Mathematics • Year 9 • Unit 4 • Lesson 20
Year 9 Capstone — Mixed Challenge
The final worksheet of Year 9 Maths. Six mixed problems pulling from probability (L16), trigonometry (L17), measurement (L18) and data (L19). One "find the mistake" on a multi-step solution, plus an open-ended STEM design challenge.
1. Mixed problems — every Unit 4 topic represented
Each question pulls from a different lesson in Unit 4. Identify the topic, pick the right tool, show working. 3 marks each
1.1 (L16 — probability) A die is rolled 1 800 times. Theoretical P(6) = 1/6. Actually, 6 appears 350 times. (a) How many sixes were expected? (b) Find experimental P(6). (c) Is there evidence of bias?
1.2 (L17 — trigonometry) A bushfire spotting tower is on top of a hill 80 m above the surrounding plain. The platform sits 12 m above the top of the hill. From the platform, a fire is spotted at an angle of depression of 4°. How far is the fire from the base of the hill (horizontal distance, 1 d.p.)?
1.3 (L18 — measurement) Two similar cones have heights 4 cm and 10 cm. The smaller has surface area 25π cm². Find the larger cone's surface area.
1.4 (L19 — data) Year 9 test scores out of 50: 22, 28, 31, 34, 36, 38, 40, 42, 45, 48. Find the median, Q1, Q3 and IQR.
1.5 (L16 + L19 — probability rules) A bag has 4 red, 3 blue and 5 yellow marbles. (a) Find P(red OR blue) on a single draw. (b) Two are drawn without replacement. Find P(both yellow).
1.6 (synthesis — L17 + L18) A square-based pyramid has base side 6 cm and slant height (from base edge midpoint to apex along the face) 5 cm. (a) Find the vertical height of the pyramid (use Pythagoras: slant² = height² + half-base²). (b) Find the volume of the pyramid using V = (1/3) × base area × height. (c) A similar pyramid has base side 18 cm. Use the scale factor to find its volume.
2. Find the mistake
A student tackles this multi-step problem: "From a 30 m cliff, the angle of depression to a swimmer is 12°. Find the slant (line-of-sight) distance from the observer to the swimmer." Their working is below. Exactly one line contains a mistake. Spot it, explain, then redo. 3 marks
Student's working:
Line 1: Angle of depression 12° = angle of elevation 12° at the swimmer (alternate angles).
Line 2: At the swimmer, opposite = 30 m, hypotenuse = slant distance (unknown, call it s).
Line 3: cos 12° = 30 ÷ s → s = 30 ÷ cos 12° ≈ 30.7 m.
Line 4: So the slant distance ≈ 30.7 m.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (refer to SOH CAH TOA and which side is opposite).
(c) Re-do the working correctly, including the corrected slant distance to 1 d.p. Then do a reasonableness check.
Stuck? Opposite + hypotenuse → use SIN, not cos. (Cosine uses adjacent + hypotenuse.)3. Open-ended challenge — STEM design brief
This is the capstone task for Year 9 Maths. Many valid answers exist. 5 marks
3.1 You are a junior civil engineer asked to design a cylindrical rainwater tank for a single-family home in Sydney. The brief: the tank must hold enough water for the family to survive a 60-day dry spell using 200 L/day, AND it must be no taller than 2.5 m so it doesn't block the neighbour's view.
Your answer must include:
(i) Calculate the minimum required capacity in litres, then convert to m³.
(ii) Decide on a height (≤ 2.5 m) and calculate the required radius using V = πr²h. Show the rearrangement.
(iii) Estimate the cost of the tank in materials: assume steel costs $80 per m² of surface area (SA = 2πr² + 2πrh). Show the calculation.
(iv) Build a 1 : 10 scale model. Give the model's dimensions in cm, and the model's capacity in litres (use the volume scale rule, k³).
(v) One sentence on which Lesson 20 reasonableness check you'd apply to verify your design before submitting it.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Die rolled 1 800 times
(a) Expected = 1 800 ÷ 6 = 300 sixes.
(b) Experimental P(6) = 350 ÷ 1 800 ≈ 0.194.
(c) Difference of 50 extra sixes from expected (350 vs 300) over 1 800 trials is larger than typical random variation — this gives some evidence of bias, and the die should be tested with even more rolls before a final judgement.
1.2 — Bushfire spotting tower
Total observer height above plain = 80 + 12 = 92 m.
Angle of depression 4° = angle of elevation 4° at the fire.
tan 4° = 92 ÷ d → d = 92 ÷ tan 4° ≈ 92 ÷ 0.0699 ≈ 1 315.6 m ≈ 1.32 km.
1.3 — Similar cones, surface area
Length scale k = 10 ÷ 4 = 2.5. Surface area scales by k² = 6.25.
Larger SA = 6.25 × 25π = 156.25π cm² ≈ 490.9 cm².
1.4 — Year 9 test scores
n = 10. Median = average of 5th and 6th values = (36 + 38) ÷ 2 = 37.
Lower half (5 values): 22, 28, 31, 34, 36 → Q1 = 31 (middle = 3rd).
Upper half (5 values): 38, 40, 42, 45, 48 → Q3 = 42.
IQR = 42 − 31 = 11.
1.5 — Marble bag
Total = 4 + 3 + 5 = 12.
(a) Red and blue are mutually exclusive (can't be both colours): P(red OR blue) = 4/12 + 3/12 = 7/12 ≈ 0.583.
(b) P(both yellow without replacement) = (5/12) × (4/11) = 20/132 = 5/33 ≈ 0.152.
1.6 — Square pyramid synthesis
(a) Half-base = 3 cm; slant = 5 cm. height² = 5² − 3² = 25 − 9 = 16; height = 4 cm.
(b) Base area = 6² = 36 cm². V = (1/3) × 36 × 4 = 48 cm³.
(c) Length scale k = 18 ÷ 6 = 3, so volume scales by k³ = 27. V_large = 27 × 48 = 1 296 cm³.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The 30 m cliff is the side OPPOSITE the 12° angle at the swimmer, and the slant distance is the hypotenuse. The correct ratio is sine (SOH: sin = opposite ÷ hypotenuse), not cosine (CAH: cos = adjacent ÷ hypotenuse). The student picked cosine, which is what you'd use if 30 m were the adjacent side, not the opposite.
(c) Corrected working:
sin 12° = 30 ÷ s
s = 30 ÷ sin 12°
s ≈ 30 ÷ 0.2079 ≈ 144.3 m.
Reasonableness check: 144.3 m > 30 m hypotenuse-vs-leg makes sense (hypotenuse is always longest). The student's answer of 30.7 m was barely bigger than the 30 m cliff — that was the give-away clue that it was wrong.
3 — Open-ended capstone (sample solution)
(i) Capacity required = 60 days × 200 L = 12 000 L = 12 m³.
(ii) Pick h = 2.5 m (the max). r² = V ÷ (π × h) = 12 ÷ (π × 2.5) ≈ 12 ÷ 7.854 ≈ 1.528, so r ≈ √1.528 ≈ 1.24 m. Check: π × 1.24² × 2.5 ≈ 12.07 m³. ✓
(iii) SA = 2π(1.24)² + 2π(1.24)(2.5) = 2π(1.538) + 2π(3.10) = 3.075π + 6.20π = 9.275π ≈ 29.14 m². Cost ≈ 29.14 × $80 ≈ $2 331.
(iv) 1 : 10 model: height = 25 cm, radius = 12.4 cm. Model capacity = 12 000 L ÷ 10³ = 12 000 ÷ 1 000 = 12 L.
(v) Reasonableness check: a tank 2.5 m tall and 2.5 m wide is roughly the size of a garden shed — realistic for a backyard. Cost of ≈ $2 300 is consistent with real rainwater tanks of similar size, confirming our calculations.
Marking: 1 mark per part (i)–(v). Award full marks for any reasonable design where: capacity is at least 12 m³, height ≤ 2.5 m, the radius calculation rearranges V = πr²h correctly, the cost uses SA correctly, and the model is k³ smaller in volume.