Similarity and Enlargement
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective
Understand the fundamental geometric properties of enlargement, scale factor (\(k\)), centre of enlargement, and how linear dimensions and areas transform under similarity.
(a) Concrete Scenario
Consider an artist in Nairobi painting a mural of a Kenyan crest onto a community hall wall from an A4 paper draft. To ensure that every emblem remains identical in proportion, the artist measures rays extending outward from a reference point (the centre of enlargement) and scales every distance by the same multiplier \(k\).
(b) Geometric Insight
Two shapes are similar if corresponding angles are equal and corresponding side lengths are proportional. Under an enlargement with scale factor \(k\):
- Angles remain invariant (completely unchanged).
- Linear lengths change by factor \(k\).
- Surface areas change by factor \(k^2\).
(c) Scale Factor Regimes
The scale factor \(k\) defines the transformation:
- If \(k > 1\): The shape is enlarged (magnified).
- If \(0 < k < 1\): The shape is reduced (demagnified).
- If \(k = 1\): The image is congruent to the object.
- If \(k < 0\): The image is inverted through the centre of enlargement.
Key Formulas
Linear Scale Factor (\(k\))
\[k = \frac{\text{Image Length}}{\text{Object Length}} = \frac{A'B'}{AB} = \frac{B'C'}{BC}\]Perimeter Transformation
\[\text{Perimeter}_{\text{image}} = k \times \text{Perimeter}_{\text{object}}\]Area Transformation
\[\text{Area}_{\text{image}} = k^2 \times \text{Area}_{\text{object}}\]Area is 2-dimensional (length \(\times\) width), so both dimensions scale by \(k\), producing a net factor of \(k \times k = k^2\).
Coordinate Transformation about the Origin \((0,0)\)
\[(x, y) \xrightarrow{\text{Scale Factor } k} (k \cdot x,\; k \cdot y)\]Worked Examples
Example 1 (Easy): Finding an Enlarged Length
Problem: A triangular banner has a base of \(8\text{ cm}\). The printer enlarges the design by a linear scale factor of \(k = 3.5\). What is the base length of the enlarged banner?
- Identify the formula: \(\text{Length}_{\text{image}} = k \times \text{Length}_{\text{object}}\)
- Substitute values: \(\text{Length} = 3.5 \times 8\)
- Compute: \(3.5 \times 8 = 28\text{ cm}\)
Final Answer: The enlarged base is \(28\text{ cm}\).
Example 2 (Medium): Area Transformation
Problem: A survey map is drawn to a scale where \(1\text{ cm}\) represents \(20\text{ m}\) on the ground (linear scale factor \(k = 20\)). If a parcel of land has an area of \(15\text{ cm}^2\) on the map, what is its actual area on the ground in \(\text{m}^2\)?
- Identify the area scale factor: Area factor \(= k^2 = 20^2 = 400\)
- Multiply object area by \(k^2\): \(\text{Actual Area} = 15 \times 400\)
- Calculate: \(15 \times 400 = 6000\text{ m}^2\)
Final Answer: The actual land area is \(6000\text{ m}^2\).
Example 3 (Hard): Intercept Theorem / Similar Triangles
Problem: In \(\triangle ABC\), line \(DE\) is drawn parallel to base \(BC\), with \(D\) on \(AB\) and \(E\) on \(AC\). Given \(AD = 6\text{ cm}\), \(DB = 4\text{ cm}\), and \(BC = 15\text{ cm}\), find the length of \(DE\).
- Establish similarity: Since \(DE \parallel BC\), \(\angle ADE = \angle ABC\) and \(\angle AED = \angle ACB\). Thus \(\triangle ADE \sim \triangle ABC\).
- Find the total length of \(AB\): \(AB = AD + DB = 6 + 4 = 10\text{ cm}\).
- Set up the side length ratio: \[\frac{DE}{BC} = \frac{AD}{AB} \implies \frac{DE}{15} = \frac{6}{10}\]
- Solve for \(DE\): \(DE = 15 \times \frac{6}{10} = 15 \times 0.6 = 9\text{ cm}\).
Final Answer: The length of \(DE\) is \(9\text{ cm}\).
Common Mistakes
Misconception 1: Adding the Scale Factor Instead of Multiplying
Misconception 2: Scaling Area by \(k\) Instead of \(k^2\)
Misconception 3: Believing Angles Enlarge with Side Lengths
Real World
Agricultural Plot Planning (Shamba Surveying)
Cadastral surveyors across Kenya create title deed deed-plans using similarity. A boundary measuring \(5\text{ cm}\) on a \(1:2500\) survey plan corresponds to an actual farm fence line of \(5 \times 2500\text{ cm} = 125\text{ m}\). Accurate scaling ensures landowners avoid land boundary disputes.
Shadow Reckoning (Thales' Method)
Engineers and tree fellers measure tall structures (like cellular towers or high-voltage electric pylons) without climbing them. By measuring the shadow of a simple \(1\text{-metre}\) vertical rod and the shadow of the tall tower at the same time of day, similar right-angled triangles provide the exact height instantly via the ratio \(\frac{H_{\text{tower}}}{L_{\text{shadow}}} = \frac{H_{\text{rod}}}{l_{\text{shadow}}}\).
Practice