes its volume but always increases its total surface.
  2. A cube has 1 cm edges. Find its surface for each cubic unit measured in centimeters, and again in millimeters (its edges are 10 mm). Explain why the rate should be compared only between solids measured in the same unit.
Answers
  1. 302\ \text{cm}^2. The lids are 2 \times 55 = 110; the strip is 2 \times 66 + 2 \times 30 = 192; and 110 + 192 = 302.
  2. 294\ \text{cm}^2: six squares of 49, and 6 \times 49 = 294.
  3. 98\ \text{cm}^2. The faces are \displaystyle 2\frac{1}{2} \times 4 = 10, \displaystyle 2\frac{1}{2} \times 6 = 15 and 4 \times 6 = 24, two of each: 2 \times 49 = 98.
  4. 612\ \text{cm}^2. The legs meet at a right angle, so each end is \displaystyle \frac{1}{2} \times 9 \times 12 = 54, both 108. The rectangles are 126 + 168 + 210 = 504. Then 108 + 504 = 612.
  5. 203\ \text{cm}^2. The base is 49; each triangle is \displaystyle \frac{1}{2} \times 7 \times 11 = 38.5, four make 154; 49 + 154 = 203.
  6. 632\ \text{cm}^2. Each end is \displaystyle \frac{1}{2} \times (6 + 12) \times 4 = 36, both 72. The rectangles are 120 + 240 + 200 = 560. Then 72 + 560 = 632.
  7. Yes. The surface is 6 \times 12 \times 12 = 864\ \text{cm}^2 and the sheet 50 \times 20 = 1{,}000\ \text{cm}^2, with 136\ \text{cm}^2 to spare, though folding and overlap use some of it.
  8. 1{,}144\ \text{cm}^2. One coat covers the top 180, front and back 252 and ends 140: 572. Two coats: 2 \times 572 = 1{,}144.
  9. 585\ \text{cm}^2: the bottom 225 and four walls 4 \times 15 \times 6 = 360.
  10. Two rolls. One box is 2 \times (200 + 100 + 50) = 700\ \text{cm}^2, six are 4{,}200\ \text{cm}^2; one roll is too little, two give 6{,}000.
  11. 1{,}500\ \text{cm}^2: four triangles of \displaystyle \frac{1}{2} \times 30 \times 25 = 375, and no base.
  12. 660\ \text{cm}^2. The lid hides the cube’s top, so five faces show: 5 \times 100 = 500. The lid adds four triangles of \displaystyle \frac{1}{2} \times 10 \times 8 = 40, so 160. Then 500 + 160 = 660.
  13. Unfolding only bends the sheet at the folds, so each piece keeps the size and shape of its face, and the net is all the faces once each. Any net of the box is made of the same six faces, so its area is the same sum.
  14. A triangle’s height must be measured within the triangle, at right angles to its base. For a face of a pyramid that is the slant height. The pyramid’s height runs inside the solid, not in any face.
  15. (a) Surface area. (b) Volume. (c) Surface area. (d) Volume.
  16. Volume 4 \times 5 \times 6 = 120\ \text{cm}^3; surface area 2 \times (20 + 24 + 30) = 148\ \text{cm}^2. One counts unit cubes, the other unit squares.
  17. 1 \times 1 \times 18: 74\ \text{cm}^2; 1 \times 2 \times 9: 58\ \text{cm}^2; 1 \times 3 \times 6: 54\ \text{cm}^2; 2 \times 3 \times 3: 42\ \text{cm}^2, since 2 \times (6 + 6 + 9) = 42. The 2 \times 3 \times 3 box takes the least.
  18. Both are 52\ \text{cm}^2: 2 \times (2 + 8 + 16) = 52 and 2 \times (6 + 8 + 12) = 52. The volumes are 16\ \text{cm}^3 and 24\ \text{cm}^3, so the 2 \times 3 \times 4 box holds more.
  19. Both hold 64\ \text{cm}^3. The cube takes 6 \times 16 = 96\ \text{cm}^2 and the box 2 \times (8 + 16 + 32) = 112\ \text{cm}^2, so the cube takes less wood.
  20. 0.6\ \text{cm}^2 for each \text{cm}^3: 600 \div 1{,}000 = 0.6.
  21. \displaystyle 2\frac{1}{15}\ \text{cm}^2 for each \text{cm}^3: 2 \times (6 + 10 + 15) = 62 and \displaystyle 62 \div 30 = 2\frac{1}{15}. The cube has 54 \div 27 = 2, so the box has a little more.
  22. Before: 2 \times (16 + 8 + 8) = 64\ \text{cm}^2 over 32\ \text{cm}^3, so 2. After: 32 cubes of 6 make 192\ \text{cm}^2, and 192 \div 32 = 6. The surface is 192 \div 64 = 3 times as much.
  23. The eight cubes. Both hold 512\ \text{cm}^3. One cube: 384\ \text{cm}^2, and 384 \div 512 = 0.75. Eight cubes: 8 \times 96 = 768\ \text{cm}^2, and 768 \div 512 = 1.5, twice as much.
  24. Every bit of the solid ends up in exactly one piece, so the pieces’ volumes add back to the whole. The old outside is still outside, and each cut makes two new faces, one on each side of the cut.
  25. In centimeters, 6\ \text{cm}^2 for each \text{cm}^3. In millimeters the surface is 6 \times 10 \times 10 = 600\ \text{mm}^2 and the volume 10 \times 10 \times 10 = 1{,}000\ \text{mm}^3, so 0.6\ \text{mm}^2 for each \text{mm}^3. The cube is the same; only the unit changed, so numbers in different units say nothing about which solid has more surface.
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