7.3 Scale Drawings, Classroom Floor Plans, Map Ratios, and Area Scaling

Understanding Ratio Scales, Statement Scales, and Linear Scale Bars

Teachers interpret scale drawings when planning classroom furniture layouts, reading school master plans, and organising bushwalking or city excursions. A scale expresses the constant multiplicative ratio between a linear measurement on a plan or map and the corresponding real-world ground distance.

Scales are presented in three standard formats:

1. Ratio Scales ($1 : n$): Both numbers share the same unit. A scale of $1 : 100$ means $1\text{ cm}$ on the drawing represents $100\text{ cm}$ ($= 1\text{ m}$) in reality, and $1\text{ mm}$ on the drawing represents $100\text{ mm}$ ($= 0.1\text{ m} = 10\text{ cm}$) in reality.
2. Statement (Word) Scales: State the mixed-unit equivalence directly in words, such as "$1\text{ cm represents }5\text{ m}$" or "$2\text{ cm to }1\text{ km}$".
3. Linear (Graphic) Scale Bars: A ruled visual bar printed on the map showing real-world distances. If a map is enlarged or reduced on a photocopier or projector screen, a printed ratio like $1 : 25{,}000$ becomes invalid, whereas the linear scale bar shrinks or stretches at the exact same rate as the map and remains accurate.

Fast Metric Conversion Rules for Ratio Scales

Because $1\text{ m} = 100\text{ cm}$ and $1\text{ km} = 1{,}000\text{ m} = 100{,}000\text{ cm}$, you can convert any $1 : n$ scale into a practical unit rate in seconds:

• Centimetres to Metres: Divide $n$ by $100$ (move the decimal point $2$ places left). At $1 : 250$, $1\text{ cm}$ on the plan represents $2.5\text{ m}$.
• Centimetres to Kilometres: Divide $n$ by $100{,}000$ (move the decimal point $5$ places left). At $1 : 50{,}000$, $1\text{ cm}$ on the map represents $0.5\text{ km}$ ($500\text{ m}$).

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Converting Between Map Distances, Ground Distances, and Travel Time

1. From Drawing Measurement to Real-World Distance (Multiply)

First write down the unit equivalence for $1\text{ cm}$ (or $1\text{ mm}$), then multiply by the map measurement:

$$\text{Real-World Distance} = (\text{Map Distance in cm}) \times (\text{Real-World Distance per cm})$$

2. From Real-World Distance to Drawing Measurement (Divide)

Divide the real-world length by the real-world distance represented by $1\text{ cm}$:

$$\text{Drawing Length (cm)} = \frac{\text{Real-World Distance}}{\text{Real-World Distance per cm}}$$

For example, if a $30\text{ m}$ swimming pool is drawn on a $1 : 200$ campus plan (where $1\text{ cm} = 2\text{ m}$), its length on paper is $\frac{30\text{ m}}{2\text{ m/cm}} = 15\text{ cm}$.

Worked Example 1: Bushwalking Excursion Distance and Arrival Time

A Year 9 outdoor education class uses a topographic map with a scale of $1 : 25{,}000$. The walking track from the picnic ground to the waterfall measures $18\text{ cm}$ along the trail on the map. If the group departs at $9\text{:}15\text{ am}$, walks at an average speed of $3.6\text{ km/h}$, and takes a $20\text{-minute}$ morning tea break along the way, at what time will they reach the waterfall?

1. Convert the ratio scale: At $1 : 25{,}000$, $1\text{ cm}$ represents $25{,}000\text{ cm} = 250\text{ m} = 0.25\text{ km}$.
2. Find the real-world trail distance:

$$d = 18 \times 0.25\text{ km} = 4.5\text{ km}$$

1. Find the walking time: Using $t = \frac{d}{s}$:

$$t = \frac{4.5\text{ km}}{3.6\text{ km/h}} = \frac{45}{36}\text{ hours} = \frac{5}{4}\text{ hours} = 1.25\text{ hours} = 1\text{ hour }15\text{ minutes}$$

1. Add the rest stop and compute arrival time: Total elapsed time is $1\text{ hour }15\text{ minutes} + 20\text{ minutes} = 1\text{ hour }35\text{ minutes}$. Starting from $9\text{:}15\text{ am}$, the group arrives at $10\text{:}50\text{ am}$.

Worked Example 2: Choosing the Largest Scale That Fits a Printable Page

A teacher wants to draw a scale plan of a rectangular school assembly quadrangle measuring $54\text{ m}$ long by $36\text{ m}$ wide inside a printable worksheet box measuring $28\text{ cm}$ wide by $19\text{ cm}$ high. Of the standard scales $1 : 100$, $1 : 200$, and $1 : 500$, which gives the largest drawing that fits completely inside the box?

1. Test $1 : 100$ ($1\text{ cm} = 1\text{ m}$): Drawing dimensions would be $54\text{ cm} \times 36\text{ cm}$, which is too large for the $28\text{ cm} \times 19\text{ cm}$ box.
2. Test $1 : 200$ ($1\text{ cm} = 2\text{ m}$): Drawing length $= 54 \div 2 = 27\text{ cm}$ ($\le 28\text{ cm}$) and drawing width $= 36 \div 2 = 18\text{ cm}$ ($\le 19\text{ cm}$). Both dimensions fit with room to spare.
3. Therefore, $1 : 200$ produces the largest drawing that fits inside the printable area.

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The Area Scale Factor Trap: Scaling Linear Dimensions vs. Area

One of the most common numeracy errors occurs when candidates calculate an area in $\text{cm}^2$ on a floor plan and multiply by the linear scale factor $k$ instead of $k^2$.

When every linear dimension is multiplied by a scale factor $k$, area is multiplied by $k^2$:

$$\text{Real-World Area} = \text{Drawing Area} \times k^2$$

Consider a $1\text{ cm} \times 1\text{ cm}$ square ($1\text{ cm}^2$) on a $1 : 200$ floor plan:

• Each $1\text{ cm}$ side represents $2\text{ m}$ ($200\text{ cm}$) in reality.
• The real-world area of that $1\text{ cm}^2$ square is $2\text{ m} \times 2\text{ m} = 4\text{ m}^2$ (which equals $200\text{ cm} \times 200\text{ cm} = 40{,}000\text{ cm}^2$, not $200\text{ cm}^2$).

Golden Rule for Floor-Plan Area Problems: Never compute area in $\text{cm}^2$ first. Always convert each linear length and width on the plan into real-world metres first, and then multiply the real-world metres together to obtain the area directly in $\text{m}^2$.

Worked Example 3: Flooring Cost from a School Library Plan

On an architectural plan drawn at a scale of $1 : 200$, a rectangular school library measures $9.5\text{ cm}$ long by $6.0\text{ cm}$ wide. Commercial carpet tiles cost $40 per square metre ($\text{m}^2$). What is the total cost to carpet the library floor?

1. Convert the linear dimensions to real-world metres first: At $1 : 200$, $1\text{ cm}$ represents $2\text{ m}$.
   • Real-world length $= 9.5 \times 2\text{ m} = 19\text{ m}$.
   • Real-world width $= 6.0 \times 2\text{ m} = 12\text{ m}$.
2. Calculate the real-world floor area:

$$\text{Area} = 19\text{ m} \times 12\text{ m} = 228\text{ m}^2$$

1. Calculate the carpeting cost:

$$\text{Cost} = 228 \times 40 = 9{,}120\text{ dollars}$$