Mathematics • Year 8 • Unit 3 • Lesson 20
Similarity — Mixed Challenge
Pull everything from Lesson 20 together: finding k, missing sides in similar figures, area ratio k², perimeter ratio k, and the link between similarity and congruence. Six mixed problems, one "find the mistake", and one open-ended scale-factor design task.
1. Mixed problems — choose the right move
Each question pulls a different idea from Lesson 20. Decide whether you need k, k² (area) or both before you start writing. Show your working. 3 marks each
1.1 △ABC ∼ △DEF. AB = 8 cm and DE = 12 cm. Find k, then find DF given AC = 5 cm.
1.2 Two similar rectangles have scale factor k = 3. The smaller rectangle has area 18 cm². Find the area of the larger rectangle.
1.3 A pentagon has perimeter 30 cm and area 60 cm². It is enlarged with k = 1.5. Find the perimeter and area of the enlargement.
1.4 Two similar triangles have areas 50 cm² and 200 cm². Find the linear scale factor k between them (large to small). (Hint: work backwards from k².)
1.5 △PQR ∼ △XYZ with k = 0.4 (image is the SMALLER one). The image triangle has PQ = 6 cm. Find the corresponding side XY in the original (larger) triangle.
1.6 Two similar cones have radii in ratio 2 : 5. State (a) the ratio of their curved surface areas, and (b) the ratio of their volumes. (Recall: surface area scales as k² and volume scales as k³.)
2. Find the mistake
A Year 8 student has tried to find the area of an enlarged shape. 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 — Question: Two similar triangles have k = 3. The smaller has area 24 cm². Find the area of the larger.
Line 1: Scale factor: k = 3.
Line 2: Area scales by k.
Line 3: Large area = 24 × 3 = 72 cm².
Line 4: So the larger triangle has area 72 cm².
(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 § Card 5 (Spot the Trap) — area scales by k², not k. With k = 3, area multiplier is 9.3. Open-ended challenge — design a scale model
This question has more than one valid answer. 4 marks
3.1 A real basketball court is 28 m long and 15 m wide, with a total floor area of 420 m².
(a) Choose a scale (e.g. 1 : 50, 1 : 100, 1 : 200) so that the model fits onto a single A4 sheet (roughly 29 cm × 21 cm). State your chosen scale and the corresponding k (real → model).
(b) Using your chosen k, calculate the model's length and width in cm. Confirm both fit on A4.
(c) Calculate the area of the model floor in cm². Verify your answer using both methods: (i) model length × model width, and (ii) real area × k².
(d) If a student instead chose a scale of 1 : 1000, explain in one sentence why the model would be too small to read.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — △ABC ∼ △DEF
k = DE ÷ AB = 12 ÷ 8 = 1.5.
DF = AC × k = 5 × 1.5 = 7.5 cm.
1.2 — Area scaling, k = 3
Area ratio = k² = 9. Large area = 18 × 9 = 162 cm².
1.3 — Pentagon, k = 1.5
New perimeter = 30 × k = 30 × 1.5 = 45 cm (linear, × k).
New area = 60 × k² = 60 × 2.25 = 135 cm² (× k²).
1.4 — Areas 200 and 50
Area ratio = 200 ÷ 50 = 4. So k² = 4, giving k = √4 = 2 (linear scale factor, large to small).
1.5 — △PQR ∼ △XYZ, k = 0.4
Image is SMALLER, so image side = object side × k. We are given the image side PQ = 6 cm and need the object side XY.
XY = PQ ÷ k = 6 ÷ 0.4 = 15 cm.
1.6 — Cones, radii 2 : 5
k = 5/2 (or 2.5).
(a) Surface area ratio = k² = 6.25 = 25 : 4.
(b) Volume ratio = k³ = 15.625 = 125 : 8.
2 — Find the mistake
(a) The mistake is on Line 2 (carried into Line 3 and 4).
(b) The student wrote "area scales by k", but areas actually scale by k². Area is "length × length", so if every length is multiplied by k, the area is multiplied by k × k = k², not k.
(c) Corrected working:
Scale factor: k = 3.
Area scales by k² = 9.
Large area = 24 × 9 = 216 cm². ✓
Sanity check: with k = 3, every length triples; tripling both length AND width of a rectangle (or both dimensions of any shape) multiplies the area by 9, not 3.
3 — Scale model basketball court (sample solution)
Many scales work. Best choice: 1 : 100, giving k = 1/100 (real → model).
(a) Scale chosen: 1 : 100, so k = 1/100 = 0.01.
(b) Model dimensions: Length = 28 m × 0.01 = 0.28 m = 28 cm. Width = 15 m × 0.01 = 0.15 m = 15 cm. Both fit on A4 (which is ~29 cm × 21 cm). ✓
(c) Model area:
(i) Model length × model width = 28 × 15 = 420 cm².
(ii) Real area × k² = 420 m² × (1/100)² = 420 m² × 1/10 000 = 0.042 m² = 0.042 × 10 000 cm² = 420 cm². ✓
Both methods agree.
(d) At 1 : 1000, k = 1/1000, so 28 m → 2.8 cm and 15 m → 1.5 cm. The model would be smaller than a credit card — too small to mark out player positions or read the dimensions clearly.
Other valid scales: 1 : 50 gives a 56 cm × 30 cm model — too big for A4. 1 : 200 gives a 14 cm × 7.5 cm model — fits A4 comfortably but is a bit small. 1 : 100 is the cleanest choice.
Marking: 1 mark for a sensible scale + correct k; 1 mark for correctly computing the two model dimensions in cm; 1 mark for the area in cm² verified by BOTH methods (× k² and direct multiplication); 1 mark for a clear sentence in (d) about why too-small a scale is unusable.