Mathematics • Year 9 • Unit 4 • Lesson 8
Similarity and Scale Factors
Build fluency with the three scale rules: lengths scale by $k$, areas by $k^2$, and volumes by $k^3$ — one step at a time, from a fully worked example through guided practice to independent problems.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Two similar triangles have sides in the ratio $2 : 5$. The smaller triangle has area $12$ cm². Find the area of the larger triangle.
Step 1 — Identify the scale factor $k$ (length).
$k = \dfrac{\text{new length}}{\text{original length}} = \dfrac{5}{2} = 2.5$
Reason: the lesson defines $k$ as the ratio of corresponding sides — new over original. Here "new" = larger.
Step 2 — Spot the rule: area scales as $k^2$.
$\text{New area} = k^2 \times \text{old area}$
Reason: when every length is scaled by $k$, area (which is length × length) is scaled by $k \times k = k^2$.
Step 3 — Compute $k^2$.
$k^2 = (2.5)^2 = 6.25$
Reason: square the linear scale factor, not the area itself.
Step 4 — Multiply.
$\text{Larger area} = 6.25 \times 12 = 75$ cm²
Answer: $\mathbf{75}$ cm².
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Two similar cubes have side lengths in the ratio $1 : 4$. The smaller cube has volume $5$ cm³. Find the volume of the larger cube.
Step 1 — Identify the scale factor: $k = \dfrac{\text{new}}{\text{old}} = \dfrac{\_\_\_}{\_\_\_} = \_\_\_\_$.
Step 2 — Spot the rule: volume scales as $k^{\_\_}$.
Step 3 — Compute $k^{\_\_}$:
$k^{\_\_} = \_\_\_^{\_\_} = \_\_\_\_$
Step 4 — Multiply by the old volume:
$\text{New volume} = \_\_\_\_ \times 5 = \_\_\_\_$ cm³
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (find $k$ or apply once). The middle two are standard (combine ideas). The last two are extension (solve backwards or with maps).
Foundation — single application
3.1 Find the scale factor $k$ if a $4$ cm line is enlarged to $10$ cm. 1 mark
3.2 Two similar squares have sides in ratio $1 : 3$. If the smaller square has area $5$ cm², find the area of the larger. 1 mark
3.3 Two similar cubes have sides in ratio $1 : 2$. The smaller has volume $7$ cm³. Find the larger volume. 1 mark
3.4 A rectangle is enlarged by scale factor $k = 4$. By what factor is its area enlarged? 1 mark
Standard — combine two ideas
3.5 Two similar cones have heights $3$ cm and $9$ cm. The smaller cone has volume $20$ cm³. Find the larger cone's volume. 2 marks
3.6 A photograph is enlarged so that each length is $2.5\times$ the original. If the original area is $48$ cm², find the new area. 2 marks
Extension — push your thinking
3.7 Two similar shapes have areas $16$ cm² and $49$ cm². Find the ratio of their corresponding sides (i.e. find $k$). 2 marks
3.8 A model car is built at scale $1 : 24$ (model : real). The real car has length $4.8$ m. Find (a) the model's length in cm, and (b) the ratio of model volume to real volume. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (cubes ratio $1:4$)
Step 1: $k = \dfrac{\mathbf{4}}{\mathbf{1}} = \mathbf{4}$.
Step 2: volume scales as $k^{\mathbf{3}}$.
Step 3: $k^{\mathbf{3}} = \mathbf{4}^{\mathbf{3}} = \mathbf{64}$.
Step 4: New volume $= \mathbf{64} \times 5 = \mathbf{320}$ cm³.
3.1 — Scale factor for 4 → 10
$k = \dfrac{10}{4} = \mathbf{2.5}$.
3.2 — Squares, ratio 1:3, smaller area 5
$k = 3$, so area $\times k^2 = 9$. Larger area $= 9 \times 5 = \mathbf{45}$ cm².
3.3 — Cubes, ratio 1:2, smaller volume 7
$k = 2$, so volume $\times k^3 = 8$. Larger volume $= 8 \times 7 = \mathbf{56}$ cm³.
3.4 — Enlarged by $k = 4$, area factor
Area scales as $k^2 = 4^2 = \mathbf{16\times}$.
3.5 — Cones, heights 3 and 9
$k = 9/3 = 3$, so $k^3 = 27$. Larger volume $= 27 \times 20 = \mathbf{540}$ cm³.
3.6 — Photograph enlarged by 2.5
Area $\times k^2 = (2.5)^2 = 6.25$. New area $= 6.25 \times 48 = \mathbf{300}$ cm².
3.7 — Areas 16 and 49, find $k$
Area ratio $k^2 = \dfrac{49}{16}$, so $k = \sqrt{\dfrac{49}{16}} = \dfrac{7}{4}$. Ratio of sides: $\mathbf{4 : 7}$ (or equivalently $k = 1.75$).
3.8 — Model car 1:24
(a) Model length $= 4.8 \div 24 = 0.2$ m $= \mathbf{20}$ cm.
(b) Volume ratio $= 1^3 : 24^3 = \mathbf{1 : 13\,824}$.
The real car has $\approx 13\,800$ times the model's volume — which is why a tiny die-cast car weighs so little compared with the real thing.