Math 7

Scale factor, area and volume

Make a rug twice as long and twice as wide. Is it twice as much rug?

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Scale factor multiplies every lengthMath 7 Surface area, and how it compares with volumeMath 6

Where you are 3 in Scale and size

Measurement12

Scale and size3

Measuring what you cannot reach Scale factor multiplies every length

Scale factor, area and volume

Why this exists

Plans, models and photographs are scale copies: one number, the scale factor k, multiplies every length, each side, each diagonal and the perimeter. But the questions people ask about a plan or a model are often about something else. How much carpet covers the room on the plan? How much batter goes into a cake twice the size? How heavy would a giant be?

Carpet and paint are measured by area, batter and weight by volume, and neither follows the scale factor. A rug twice as long and twice as wide is four times as much rug, and a cake twice as wide and twice as tall takes eight times the batter. This lesson finds the rule for area, then for volume, and then uses it to show why a giant built exactly like a person could not stand up.

Definition

A scale copy at scale factor k, where k > 0, has every length equal to k times the matching length of the original. When a shape is copied at scale factor k:

its area is multiplied by k^2 = k \times k, the area factor;
if it is a solid, its volume is multiplied by k^3 = k \times k \times k, the volume factor, and its surface area, which is an area, by k^2.

Area here is the number of unit squares that cover a shape, and volume the number of unit cubes that fill a solid. So the copy takes k^2 times as many squares to cover it and k^3 times as many cubes to fill it.

Intuition

Area. In Figure 1 a rectangle 3 squares long and 2 high is copied at scale factor 2. The copy is 6 long and 4 high.

Figure 1Doubling the length and the width makes four copies of the original, so four times the area.

The original is covered by 3 \times 2 = 6 squares and the copy by 6 \times 4 = 24: four times as many, not two, and the dashed lines show four rectangles the size of the original. Each row of the copy has twice as many squares as a row of the original, and there are twice as many rows. The scale factor is used once for the length and once more for the height, so the area is multiplied by 2 \times 2 = 4.

Volume. A solid has a length, a width and a height, and a scale copy multiplies all three. Figure 2 shows a unit cube copied at scale factors 2 and 3.

Figure 2An edge of 2 gives 8 cubes, and an edge of 3 gives 27.

The cube with edge 2 has 2 \times 2 = 4 cubes in each layer and 2 layers, so 2 \times 2 \times 2 = 8 cubes. The cube with edge 3 has 3 \times 3 = 9 in each layer and 3 layers, so 27 cubes. The scale factor is used three times.

So the rule counts the lengths a measurement is built from: a length uses the scale factor once, an area (a length times a length) twice, and a volume (three lengths multiplied) three times.

Results

Area. A copy at scale factor k of a shape with area A has area

A' = k^2 A

Volume and surface area. A copy at scale factor k of a solid with volume V and surface area S has

V' = k^3 V \qquad S' = k^2 S

Back to the scale factor. If the copy has n times the area of the original, then k^2 = n, so k is the positive number whose square is n. If it has n times the volume, then k^3 = n.

Load on a cross-section. If a solid made of the same material throughout is copied at scale factor k, its weight is multiplied by k^3 and the area of any cross-section by k^2, so the weight resting on each square unit of a cross-section is multiplied by

\frac{k^3}{k^2} = k

Why it is true

A square. A unit square copied at scale factor k is a square with side k, so its area is k \times k = k^2. For a whole number k that is k rows of k unit squares; for a k such as \tfrac{1}{3} or 2.5, the area of a square is still side times side.

A rectangle. A rectangle with length \ell and width w has area \ell w. Its copy has length k\ell and width kw, so its area is

(k\ell)(kw) = (k \times k)(\ell w) = k^2 \ell w

where the commutative and associative properties of multiplication allow the factors to be reordered and regrouped.

Any shape. Cover the shape with small squares. A square of side s becomes a square of side ks, with area k^2s^2, which is k^2 times its old area. Every piece is multiplied by k^2, so by the distributive property their total is too. A curved shape is covered more and more closely as the squares get smaller, with the factor k^2 at every stage. The careful version, for any pair of similar figures, is in scale factor and perimeter, area and volume.

Volume. A box \ell by w by h becomes a box k\ell by kw by kh, with volume

(k\ell)(kw)(kh) = k^3 \ell wh

and any solid, filled with small cubes, is multiplied by k^3 piece by piece. The surface of a solid is made of faces, each an area, so each face, and their total, is multiplied by k^2.

The load. If every cubic centimeter weighs the same, weight W is proportional to volume, so it is multiplied by k^3. A cross-section, the flat face left by cutting the solid straight across, is an area C, multiplied by k^2. The weight per square unit of cross-section becomes

\frac{k^3 W}{k^2 C} = k \cdot \frac{W}{C}

Worked examples

Examples 1 to 4 are the first day, on area. Examples 5 to 7 are the second: volume, area and volume together, and giants.

Example 1

Counting the squares in a copy

Figure 3 shows a shape made of 5 unit squares and its copy at scale factor 3. Find the area of the copy, first by counting and then with the area factor. Compare the perimeters too.

Figure 3Every unit square of the original becomes a block of 3 \times 3 = 9 unit squares.

Solution

Each unit square of the original becomes a square 3 units on each side, which holds 3 \times 3 = 9 unit squares. The original is 5 unit squares, so the copy is 5 blocks of 9:

5 \times 9 = 45 \text{ square units}

The area factor gives the same at once: k^2 = 3^2 = 9, and 9 \times 5 = 45.

Check. Cut the copy differently. Along the bottom is a strip 9 long and 3 high, 27 squares, and standing on its left end is a column 3 wide and 6 high, 18 squares. 27 + 18 = 45.

The perimeter of the original, going round from the bottom left corner, is 3 + 1 + 2 + 2 + 1 + 3 = 12 units, and the copy’s is 3 \times 12 = 36 units, since a perimeter is a length. The perimeter was tripled while the area was multiplied by 9.

Example 2

A rug on a plan

On Isolde’s plan of her sitting room, a rug is a rectangle 5 cm by 3.5 cm. On the plan, 1 cm stands for 40 cm in the room. Find the area of the real rug in square meters.

Solution

Every real length is 40 times its length on the plan, so the room is a scale copy of the plan with k = 40 (see reading a map or plan). On the plan the rug’s area is

5 \times 3.5 = 17.5\text{ cm}^2

The area factor is k^2 = 40^2 = 1600, so the real area is

17.5 \times 1600 = 28\,000\text{ cm}^2

A square meter is a square 100 cm on each side, so it holds 100 \times 100 = 10\,000 square centimeters, the area rule again with k = 100 (see converting square units). The real rug covers 28\,000 \div 10\,000 = 2.8\text{ m}^2.

Check. Work with lengths first. The real rug is 5 \times 40 = 200 cm = 2 m long and 3.5 \times 40 = 140 cm = 1.4 m wide, and 2 \times 1.4 = 2.8. The factor 40 appears twice, once for the length and once for the width.

Example 3

Tiles for a floor of any shape

Kunigunde’s kitchen floor is L-shaped. On a plan drawn at scale 1 : 30, the floor has an area of 160\text{ cm}^2. She will cover it with square tiles 40 cm on each side. Find the real area of the floor, and the number of tiles whose total area equals it.

Solution

A scale of 1 : 30 means each real length is 30 times its length on the plan, so k = 30. None of the floor’s lengths is given, and none is needed: the area factor works for a shape of any outline. It is k^2 = 30^2 = 900, so the real floor has area

160 \times 900 = 144\,000\text{ cm}^2

which is 144\,000 \div 10\,000 = 14.4\text{ m}^2. Each tile has area 40 \times 40 = 1600\text{ cm}^2, so the number of tiles is

144\,000 \div 1600 = 90

Check. In square meters, a tile is 0.4 \times 0.4 = 0.16\text{ m}^2, and 14.4 \div 0.16 = 90. Tiles cut at the edges may mean buying more, but 90 is the number whose areas add up to the floor’s.

Example 4

How wide for four times the pizza

Lorenzo’s pizza is a circle 24 cm across. (a) How wide must a pizza of the same thickness be to hold 4 times as much? (b) How many times as much is in a pizza 36 cm across?

Solution

With the same thickness, the amount of pizza goes with the area of its top. Any two circles are scale copies of each other, with the scale factor equal to the ratio of their widths.

(a) Let k be the scale factor from Lorenzo’s pizza to the larger one. The area must be multiplied by 4, so

k^2 = 4

A scale factor is positive, and 2 \times 2 = 4, so k = 2. The pizza must be 24 \times 2 = 48 cm across: twice as wide, not four times.

(b) Here k = \tfrac{36}{24} = \tfrac{3}{2}, so the area is multiplied by

k^2 = \tfrac{3}{2} \times \tfrac{3}{2} = \tfrac{9}{4} = 2.25

Half as wide again holds 2.25 times as much: more than twice the pizza.

Check. With the area of a circle, \pi r^2: Lorenzo’s pizza has radius 12 and area 144\pi. A radius of 24 gives 576\pi = 4 \times 144\pi, and a radius of 18 gives 324\pi, and 324 \div 144 = 2.25.

Example 5

Building blocks

Jerome builds a solid box of building blocks 3 blocks long, 2 wide and 2 high. He then builds copies at scale factor 2 and at scale factor 3, with blocks of the same size. How many blocks does each one take?

Solution

The original takes 3 \times 2 \times 2 = 12 blocks.

At scale factor 2 the box is 6 by 4 by 4, which is 6 \times 4 \times 4 = 96 blocks. Each block of the original has become a cube of 2 \times 2 \times 2 = 8 blocks, and 8 \times 12 = 96.

At scale factor 3 the box is 9 by 6 by 6, which is 9 \times 6 \times 6 = 324 blocks, and k^3 = 27 gives 27 \times 12 = 324.

Compare the top face, an area. It is 3 \times 2 = 6 blocks on the original, 6 \times 4 = 24 on the first copy and 9 \times 6 = 54 on the second: multiplied by 4 and by 9, while the number of blocks was multiplied by 8 and by 27.

Example 6

A bigger cake

Melisande bakes a round cake 18 cm across and 8 cm tall. It takes 600 g of batter, and 200 g of icing spread evenly over its top and sides. She wants a cake of exactly the same shape that is 27 cm across. How tall is it, how much batter does it take, and how much icing at the same thickness?

Solution

The scale factor is k = \tfrac{27}{18} = \tfrac{3}{2}. The height is a length, so the new cake is 8 \times \tfrac{3}{2} = 12 cm tall.

Every cubic centimeter of the same batter weighs the same, so the batter goes with the volume. The volume factor is k^3 = \tfrac{27}{8}:

600 \times \tfrac{27}{8} = 2025\text{ g}

The icing is a layer of the same thickness over the top and sides, so it goes with the surface area. The area factor is k^2 = \tfrac{9}{4}:

200 \times \tfrac{9}{4} = 450\text{ g}

Check. With the volume of a cylinder, the small cake is \pi \times 9^2 \times 8 = 648\pi\text{ cm}^3 and the large one \pi \times 13.5^2 \times 12 = 2187\pi\text{ cm}^3, and 2187 \div 648 = 3.375 = \tfrac{27}{8}. The iced surface of the small cake is the top, 81\pi, and the side, 18\pi \times 8 = 144\pi, together 225\pi; the large cake’s is 182.25\pi + 27\pi \times 12 = 506.25\pi, and 506.25 \div 225 = 2.25. The batter more than triples while the icing only a little more than doubles.

Example 7

Why a giant could not stand

In a story, a giant is shaped exactly like Jerome but 10 times as tall, so every length of the giant is 10 times the matching length of Jerome. Jerome weighs 70 kg. Compare the giant’s weight, and the strength of its leg bones, with Jerome’s. How many times as wide as Jerome’s would the giant’s leg bones have to be to carry their load as easily as his do?

Solution

The giant is the same flesh and bone as Jerome all the way through, so its weight goes with its volume. The volume factor is k^3 = 10^3 = 1000, so the giant weighs

70 \times 1000 = 70\,000\text{ kg}

How much a bone can hold up depends on the area of its cross-section, the flat face left if the bone were cut straight across. That is an area, so the giant’s bones are k^2 = 100 times as strong as Jerome’s. With 1000 times the weight on bones 100 times as strong, each square centimeter of its bone carries

\frac{1000}{100} = 10

times what Jerome’s does, as if Jerome stood with nine more people of his weight on his shoulders.

For each square centimeter to carry no more than Jerome’s, the cross-section would have to grow by 1000, like the weight. Bones made w times as wide, keeping their shape, have their cross-section multiplied by w^2, so

w^2 = 1000

Since 31^2 = 961 and 32^2 = 1024, w is between 31 and 32, about 31.6. Leg bones about 32 times as wide on a body only 10 times as tall make thick, stumpy legs, so a giant that could stand would not be a scale copy of a person. Galileo Galilei made an argument of this kind in Two New Sciences (1638).

Boundaries

The copy must be a scale copy. The rules need every length multiplied by the same k. A rug made twice as long but no wider has twice the area, not four times; when length and width change by different factors, the area is multiplied by their product, as in scale factor and perimeter, area and volume.

The power counts what is measured, not the object. The fringe around a rug is a length, so it is multiplied by k. The surface of a cake is an area, so it is multiplied by k^2, though the cake is a solid.

The scale factor is a plain number. A plan where 1 cm stands for 0.5 m has k = 50, not k = 0.5: both lengths must be in the same unit before they are divided.

Amounts follow area or volume only when they are spread evenly. Paint and icing follow area when the layer is as thick on the copy as on the original; batter and weight follow volume when every cubic centimeter is the same stuff. A hollow cake, or a model painted twice as thick, breaks the link.

Geometry gives the ratios; the rest is about materials. The factors k^3 for weight and k^2 for cross-section are exact. That a bone’s strength goes with its cross-section is a fact about bone, not about scale copies, and the conclusion about giants rests on it.

Where this goes next

The same rule holds for every pair of similar figures and similar solids; scale factor and perimeter, area and volume proves it there and runs it backwards with cube roots. Finding k from k^2 = n or k^3 = n is the job of square and cube roots.

Changing units is a scaling too: a square meter is 100^2 = 10\,000 square centimeters and a cubic meter is 100^3 = 1\,000\,000 cubic centimeters, the subject of converting square and cubic units.

A copy has k^2 times the surface but k^3 times the volume, so the larger it is, the less surface it has for each cubic centimeter inside; surface area and how it compares with volume follows that thread, and weight from volume is density.

Exercises

Exercises 1 to 14 go with the first day of the lesson (Examples 1 to 4), and Exercises 15 to 25 with the second (Examples 5 to 7).

Part 1: area from the scale factor

A rectangle 4 units by 5 units is drawn on a grid and copied at scale factor 3. Find the length and width of the copy, the area of each rectangle, and how many times the original’s area the copy has.
A shape has area 11\text{ cm}^2. Find the area of its copy at scale factor 2, and of its copy at scale factor 5.
A triangle has area 18\text{ cm}^2. Find the area of its copy at scale factor \tfrac{1}{3}.
A shape has area 8\text{ cm}^2. Find the area of its copy at scale factor 2.5.
A shape made of 7 unit squares is copied at scale factor 4. How many unit squares cover the copy?

Part 1: plans, rugs, tiles and paint

On a plan where 1 cm stands for 60 cm, a rug is 4 cm by 2.5 cm. Find the area of the real rug in square meters.
A room on a plan drawn at scale 1 : 100 has an area of 14\text{ cm}^2. Find the area of the real room in square meters.
Hieronymus’s hall floor has an area of 270\text{ cm}^2 on a plan drawn at scale 1 : 20. Find the real area, and the number of square tiles 30 cm on each side whose total area equals it.
A wall of a doll’s house built at scale 1 : 12 takes 5 mL of paint. How much paint does the matching wall of the real house take, painted to the same thickness?

Part 1: finding the scale factor from areas

A photograph 6 cm by 4 cm is enlarged so that its area is 25 times as large. Find the size of the enlargement.
Isolde’s square rug has sides of 1.2 m. She wants a rug of the same shape with 4 times the area. How long are its sides?
A shape has area 50\text{ cm}^2 and a scale copy of it has area 2\text{ cm}^2. Find the scale factor.

Part 1: reasoning

Hieronymus says that a rectangle of area 12\text{ cm}^2, copied at scale factor 2, has area 24\text{ cm}^2, because every length was doubled. Explain what the doubling does to the area, and find the copy’s area.
Two rugs have the same shape, and one is 3 times as long as the other. A fringe runs all the way round the edge of each, and each is woven to the same thickness. How many times as much fringe, and how many times as much wool, does the larger rug take? Explain why the two answers differ.

Part 2: volume from the scale factor

A box of building blocks is 2 by 3 by 4 blocks. It is copied at scale factor 2 with blocks of the same size. Find the size of the copy, the number of blocks in it, and how many times the original number that is.
A solid has volume 7\text{ cm}^3. Find the volume of its copy at scale factor 3.
A cube with edge 4.5 cm is built from sugar cubes with edge 1.5 cm. How many sugar cubes does it take?
A clay model of a statue is made at scale factor \tfrac{1}{4}. The statue has volume 3200\text{ cm}^3. Find the volume of the model.

Part 2: area and volume together

A model house is built at scale 1 : 20. Its walls take 40 mL of paint, and its rooms hold 12 liters of air. How much paint do the real walls take, at the same thickness, and how many liters of air do the real rooms hold?
A box measures 1 cm by 2 cm by 3 cm. Find its surface area and volume, and then the surface area and volume of its copy at scale factor 5.
A large cube is built from 125 small cubes. How many small cubes lie along each edge? How many faces of small cubes show on the outside of the large cube, counting all six of its faces?

Part 2: giants

A giant has exactly Lorenzo’s shape and is 3 times as tall. Lorenzo weighs 60 kg. Find the giant’s weight, the factor by which the cross-section of its bones is multiplied, and the factor by which the load on each square centimeter of bone is multiplied.
A toy figure is made of the same stuff as Melisande, all the way through, and is exactly her shape at \tfrac{1}{20} of her height. By what factor are its weight, the cross-section of its bones, and the load on each square centimeter of bone multiplied?

Part 2: reasoning

A giant is 4 times as tall as a person, in every direction. For each square centimeter of its bones to carry no more than the person’s do, its bones’ cross-section must be multiplied by the same factor as its weight. How many times as wide must its bones be? Explain why the giant is then not a scale copy of the person.
A cube with edge k is a copy of the unit cube at scale factor k. Find its surface area and its volume for k = 2, k = 6 and k = 10. For which k are the two numbers equal, and why does the volume overtake the surface area as k grows?

Answers

The copy is 12 by 15. The areas are 20 and 180, so the copy has 9 times the area, which is 3^2.
44\text{ cm}^2 and 275\text{ cm}^2. The area factors are 2^2 = 4 and 5^2 = 25, and 11 \times 4 = 44, 11 \times 25 = 275.
2\text{ cm}^2. The area factor is \left(\tfrac{1}{3}\right)^2 = \tfrac{1}{9}, and 18 \times \tfrac{1}{9} = 2.
50\text{ cm}^2. The area factor is 2.5^2 = 6.25, and 8 \times 6.25 = 50.
112. Each unit square becomes a 4 by 4 block of 16, and 7 \times 16 = 112.
3.6\text{ m}^2. Here k = 60, the plan area is 4 \times 2.5 = 10\text{ cm}^2, and 10 \times 3600 = 36\,000\text{ cm}^2 = 3.6\text{ m}^2. Check: the rug is 2.4 m by 1.5 m, and 2.4 \times 1.5 = 3.6.
14\text{ m}^2. The area factor is 100^2 = 10\,000, so 14 \times 10\,000 = 140\,000\text{ cm}^2 = 14\text{ m}^2. At this scale each square centimeter of the plan stands for one square meter.
10.8\text{ m}^2 and 120 tiles. The area factor is 20^2 = 400, so the floor is 270 \times 400 = 108\,000\text{ cm}^2 = 10.8\text{ m}^2. A tile is 30 \times 30 = 900\text{ cm}^2, and 108\,000 \div 900 = 120.
720 mL. Paint covers an area, and the area factor is 12^2 = 144: 5 \times 144 = 720.
30 cm by 20 cm. k^2 = 25 and k > 0, so k = 5, and 6 \times 5 = 30, 4 \times 5 = 20.
2.4 m. k^2 = 4 gives k = 2, and 1.2 \times 2 = 2.4.
\tfrac{1}{5}. k^2 = \tfrac{2}{50} = \tfrac{1}{25}, and the positive number whose square is \tfrac{1}{25} is \tfrac{1}{5}.
48\text{ cm}^2. The area is length times width, and both are doubled, so the area is multiplied by 2 \times 2 = 4: 12 \times 4 = 48. Doubling one length alone would double the area; doubling both doubles it twice.
3 times the fringe and 9 times the wool. The fringe runs along the perimeter, a length, so it is multiplied by k = 3. The wool covers the area, which is multiplied by k^2 = 9. The answers differ because a perimeter uses the scale factor once and an area uses it twice, once for the length and once for the width.
The copy is 4 by 6 by 8 blocks, which is 192 blocks, 8 times the original 24.
189\text{ cm}^3. The volume factor is 3^3 = 27, and 7 \times 27 = 189.
27. The large cube is a copy of a sugar cube at k = \tfrac{4.5}{1.5} = 3, so it takes 3^3 = 27 sugar cubes: 3 layers of 3 \times 3.
50\text{ cm}^3. The volume factor is \left(\tfrac{1}{4}\right)^3 = \tfrac{1}{64}, and 3200 \div 64 = 50.
16 liters of paint and 96\,000 liters of air. Here k = 20. Paint follows area: 40 \times 20^2 = 40 \times 400 = 16\,000 mL = 16 liters. Air follows volume: 12 \times 20^3 = 12 \times 8000 = 96\,000 liters.
The box has surface area 2(2 + 3 + 6) = 22\text{ cm}^2 and volume 6\text{ cm}^3. The copy has surface area 22 \times 25 = 550\text{ cm}^2 and volume 6 \times 125 = 750\text{ cm}^3. Check: the copy is 5 by 10 by 15, with surface area 2(50 + 75 + 150) = 550 and volume 5 \times 10 \times 15 = 750.
5 along each edge, and 150 faces show. The large cube is a copy of one small cube with k^3 = 125, so k = 5. Each face of the large cube is 5 \times 5 = 25 small faces, and 6 \times 25 = 150, which is k^2 = 25 times the 6 faces of one small cube.
1620 kg; the cross-section is multiplied by 9; the load per square centimeter by 3. The weight follows volume, 60 \times 3^3 = 60 \times 27 = 1620. The cross-section is an area, 3^2 = 9, and \tfrac{27}{9} = 3.
Weight by \tfrac{1}{8000}, cross-section by \tfrac{1}{400}, load per square centimeter by \tfrac{1}{20}. These are k^3, k^2 and \tfrac{k^3}{k^2} = k with k = \tfrac{1}{20}.
8 times as wide. The weight is multiplied by 4^3 = 64, so the cross-section must be too. If the bones are w times as wide, their cross-section is multiplied by w^2, and w^2 = 64 gives w = 8. The bones are then 8 times as wide while the body is 4 times as tall, so not every length has been multiplied by the same factor, and the giant is not a scale copy.
Surface area 6k^2 and volume k^3. For k = 2: 24 and 8. For k = 6: 216 and 216. For k = 10: 600 and 1000. They are equal when k^3 = 6k^2; dividing by k^2, which is not zero, gives k = 6. The volume has one more factor of k than the surface area, so for k above 6 that extra factor is more than 6 and the volume is the larger number. (Only the numbers are compared: one is in square units, the other in cubic units.)