Mathematics • Year 9 • Unit 4 • Lesson 8

Scale Factors in the Real World

Use the three scale rules — length × $k$, area × $k^2$, volume × $k^3$ — in everyday contexts: bushwalking maps, scaling up a recipe, an architect's model, a film prop and resizing a phone screenshot. Then explain your method in your own words.

Apply · Real-World Maths

1. Word problems

Each problem uses a scale rule from Lesson 8: lengths scale by $k$, areas by $k^2$, volumes by $k^3$. Show your working — a single final answer with no working only earns half marks.

1.1 — Bushwalking map. A NSW bushwalking map has scale $1 : 25\,000$ (so $1$ cm on the map represents $25\,000$ cm in reality).

(a) A track measures $8$ cm on the map. What is the real track length in metres? In kilometres?
(b) A small lake is $2$ cm × $3$ cm on the map. What is its real area, in km²? 4 marks

Stuck on (b)? Find the real width and length first, then multiply. Don't be tempted to multiply $6$ cm² by $25\,000$ — area scales by $k^2$, not $k$.

1.2 — Scaling up a recipe. A pizza recipe uses a $25$ cm diameter base for $4$ people. You want to make a larger pizza that is similar in shape (a circle) but feeds $9$ people, where the amount of base needed is proportional to the area of the pizza.

(a) What is the area scale factor needed?
(b) What is the linear scale factor $k$ for the diameter?
(c) What diameter should the larger pizza be? 3 marks

Stuck? Servings $\propto$ area $\propto k^2$. So $k^2 = 9/4$ and $k = 3/2$.

1.3 — Architect's model. An architect builds a $1 : 50$ scale model of a house. The model's footprint is $30$ cm × $24$ cm, and the model contains $0.012$ m³ of plaster (approximating the volume of the solid walls).

(a) What are the real-life dimensions of the footprint, in metres?
(b) What is the real-life floor area, in m²?
(c) Estimate the real-life volume of concrete needed for the walls, in m³. 4 marks

Stuck? Volume scales by $k^3 = 50^3 = 125\,000$.

1.4 — Film prop "giant". In a film, a prop spider is built at scale $5 : 1$ (the prop is $5\times$ larger than a real spider in every dimension). A real spider weighs $2$ grams. Assuming the prop is made of the same density of material as the real spider, what would the prop weigh? Give your answer in kg, to 1 decimal place.

3 marks

Stuck? Weight $\propto$ volume $\propto k^3 = 125$.

1.5 — Phone screenshot. A screenshot is $1080$ pixels wide × $1920$ pixels tall. A student resizes it so the new width is $1620$ pixels and the new height is $2880$ pixels (each side scaled by the same factor).

(a) Find the scale factor $k$ used.
(b) Find the ratio of new pixel area to original pixel area.
(c) If the original was a $500$ KB file and file size is roughly proportional to pixel area, predict the new file size in KB. 3 marks

Stuck? $k = 1620/1080 = 1.5$. Area ratio $= k^2 = 2.25$.

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate doubles every side length of a fish tank and says "the new tank holds twice as much water." In your own words, explain (i) what mistake they have made, (ii) which rule from Lesson 8 they have ignored, and (iii) how much water the new tank actually holds compared with the original. Refer to "$k^3$" somewhere in your explanation.

Stuck? Revisit lesson § "Common Misconceptions" — the second misconception listed is exactly this.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Bushwalking map (1 : 25 000)

(a) Real length $= 8 \times 25\,000 = 200\,000$ cm $= \mathbf{2000}$ m $= \mathbf{2}$ km.
(b) Real width $= 2 \times 25\,000 = 50\,000$ cm $= 500$ m $= 0.5$ km. Real length $= 3 \times 25\,000 = 75\,000$ cm $= 0.75$ km. Area $= 0.5 \times 0.75 = \mathbf{0.375}$ km².
Check via area scale factor: map area $= 6$ cm² $\times k^2 = 6 \times 25\,000^2$ cm² $= 3.75 \times 10^9$ cm² $= 0.375$ km². ✓

1.2 — Scaling up a pizza

(a) Area scale factor $= 9/4 = \mathbf{2.25}$.
(b) $k = \sqrt{2.25} = \mathbf{1.5}$.
(c) New diameter $= 1.5 \times 25 = \mathbf{37.5}$ cm.
This is why pizza shops use a small number of sizes — going from 25 cm to 37.5 cm gives more than double the area, even though the diameter only increased by 50%.

1.3 — Architect's model (1 : 50)

(a) Real footprint $= (30 \times 50)$ cm × $(24 \times 50)$ cm $= 1500$ cm × $1200$ cm $= \mathbf{15}$ m × $\mathbf{12}$ m.
(b) Floor area $= 15 \times 12 = \mathbf{180}$ m².
(c) Real volume $= 0.012 \times 50^3 = 0.012 \times 125\,000 = \mathbf{1500}$ m³ of concrete.
That's a LOT — which is why architects build models from light materials like foamboard, not solid plaster.

1.4 — Film prop spider

$k = 5$, so volume (and mass) scale by $k^3 = 125$.
Prop mass $= 2 \times 125 = 250$ g $= \mathbf{0.3}$ kg (to 1 d.p.).
This is the "square-cube law" — why giant animals in films would actually collapse under their own weight. Their bones grow with $k^2$ cross-section but their weight grows with $k^3$ volume.

1.5 — Phone screenshot

(a) $k = 1620/1080 = \mathbf{1.5}$. (Check: $1920 \times 1.5 = 2880$ ✓.)
(b) Area ratio $= k^2 = (1.5)^2 = \mathbf{2.25}$.
(c) New file size $\approx 500 \times 2.25 = \mathbf{1125}$ KB ($\approx 1.1$ MB).

2.1 — Explain your thinking (sample response)

My classmate has confused the way volume scales. Doubling every length means $k = 2$, but the new volume is multiplied by $\mathbf{k^3 = 8}$, not by $k$. The rule from Lesson 8 they have ignored is "volume scales as $k^3$" for similar solids. So the new fish tank holds $\mathbf{8\times}$ as much water as the original, not just $2\times$. This is why a slightly bigger fish tank can be much heavier than expected — the increase in volume (and weight of water) grows much faster than the increase in length.

Marking: 1 for naming the mistake (linear vs cubic scaling); 1 for naming the rule "$k^3$"; 1 for the correct factor of 8; 1 for a clear, full-sentence explanation referring to $k^3$.