Scale Factors, Scale Diagrams and Maps

A scale tells you how the lengths on a drawing, map or model compare with the lengths in real life. It is written as a ratio.

Scale ratio

$\text{drawing length} : \text{real length}$

A scale of $1 : 200$ means that $1$ unit on the drawing represents $200$ of the same units in real life.
$1$ cm on the drawing $\to$ $200$ cm in real life.
$1$ mm on the drawing $\to$ $200$ mm in real life.
The units on both sides are always the same, which is why a scale ratio has no units written after it.

Only two things can happen, and it is worth being certain which one you need.

You know…	You want…	Do this
Map / drawing length	Real length	Multiply by the scale number
Real length	Map / drawing length	Divide by the scale number

Read the scale and identify the multiplier, e.g. $1 : 25\,000 \Rightarrow$ multiplier $25\,000$.
Decide the direction (map $\to$ real is $\times$; real $\to$ map is $\div$).
Do the multiplication or division, keeping the same unit throughout.
Convert to the unit the question asks for (usually cm $\to$ m $\to$ km).
Sense check: real distances should be big, map distances small.

Worked Example 1 — Map to real distance

On a map with scale $1 : 25\,000$, a footpath is $8.4$ cm long. Find its real length in kilometres.

①Map $\to$ real, so multiply: $8.4 \times 25\,000 = 210\,000$ cm

②cm $\to$ m: $210\,000 \div 100 = 2100$ m

③m $\to$ km: $2100 \div 1000 = 2.1$ km

Answer: $2.1$ km

Worked Example 2 — Real distance to map

Two villages are $6.5$ km apart. How far apart are they on a $1 : 50\,000$ map?

①Convert to cm first: $6.5$ km $= 6.5 \times 1000 = 6500$ m $= 650\,000$ cm

②Real $\to$ map, so divide: $650\,000 \div 50\,000 = 13$

Answer: $13$ cm on the map

Shortcut for $1 : 100\,000$-style scales: on a $1 : 25\,000$ map, $1$ cm $= 25\,000$ cm $= 250$ m $= 0.25$ km. So "cm on the map $\times 0.25 =$ km in real life". On a $1 : 50\,000$ map, $1$ cm $= 0.5$ km.

Some scales are written in words, such as "$1$ cm represents $4$ km". These are easier to use, but you may be asked to rewrite them as a ratio.

Worked Example 3 — Turning a worded scale into a ratio

A map scale is "$1$ cm represents $4$ km". Write this as a ratio in the form $1 : n$.

①Both sides must use the same unit, so change km into cm.

②$4$ km $= 4 \times 1000 = 4000$ m $= 4000 \times 100 = 400\,000$ cm

③Ratio $= 1\text{ cm} : 400\,000\text{ cm} = 1 : 400\,000$

Answer: $1 : 400\,000$

Worked Example 4 — The other way round

A model railway is built to a scale of $1 : 76$. Complete: "$1$ cm on the model represents ___ cm in real life", and find the real length of a carriage that is $23$ cm long on the model.

①$1$ cm on the model represents $76$ cm in real life.

②Real length $= 23 \times 76 = 1748$ cm

③$1748 \div 100 = 17.48$ m

Answer: $17.48$ m

If you know a pair of matching lengths, you can work out the scale by writing them as a ratio and simplifying to $1 : n$.

Method

$\text{scale} = 1 : \dfrac{\text{real length}}{\text{drawing length}}$ (same units!)

Worked Example 5 — Working out the scale

A plan of a school hall is drawn so that the hall, which is really $30$ m long, appears as $12$ cm on the plan. Find the scale in the form $1 : n$.

①Same units: $30$ m $= 3000$ cm

②Ratio $= 12 : 3000$

③Divide both sides by $12$: $\;1 : 250$

Answer: scale $= 1 : 250$

A scale drawing is an accurate diagram in which every length has been reduced (or enlarged) by the same scale factor, and every angle is kept exactly the same.

Bearings are angles used for directions. They are:
• measured from North
• measured clockwise
• always written with three figures, e.g. $072^\circ$, $145^\circ$, $310^\circ$.

Choose or read the scale (e.g. $1$ cm represents $5$ km).
Convert every real distance into a drawing length by dividing.
Draw a North line at the starting point.
Measure the bearing clockwise from North with a protractor.
Measure the drawing length along that direction with a ruler.
To answer, measure the drawing and multiply back up to real life.

Worked Example 6 — Constructing a scale drawing

A ship sails $36$ km on a bearing of $110^\circ$, then $24$ km on a bearing of $200^\circ$. Using a scale of $1$ cm : $6$ km, describe how to find the distance and bearing of the ship from its start.

①First leg on paper: $36 \div 6 = 6$ cm at $110^\circ$ from North.

②Draw a new North line at the end of that leg.

③Second leg on paper: $24 \div 6 = 4$ cm at $200^\circ$.

④Join the start to the finish and measure: about $7.2$ cm.

⑤Real distance $= 7.2 \times 6 = 43.2$ km, and measure the bearing of that line from the original North line (about $144^\circ$).

Note: the two legs meet at $90^\circ$ (since $200 - 110 = 90$), so we can check with Pythagoras: $\sqrt{36^2 + 24^2} = \sqrt{1296+576} = \sqrt{1872} = 43.3$ km ✓

Watch out: in scale-drawing questions the marks are for accuracy. Use a sharp pencil, and answers are usually accepted within $\pm 2$ mm and $\pm 2^\circ$.

If lengths are multiplied by $n$, areas are multiplied by $n^2$.

Worked Example 7 — Area from a plan

On a plan with scale $1 : 200$, a garden is drawn as a rectangle $9$ cm by $6$ cm. Find the real area in $\text{m}^2$.

Method 1 — convert the lengths first (safest):

①Real lengths: $9 \times 200 = 1800$ cm $= 18$ m, and $6 \times 200 = 1200$ cm $= 12$ m.

②Real area $= 18 \times 12 = 216\text{ m}^2$

Method 2 — use the area scale factor:

③Plan area $= 9 \times 6 = 54\text{ cm}^2$

④Area factor $= 200^2 = 40\,000$, so real area $= 54 \times 40\,000 = 2\,160\,000\text{ cm}^2$

⑤$2\,160\,000 \div 10\,000 = 216\text{ m}^2$ ✓

Reading a scale

$1 : n$ means "drawing : real". Same units on both sides, so no units are written.

Map → real

Multiply by $n$, then convert cm → m → km.

Real → map

Convert to cm first, then divide by $n$.

Handy facts

$1:25\,000$: $1$ cm $=0.25$ km. $1:50\,000$: $1$ cm $=0.5$ km.

Finding a scale

Write drawing : real in the same units, then divide both by the drawing length.

Bearings

From North, clockwise, three figures. Draw a fresh North line at each new point.

Areas

Length factor $n$ → area factor $n^2$. Safer to convert lengths first.

Angles

Scale drawings preserve angles exactly — only lengths change.

Question 1

A map has a scale of $1 : 20\,000$. Two churches are $5$ cm apart on the map. Find the real distance in metres.

▶ Show solution

Map → real, so multiply: $5 \times 20\,000 = 100\,000$ cm.

$100\,000 \div 100 = 1000$ m.

Answer: $1000$ m (or $1$ km)

Question 2

On a $1 : 25\,000$ map, how long is a $3.5$ km road?

▶ Show solution

$3.5$ km $= 3500$ m $= 350\,000$ cm.

Real → map, so divide: $350\,000 \div 25\,000 = 14$ cm.

Answer: $14$ cm

Question 3

Write the scale "$1$ cm represents $250$ m" in the form $1 : n$.

▶ Show solution

Convert $250$ m into cm: $250 \times 100 = 25\,000$ cm.

Ratio $= 1 : 25\,000$.

Question 4

A model aeroplane is $42$ cm long. The real aeroplane is $33.6$ m long. Find the scale of the model in the form $1 : n$.

▶ Show solution

Same units: $33.6$ m $= 3360$ cm.

Ratio $= 42 : 3360$.

Divide both by $42$: $\;1 : 80$.

Answer: $1 : 80$

Question 5

A floor plan uses a scale of $1 : 50$. A kitchen wall is $6.4$ m long in real life. How long is it on the plan?

▶ Show solution

$6.4$ m $= 640$ cm.

$640 \div 50 = 12.8$ cm.

Answer: $12.8$ cm

Question 6

A map uses a scale of $1 : 100\,000$. A lake is drawn with area $6\text{ cm}^2$. Find the real area of the lake in $\text{km}^2$.

▶ Show solution

$1$ cm on the map $= 100\,000$ cm $= 1000$ m $= 1$ km.

So $1\text{ cm}^2$ on the map $= 1\text{ km}^2$ in real life.

Answer: $6\text{ km}^2$

Question 7

Town B is $45$ km due East of town A. Town C is $60$ km due North of town B. On a scale drawing using $1$ cm : $15$ km, (a) what lengths would you draw, and (b) what is the real straight-line distance from A to C?

▶ Show solution

(a) AB $= 45 \div 15 = 3$ cm. BC $= 60 \div 15 = 4$ cm.

(b) The angle at B is $90^\circ$, so by Pythagoras:

$AC = \sqrt{45^2 + 60^2} = \sqrt{2025 + 3600} = \sqrt{5625} = 75$ km.

(On the drawing this would measure $75 \div 15 = 5$ cm ✓)

Question 8

The bearing of Q from P is $065^\circ$. Find the bearing of P from Q.

▶ Show solution

A back bearing differs by $180^\circ$.

Since $065^\circ < 180^\circ$, add: $65 + 180 = 245$.

Answer: $245^\circ$

Rule: if the bearing is less than $180^\circ$, add $180^\circ$; if it is more than $180^\circ$, subtract $180^\circ$.

Question 9

A walker measures a route on a $1 : 25\,000$ map as $18.6$ cm. She walks at an average speed of $4$ km/h. How long will the walk take, in hours and minutes?

▶ Show solution

Real distance $= 18.6 \times 25\,000 = 465\,000$ cm.

$465\,000 \div 100 = 4650$ m $= 4.65$ km.

Time $= \dfrac{4.65}{4} = 1.1625$ hours.

$0.1625 \times 60 = 9.75$ minutes $\approx 10$ minutes.

Answer: about $1$ hour $10$ minutes

Question 10

A scale model of a building uses a scale of $1 : 150$. The model has a rectangular base measuring $24$ cm by $16$ cm, and its volume is $9600\text{ cm}^3$.

(a) Find the real base area in $\text{m}^2$. (b) Find the real volume in $\text{m}^3$.

▶ Show solution

(a) Real lengths: $24 \times 150 = 3600$ cm $= 36$ m; $16 \times 150 = 2400$ cm $= 24$ m.

Real area $= 36 \times 24 = 864\text{ m}^2$.

(b) Volume scale factor $= 150^3 = 3\,375\,000$.

Real volume $= 9600 \times 3\,375\,000 = 3.24 \times 10^{10}\text{ cm}^3$.

$1\text{ m}^3 = 1\,000\,000\text{ cm}^3$, so real volume $= 3.24 \times 10^{10} \div 10^6 = 32\,400\text{ m}^3$.