Enlargement, maps and scale models

A city map folded into your pocket, an architect's blueprint pinned to a wall, a toy car on a shelf — every one of these is a shrunken (or enlarged) copy of something real, and the mathematics that makes the copy trustworthy is similarity. An enlargement is a transformation that keeps every shape exactly the same, just resized by a fixed scale factor. Once you know that one number, you can convert any length on the map, model or drawing into its real-world counterpart, and back again. This section takes the similar-triangles idea you already know and turns it into a practical toolkit: reading a scale ratio, extracting a scale factor from a single known pair of lengths, and using it confidently in both directions without mixing up which way the multiplication goes.

The core insight

An enlargement multiplies every length of a figure by the same scale factor $k$ , so a map or model with scale $1:n$ converts between drawn length $\ell$ and real length $L$ by $L = n\ell$ , and this single factor — never applied to area or volume without squaring or cubing it — is the working tool behind every map, plan and scale-model problem.

The core idea

An enlargement (or reduction) produces a figure similar to the original: same shape, same angles, but every linear measurement — sides, heights, diagonals, perimeters — multiplied by one fixed number, the scale factor $k$ . A map is similar to the land it represents; a scale model is similar to the real object. This is exactly the similar-triangles idea extended to any figure at all. The scale factor is defined as the ratio of a drawn (map or model) length to the corresponding real length:

k = \frac{\text{length on map/model}}{\text{corresponding real length}}

where: the numerator and denominator are measured in the same unit, so $k$ itself is a pure number (or, written as $1:n$ , a ratio with $n = 1/k$ ). Scales are stated in different but equivalent forms — a ratio $1:n$ , a decimal like $k = 0.005$ , or a phrase like "half size." All three describe the same relationship, and being able to switch between them is half the battle in exam questions. If a map or model has scale $1:n$ , every real length $L$ and its drawn length $\ell$ are related by:

\ell = \frac{L}{n} \qquad \text{equivalently} \qquad L = n\ell

where: $\ell$ is the length measured on the map/model, $L$ is the true real-world length, and $n$ is the number after the colon in the scale $1:n$ (so the map is $n$ times smaller than reality, not $n$ times bigger).

Figure

A ruler holding a six centimetre map segment beside an arrow pointing to a three kilometre real distance, illustrating that multiplying the map length by the scale factor gives the real length

Area and volume do not scale by $k$ itself — area scales by $k^2$ and volume by $k^3$ — but that squared/cubed scaling belongs to a separate result (areas of similar triangles); here the tool is the plain linear scale factor $k$ , used the same way whichever direction you convert. There is one more subtlety worth internalising before working problems: the scale factor is only ever valid between figures that are genuinely similar. A map is similar to the flat region it depicts, and a scale model is similar to the real object it copies, so any pair of corresponding lengths on that map or model shares the same $k$ . It is a common slip to assume a scale factor found from one dimension of a shape (say, its length) automatically carries over to an unrelated or non-corresponding measurement on a different figure — it only carries over within the same similar pair, and only between lengths that genuinely correspond to one another. When a question gives you one known pair of lengths and asks for another, always check first that both belong to the same map or the same model.

Derivation AConverting a map distance into a real distance and back

The goal is to show, from the definition of scale factor alone, that a real length is found by multiplying the map length by $n$ (for a scale $1:n$ ), and that this same relationship inverts cleanly to recover a map length from a real one, which is the single piece of algebra every map/model question in this chapter reduces to.

Step 1. Start from the definition of the scale factor as a ratio of corresponding lengths, measured in the same unit: $k = \dfrac{\ell}{L}$ , where $\ell$ is the map length and $L$ is the real length.

Step 2. Write the scale as $1:n$ , which means $k = \dfrac{1}{n}$ — one unit on the map corresponds to $n$ of that same unit in reality. Substituting into Step 1 gives $\dfrac{1}{n} = \dfrac{\ell}{L}$ .

Step 3. Cross-multiply to isolate the real length: $L = n\ell$ . This says exactly what "scale $1:n$ " means in practice — take the map length and multiply by $n$ to recover the real length, provided both are in the same starting unit before you multiply.

Step 4. Invert the same equation to go the other way: dividing both sides by $n$ gives $\ell = \dfrac{L}{n}$ , the map length is the real length divided by $n$ . Both directions use the identical relationship; only the arrangement changes depending on which quantity is unknown.

Worked example

A map is drawn to a scale of $1:50{,}000$ . Two towns are $7\ \text{cm}$ apart on the map. A new housing estate is being planned exactly halfway between the two towns. Find (a) the actual distance between the towns in kilometres, and (b) the real distance from either town to the estate.

Step 1 — Interpret the scale. $1\ \text{cm}$ on the map represents $50{,}000\ \text{cm}$ in reality, so $n = 50{,}000$ .

Step 2 — Scale up the map distance. Using $L = n\ell$ : $L = 7 \times 50{,}000 = 350{,}000\ \text{cm}$ .

Step 3 — Convert to kilometres. $1\ \text{km} = 100{,}000\ \text{cm}$ , so $L = \dfrac{350{,}000}{100{,}000} = 3.5\ \text{km}$ . The two towns are $3.5\ \text{km}$ apart.

Step 4 — Halve the real distance for the estate. "Halfway between the towns" is a real-world midpoint, so it is simply half the real distance already found: $\dfrac{3.5}{2} = 1.75\ \text{km}$ from either town. The estate is $1.75\ \text{km}$ from each town.

Competitive stretch

The same map also shows a rectangular park whose longer side measures $2.4\ \text{cm}$ . Find the real length of that side, in metres, and use it alongside the earlier $7\ \text{cm}$ town-to-town distance to say how many times longer the town-to-town distance is than the park's longer side.

Step 1 — Scale up the park's side. $L = n\ell = 50{,}000 \times 2.4 = 120{,}000\ \text{cm}$ . Converting to metres ( $1\ \text{m} = 100\ \text{cm}$ ): $L = 1200\ \text{m}$ .

Step 2 — Compare the two real distances. The town-to-town distance is $3.5\ \text{km} = 3500\ \text{m}$ , so the ratio is $\dfrac{3500}{1200} \approx 2.92$ . The towns are about $2.9$ times farther apart than the length of the park. Notice this ratio could equally have been read straight off the map distances, $\dfrac{7}{2.4} \approx 2.92$ — because both figures share the same scale factor, ratios between corresponding real lengths equal the same ratios between their map lengths, so no conversion to real units is even needed for a pure comparison.

Common traps

Mixing units before applying the scale factor — convert map/model and real lengths to the same unit first, then multiply or divide.
Treating $1:n$ as "the map is $n$ times bigger" instead of smaller — the map is the small figure; multiply the map length by $n$ to reach the real length.
Applying the linear scale factor directly to an area or a volume — area needs $k^2$ and volume needs $k^3$ , never the plain $k$ .
Rounding an intermediate unit conversion too early, which shifts the final answer — keep full precision until the last step.

Ask the tutor: stuck on which way to multiply or divide in a map-scale or scale-model problem — ask the AI tutor to check your setup before you compute.

How do I know whether to multiply or divide by n when a question gives a scale of 1 to n?

Ask which length you already have. If you have the map or model length and want the real length, multiply by $n$ (the real thing is bigger, since the map is the small copy). If you have the real length and want the map length, divide by $n$ . Writing the relation $L = n\ell$ once at the start of a question and reading off which letter is unknown removes the guesswork entirely.

Why can't I just use the same scale factor for the area of a scaled shape?

Because area is a product of two lengths, not one. If every length is multiplied by $k$ , an area made of two such lengths (like length times breadth) gets multiplied by $k \times k = k^2$ , not by $k$ alone. A map at scale $1:1000$ scales a length by $\tfrac{1}{1000}$ but scales an area by $\tfrac{1}{1000^2} = \tfrac{1}{1{,}000{,}000}$ . This section only covers the linear case; the squared rule belongs to the areas-of-similar-triangles result.

If a scale is given as a decimal, like 0.02, how do I turn it into a ratio?

The decimal already is the scale factor $k$ , so write $k$ as a fraction and simplify: $0.02 = \tfrac{2}{100} = \tfrac{1}{50}$ , giving the ratio $1:50$ . Going the other way, a ratio $1:n$ becomes the decimal $k = \tfrac{1}{n}$ . Both forms describe the identical relationship between map/model length and real length, so convert to whichever form the question is asking for.

Enlargement, maps and scale models

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads