Scale Drawing
Interactive lessons, worked examples, and problem-solving techniques to help Kenyan students master KJSEA, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Master the geometry and algebra of scale drawings, three-figure bearings, and angles to represent real-world spatial relationships accurately on paper.
Concrete Scenario: Imagine mapping out a school compound in Nakuru from the main gate to the science laboratory. The true ground distance is \(60\text{ m}\) along a specific direction—such as \(060^\circ\) from North. Because a \(60\text{ m}\) sheet of paper is impossible to carry, we shrink every distance by a fixed ratio called the scale factor, while keeping all angles and directions identical.
Geometric Insight: A scale drawing produces a similar figure to the actual terrain. Dilatations preserve angles exactly; therefore, a bearing measured on the ground with a magnetic compass is drawn with the exact same angle on paper using a protractor and a reference North line.
Algebraic Principles:
- Representative Fraction (Scale): A scale of \(1 : n\) means \(1\text{ unit on paper} = n\text{ units in reality}\).
- Scale Factor: \(s = \frac{1}{n}\). Thus, \(L_{\text{draw}} = \frac{L_{\text{real}}}{n}\) and \(L_{\text{real}} = L_{\text{draw}} \times n\). (Ensure units match before multiplying or dividing!)
- Three-Figure Bearings: Angles measured strictly clockwise from True North, written with three digits (e.g., \(045^\circ\), \(090^\circ\), \(230^\circ\)).
Scale Slider Map — Gate to Admin Block (with Compass Rose)
Adjust the scale and bearing to see how the drawing length changes while the bearing stays true. The compass rose overlay lets you read the bearing directly.
Key Formulas
Used when translating field measurements to paper.
Used when reading actual distances off a blueprint or topographical map.
Measured in a clockwise direction starting from True North (\(000^\circ\)).
The direction from the destination back to the starting point.
Area scales with the square of the linear scale factor.
Worked Examples
Problem: A surveyor measures a boundary wall of length \(85\text{ m}\) on a farm. Determine the length of this wall on a blueprint with a scale of \(1 : 500\).
- Convert units to centimetres: \[ L_{\text{real}} = 85\text{ m} = 85 \times 100\text{ cm} = 8500\text{ cm} \]
- Apply the scale ratio \(1 : 500\): \[ L_{\text{draw}} = \frac{8500\text{ cm}}{500} = 17\text{ cm} \]
- Conclusion: The line drawn on the map is \(17\text{ cm}\).
Problem: On a topographical map with scale \(1 : 25\,000\), the vector from Matatu Terminal \(A\) to Market \(B\) measures \(4.8\text{ cm}\) on a bearing of \(075^\circ\). Calculate the real ground distance in kilometres, and state the bearing of \(A\) from \(B\) (the back bearing).
- Calculate real ground distance: \[ L_{\text{real}} = 4.8\text{ cm} \times 25\,000 = 120\,000\text{ cm} \]
- Convert to kilometres: \[ 120\,000\text{ cm} = \frac{120\,000}{100}\text{ m} = 1200\text{ m} = 1.2\text{ km} \]
- Find the back bearing (direction of \(A\) from \(B\)): Since the forward bearing is \(\theta = 075^\circ < 180^\circ\): \[ \text{Back Bearing} = 075^\circ + 180^\circ = 255^\circ \]
- Conclusion: The actual distance is \(1.2\text{ km}\) and the bearing of \(A\) from \(B\) is \(255^\circ\).
Problem: A triangular piece of land \(PQR\) is surveyed. From \(P\), a boundary runs due North (\(000^\circ\)) for \(60\text{ m}\) to \(Q\). From \(Q\), another boundary runs due East (\(090^\circ\)) for \(80\text{ m}\) to \(R\). Using a scale of \(1 : 1000\), determine the drawing lengths \(PQ\) and \(QR\), the length of \(PR\) on paper, and the actual land area in square metres.
- Calculate drawing lengths: \[ PQ_{\text{draw}} = \frac{60 \times 100\text{ cm}}{1000} = 6\text{ cm} \] \[ QR_{\text{draw}} = \frac{80 \times 100\text{ cm}}{1000} = 8\text{ cm} \]
- Find map hypotenuse \(PR_{\text{draw}}\) using Pythagoras: Since North and East are perpendicular (\(\angle PQR = 90^\circ\)): \[ PR_{\text{draw}} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm} \]
- Calculate the actual area: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 80\text{ m} \times 60\text{ m} = 2400\text{ m}^2 \]
Common Mistakes
Why it feels right: In Cartesian coordinate geometry, standard angles are measured anticlockwise from the positive \(x\)-axis (East).
Correction: In navigation and surveying, bearings are ALWAYS measured clockwise starting from True North (\(000^\circ\)). An angle of \(60^\circ\) East of North is written as \(060^\circ\).
Why it feels right: Learners often divide metres directly by the scale factor (e.g., dividing \(120\text{ m}\) by \(500\) to get \(0.24\)) without specifying units, leading to confusion between metres, centimetres, and millimetres.
Correction: Always convert the real-world distance to centimetres (multiply by \(100\)) before dividing by the scale denominator \(n\).
Why it feels right: If lengths scale by \(1:500\), it is tempting to assume area also scales by \(500\).
Correction: Area has two dimensions (length \(\times\) width). Therefore, area scales by the square of the linear scale factor: \(\text{Area}_{\text{real}} = \text{Area}_{\text{draw}} \times n^2\).
Real World
Practice