Mathematics • Year 9 • Unit 4 • Lesson 18
Measurement and Geometry — Mixed Challenge
Combine SA/V formulas, the scale rules (k, k², k³), and the congruence/similarity tests from the lesson. Pick the right tool, find a classmate's mistake, and design your own scaled-up storage tank.
1. Mixed problems — choose the right tool
Each question uses a different mix from Lesson 18. Decide what rule applies first. 3 marks each
1.1 A cylinder has radius 5 cm and height 12 cm. (a) Find its volume in terms of π. (b) Find its surface area in terms of π.
1.2 Two similar rectangular prisms have lengths in ratio 3 : 7. (a) Find the ratio of their surface areas. (b) Find the ratio of their volumes. (c) If the smaller has volume 54 cm³, find the larger volume.
1.3 A triangle has angles 30°, 60°, 90°, with the shortest side 4 cm. A second triangle is similar with shortest side 12 cm. Find the length scale factor and state which similarity test from the lesson justifies your claim of similarity (SSS, SAS, or AA).
1.4 A composite solid is made of a 10 cm × 6 cm × 4 cm rectangular prism with a half-cylinder (r = 2 cm, length = 10 cm) sitting on top along the 10 cm edge. Find the total volume to 1 d.p.
1.5 Two similar spheres have radii 2 cm and 6 cm. The smaller has surface area 16π cm² and volume (32π/3) cm³. (a) Find k. (b) Find the larger SA. (c) Find the larger volume.
1.6 Two equilateral triangles have side lengths 5 cm and 5 cm. Are they congruent, similar, or both? Justify with the lesson's tests (SSS, SAS, ASA, RHS) — naming the specific test that applies.
2. Find the mistake
A classmate is asked: "Two similar cones have heights 5 cm and 15 cm; the smaller has volume 20π cm³. Find the larger volume." Their working follows. Exactly one line contains a mistake. Spot it, explain, then redo. 3 marks
Student's working:
Line 1: Length scale factor k = 15 ÷ 5 = 3.
Line 2: Volume scales by k² = 3² = 9.
Line 3: V_large = 9 × 20π = 180π cm³.
Line 4: So the larger cone has volume 180π cm³.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (use the lesson's scale rule).
(c) Re-do the working correctly, including the corrected final volume.
Stuck? Lesson § "Scale Factors" — length k, area k², volume k³. The student used k² (the area rule) where they should have used k³.3. Open-ended challenge — design a scaled-up tank
This question has many valid answers. 4 marks
3.1 A small cylindrical water tank has radius 0.5 m and height 1 m, holding 0.25π m³ ≈ 785 L. A community wants a similar tank (same shape, same height-to-radius ratio) that holds about 10 000 L.
Your answer must include:
(i) The required volume in m³ (convert 10 000 L).
(ii) The volume scale factor k³ = new V ÷ old V, and from that the length scale factor k.
(iii) The new tank's radius and height (each = k × the old value).
(iv) A reasonableness check: plug your new radius and height back into V = πr²h and confirm you get close to 10 000 L (within 10%).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cylinder r=5, h=12
(a) V = π × 5² × 12 = 300π cm³ ≈ 942.5 cm³.
(b) SA = 2π(5²) + 2π(5)(12) = 50π + 120π = 170π cm² ≈ 534.1 cm².
1.2 — Similar prisms 3:7
(a) Area ratio = 3² : 7² = 9 : 49.
(b) Volume ratio = 3³ : 7³ = 27 : 343.
(c) V_large = 54 × (343/27) = 54 × 12.7037… ≈ 686 cm³.
1.3 — 30°-60°-90° triangles
Length scale factor k = 12 ÷ 4 = 3.
Justified by AA (the angles 30°, 60°, 90° are the same in both — two equal angles is enough to prove similarity). SSS would also work because all corresponding sides are in the ratio 1 : 3.
1.4 — Composite solid
Prism volume = 10 × 6 × 4 = 240 cm³.
Half-cylinder volume = (1/2) × π × 2² × 10 = (1/2)(40π) = 20π ≈ 62.83 cm³.
Total ≈ 302.8 cm³.
1.5 — Similar spheres
(a) k = 6 ÷ 2 = 3.
(b) SA scales by k² = 9, so larger SA = 9 × 16π = 144π cm².
(c) Volume scales by k³ = 27, so larger V = 27 × (32π/3) = 9 × 32π = 288π cm³.
1.6 — Two equilateral triangles, both side 5
They are both congruent AND similar. Congruent by SSS (all three sides equal, 5 = 5 = 5). Similar with scale factor k = 1.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student used the area scale rule (k²) when they should have used the volume scale rule (k³). The lesson is clear: length scales by k, AREA scales by k², VOLUME scales by k³.
(c) Corrected working:
k = 15 ÷ 5 = 3 (correct).
Volume scales by k³ = 3³ = 27.
V_large = 27 × 20π = 540π cm³ ≈ 1 696.5 cm³.
3 — Open-ended challenge (sample solution)
(i) Required volume = 10 000 L = 10 m³.
(ii) Old volume = π × 0.5² × 1 = 0.25π ≈ 0.785 m³. Volume scale factor k³ = 10 ÷ 0.785 ≈ 12.74, so k = ³√12.74 ≈ 2.33.
(iii) New radius ≈ 0.5 × 2.33 = 1.165 m; new height ≈ 1 × 2.33 = 2.33 m.
(iv) Check: V = π × 1.165² × 2.33 = π × 1.357 × 2.33 ≈ 9.93 m³ ≈ 9 930 L. Within 1% of the 10 000 L target. ✓
Marking: 1 mark per part (i)–(iv). Award full marks for any valid k, new r/h pair that on the reasonableness check is within 10% of 10 000 L.