Mathematics • Year 9 • Unit 4 • Lesson 8
Similarity and Scale — Mixed Challenge
Pull together every idea from Lesson 8: identifying similar figures, the three scale rules ($k$, $k^2$, $k^3$), map scales, and working backwards from area or volume to length. You'll choose the right tool, spot someone else's mistake, and tackle an open-ended challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of length / area / volume scaling. Decide which power of $k$ applies before you start writing. Show your working. 3 marks each
1.1 Two similar rectangles have sides in ratio $2 : 7$. The smaller has area $20$ cm². Find the area of the larger.
1.2 Two similar cylinders have volumes in ratio $1 : 27$. What is the ratio of their radii? What is the ratio of their surface areas?
1.3 A map has scale $1 : 100\,000$. A rectangular farm measures $4$ cm × $5$ cm on the map. Find the real-life area of the farm in km².
1.4 A model train at scale $1 : 87$ has a carriage volume of $50$ cm³. Find the real carriage's volume in m³ (to 1 d.p.). (Hint: $1$ m³ $= 1\,000\,000$ cm³.)
1.5 Two equilateral triangles have areas $25$ cm² and $100$ cm². Find the ratio of their side lengths, and the ratio of their perimeters.
1.6 A solid metal sphere of radius $2$ cm is melted down and recast as a solid metal sphere of radius $4$ cm. How many of the small spheres are needed to make one large sphere? Justify your answer using scale factors (no need to compute volumes directly).
2. Find the mistake
Another student has tried to find the volume of a larger cube when the linear scale factor is $k = 3$ and the smaller cube has volume $12$ cm³. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — larger cube volume, $k = 3$, smaller cube $V = 12$ cm³:
Line 1: Linear scale factor $k = 3$.
Line 2: Volume scales as $k^2 = 9$.
Line 3: New volume $= 9 \times 12$ cm³.
Line 4: New volume $= 108$ cm³.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Revisit lesson § "Common Misconceptions" — the second listed pitfall is "volume scales as $k^2$" — that's exactly the trap here. Volume scales as $k^3$.3. Open-ended challenge — the giraffe problem
This question is the famous "square-cube law" applied to biology. 4 marks
3.1 Suppose a real giraffe is scaled up by a linear factor of $k = 2$ in every direction — so every length, including bone diameter and leg length, doubles.
(i) By what factor does the giraffe's weight increase? (Weight $\propto$ volume.)
(ii) By what factor does the cross-sectional area of one leg bone increase? (This area is what bears the weight.)
(iii) Compute the ratio weight per unit bone area for the new giraffe compared with the original. Show that it is $\mathbf{k}$ times higher.
(iv) Explain in two sentences why this means real animals cannot scale up indefinitely — i.e. why giant kaiju-style creatures from films could not physically exist.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Rectangles ratio 2 : 7
$k = 7/2 = 3.5$. $k^2 = 12.25$. Larger area $= 12.25 \times 20 = \mathbf{245}$ cm².
1.2 — Cylinders, volumes 1 : 27
$k^3 = 27$, so $k = 3$. Ratio of radii $= \mathbf{1 : 3}$.
Ratio of surface areas $= k^2 : 1 = \mathbf{1 : 9}$.
1.3 — Map 1 : 100 000, farm 4 × 5 cm
Real dimensions: $4 \times 100\,000 = 400\,000$ cm $= 4$ km, and $5 \times 100\,000 = 500\,000$ cm $= 5$ km.
Real area $= 4 \times 5 = \mathbf{20}$ km².
Check via $k^2$: map area $20$ cm² $\times (100\,000)^2 = 2 \times 10^{11}$ cm² $= 20$ km². ✓
1.4 — Model train (1 : 87)
$k = 87$, so $k^3 = 658\,503$.
Real volume $= 50 \times 658\,503 = 32\,925\,150$ cm³ $\approx 32.9$ m³ — call it $\mathbf{32.9}$ m³.
1.5 — Triangles, areas 25 and 100
$k^2 = 100/25 = 4$, so $k = 2$. Ratio of side lengths $= \mathbf{1 : 2}$.
Perimeter is a length, so ratio of perimeters $= \mathbf{1 : 2}$ as well.
1.6 — Spheres radius 2 → radius 4
$k = 4/2 = 2$. Big sphere volume $= k^3 = 8\times$ small sphere volume.
Therefore $\mathbf{8}$ small spheres are needed to make one large sphere.
No need to compute $\tfrac{4}{3}\pi r^3$ at all — the scale rule does the heavy lifting.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) Volume scales as $k^3$, not $k^2$. The student has used the area rule instead of the volume rule. The lesson states this explicitly: lengths × $k$, areas × $k^2$, volumes × $k^3$.
(c) Corrected working:
Linear scale factor $k = 3$.
Volume scales as $k^3 = 27$.
New volume $= 27 \times 12$ cm³
New volume $= \mathbf{324}$ cm³.
This is exactly the second misconception listed in the lesson — confusing $k^2$ and $k^3$.
3 — Open-ended challenge (sample solution)
(i) Weight scales as $k^3 = 2^3 = \mathbf{8\times}$.
(ii) Bone cross-sectional area scales as $k^2 = 2^2 = \mathbf{4\times}$.
(iii) Weight per unit area (= pressure on the bone) scales by $\dfrac{k^3}{k^2} = k = \mathbf{2\times}$. The new giraffe's bones must endure twice the pressure that a normal giraffe's bones do.
(iv) In two sentences: "Doubling all dimensions makes the giraffe $8\times$ heavier but only quadruples the bone cross-section, so each square cm of bone must support twice as much weight. At some scale the bone cannot endure this pressure and the animal would collapse — which is why a giraffe-shaped animal cannot simply be 'scaled up' to elephant-or-larger sizes without changing its proportions."
Marking: 1 for each of (i)-(iii); 1 for a clear two-sentence biological explanation in (iv). This is real physics — known as the "square-cube law" and explains why elephants have such thick legs compared with mice.