Mathematics Standard • Year 11 • Module 2 • Lesson 19
The Trapezoidal Rule — Past-Paper Style
HSC Mathematics Standard 2-style writing: multi-mark short-answer trapezoidal rule problems plus one extended-response calculation with explicit marking criteria.
1. Short-answer questions
1.1 Widths of an irregular block, measured at 8 m intervals, are: 0, 12, 19, 16, 10, 0 (metres). Use the trapezoidal rule to estimate the area. 3 marks Band 3
1.2 A trapezoidal land block has parallel boundaries of length 28 m and 42 m, separated by a perpendicular distance of 35 m.
(a) Use the trapezoidal rule to estimate the area.
(b) Convert your answer to hectares. 3 marks Band 3–4
1.3 A surveyor records cross-sectional depths of a creek at 1.5 m intervals: 0.2, 0.8, 1.4, 1.7, 1.3, 0.9, 0.4 (metres).
(a) Estimate the cross-sectional area, in m², to 2 d.p.
(b) The creek flows at 0.5 m/s. Estimate the discharge rate (volume per second). 4 marks Band 4
2. Extended response
2.1 A council needs to estimate the area of an irregular reserve so that it can budget for fertiliser at $1.20 per m².
The reserve is bounded on one side by a straight road. A surveyor takes width measurements perpendicular to the road, at 15 m intervals from one end of the reserve to the other. The widths recorded are: 0, 22, 34, 41, 38, 27, 13, 0 (metres).
(a) State how many strips this data divides the reserve into, and the value of h.
(b) Use the trapezoidal rule to estimate the area of the reserve, in m².
(c) Express the area in hectares, to 3 decimal places.
(d) Calculate the council's fertiliser budget, to the nearest dollar, at $1.20 per m². 7 marks Band 5–6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — states 7 strips and h = 15 m.
Part (b) — 3 marks
• 1 mark — correctly identifies first, last and the interior sum.
• 1 mark — correct substitution into A ≈ (h/2)(d_f + 2d_m + d_l).
• 1 mark — correct numerical area with units (m²).
Part (c) — 1 mark
• 1 mark — correct conversion to hectares (÷ 10 000) with 3 d.p.
Part (d) — 2 marks
• 1 mark — correct multiplication of area in m² by $1.20.
• 1 mark — final budget rounded to the nearest dollar, with units and a clear conclusion sentence.
Your response:
Stuck on (b)? With 8 measurements, the first and last appear once and the 6 interior values are each doubled.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Block at h = 8 m (3 marks)
Sample response.
6 measurements → 5 strips; h = 8.
d_f = 0; d_l = 0; middle = 12 + 19 + 16 + 10 = 57.
A ≈ (8/2)(0 + 2 × 57 + 0) = 4 × 114 = 456 m².
Marking notes. 1 mark — correct identification of strips and h. 1 mark — correct first/last/middle setup. 1 mark — correct numerical answer with units.
1.2 — One-strip block 28/42 by 35 (3 marks)
(a) Sample response. One strip: A ≈ (35/2)(28 + 42) = 17.5 × 70 = 1225 m².
(b) Sample response. 1225 ÷ 10 000 = 0.1225 ha.
Marking notes. (a) 1 mark — correct h/2 = 17.5. 1 mark — correct numerical area. (b) 1 mark — correct conversion. Common error: dividing by 1000 instead of 10 000 gives 1.225 ha, which is 10× too large.
1.3 — Creek (4 marks)
(a) Sample response. 7 depths → 6 strips; h = 1.5; d_f = 0.2, d_l = 0.4; middle = 0.8 + 1.4 + 1.7 + 1.3 + 0.9 = 6.1. A ≈ (1.5/2)(0.2 + 2 × 6.1 + 0.4) = 0.75 × (0.2 + 12.2 + 0.4) = 0.75 × 12.8 = 9.60 m².
(b) Sample response. Discharge = 9.6 × 0.5 = 4.80 m³/s.
Marking notes. (a) 1 mark — correct strips/h. 1 mark — correct setup. 1 mark — correct area. (b) 1 mark — correct discharge with units (m³/s).
2.1 — Reserve area + budget, sample Band-6 (7 marks)
Sample Band-6 response.
(a) Strips and h.
8 measurements → 7 strips; h = 15 m. [1 mark.]
(b) Area.
d_f = 0; d_l = 0; middle = 22 + 34 + 41 + 38 + 27 + 13 = 175. [1 mark.]
A ≈ (15/2)(0 + 2 × 175 + 0) = 7.5 × 350. [1 mark.]
A ≈ 2625 m². [1 mark.]
(c) Hectares.
2625 ÷ 10 000 = 0.263 ha (to 3 d.p.). [1 mark.]
(d) Fertiliser budget.
Budget = 2625 × $1.20 = $3150. [1 mark — multiplication.]
Conclusion: the council should budget approximately $3150 for fertiliser, based on a reserve area estimated at 2625 m² (≈ 0.263 ha). [1 mark — clear conclusion sentence with units.]
Total: 7/7.
Band descriptors for marker.
Band 3: Identifies the number of strips and h, but doubles the wrong values (e.g. doubles d_f or d_l) and gets a numerically wrong area. ≈ 2–3 marks.
Band 4: Correct area in m²; converts to hectares but with the wrong divisor (e.g. ÷ 1000); attempts (d) with the wrong area. ≈ 4 marks.
Band 5: Full correct area and hectare conversion; correct fertiliser budget arithmetic but bare numerical answer without a conclusion sentence. ≈ 5–6 marks.
Band 6: Complete and correct throughout, with units on all answers and a clear conclusion sentence that the council can act on. 7/7.