Mathematics Standard • Year 11 • Module 2 • Lesson 19
The Trapezoidal Rule — Skill Drill
Drill the trapezoidal rule for one strip and multiple strips: correctly identify first, last and middle offsets and substitute into A ≈ (h/2)(d_f + 2d_m + d_l).
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Write the trapezoidal rule for multiple strips. Define h, d_f, d_l and d_m.
A ≈ ____________________________________
Q1.2 A surveyor records 7 measurements at equal intervals. How many strips does this produce? ____________
Q1.3 In the multi-strip formula, the first and last measurements each appear ____ time(s), while every middle measurement appears ____ time(s) inside the bracket.
2. Worked example — irregular land block (5 strips)
Every line of working has a reason on the right.
Problem. A surveyor measures widths of an irregular block at 10 m intervals: 0, 14, 22, 19, 11, 0 (metres). Use the trapezoidal rule to estimate the area.
Step 1 — Count strips and identify h.
6 measurements → 5 strips; h = 10 m.
Reason: number of strips = (number of measurements) − 1.
Step 2 — Identify first, last and middle.
d_f = 0; d_l = 0; middle = 14 + 22 + 19 + 11 = 66.
Reason: the first and last appear once; sum the four interior values for d_m.
Step 3 — Substitute into the formula.
A ≈ (10 ÷ 2) × (0 + 2 × 66 + 0) = 5 × 132
Reason: h/2 = 5; the bracket contains the first, twice each middle, and the last.
Conclusion. A ≈ 660 m².
3. Faded example — fill in the missing steps
A creek cross-section is measured at 3 m intervals. The depths (m) are 0.8, 1.6, 2.4, 1.9, 1.1. Estimate the cross-sectional area. 4 marks
Step 1 — Strips and h: ____ measurements → ____ strips; h = ____ m.
Step 2 — First / last / middle: d_f = ____; d_l = ____; middle = ____ + ____ + ____ = ____.
Step 3 — Substitute: A ≈ (____ ÷ 2) × (____ + 2 × ____ + ____) = ____ × ____ = ____.
Conclusion. A ≈ ____________ m².
4. Graduated practice — trapezoidal rule
Show your working and label all answers with units (m² or hectares as required; 1 ha = 10 000 m²).
Foundation — single strip and recall (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | A trapezium has parallel sides 30 m and 50 m, separated by 20 m. Find the area using A = ½(a + b)h. | |
| 4.2 1 | 4 measurements produce how many strips? | |
| 4.3 1 | Convert 4500 m² to hectares. | |
| 4.4 1 | If d_1 = 0, d_2 = 8, d_3 = 10, d_4 = 0 and h = 6, what is h/2 × (d_1 + 2d_2 + 2d_3 + d_4)? |
Standard — typical HSC difficulty (6 questions)
Show the substitution clearly and box the final answer with units.
4.5 A block has parallel boundaries 60 m and 80 m, 40 m apart. Use the one-strip trapezoidal rule to estimate its area. 2 marks
4.6 Widths at 5 m intervals: 0, 9, 15, 11, 0 (m). Estimate the area. 2 marks
4.7 Depths at 4 m intervals across a stream: 0.5, 1.8, 2.1, 1.4, 0.7 (m). Estimate the cross-sectional area. 2 marks
4.8 Widths at 12 m intervals: 0, 20, 32, 28, 17, 0 (m). Estimate the area. 2 marks
4.9 A paddock has parallel fence lines 130 m and 110 m, 90 m apart. Estimate the area and express it in hectares. 2 marks
4.10 Widths at 6 m intervals: 4, 10, 13, 11, 6 (m). Estimate the area. 2 marks
Extension — error/accuracy reasoning (2 questions)
4.11 A lake's width is measured at 15 m intervals: 0, 18, 26, 30, 24, 16, 0 (m). (a) Estimate the area using the trapezoidal rule. (b) Express the answer in hectares to 3 d.p. 3 marks
4.12 Two students measure the same paddock. Student A takes widths at 20 m intervals (4 measurements); Student B takes widths at 10 m intervals (7 measurements). Explain in 2–3 sentences which estimate is likely to be more accurate, and why. 3 marks
5. Self-check the easy 3
Tick once you've verified each method.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Multi-strip formula
A ≈ (h/2) × (d_f + 2d_m + d_l), where h = constant interval between measurements; d_f = first measurement; d_l = last measurement; d_m = sum of all middle (interior) measurements.
Q1.2 — Strips from 7 measurements
6 strips (always one fewer than the number of measurements).
Q1.3 — Multiplier counts
First and last each appear 1 time; every middle measurement appears 2 times.
Q3 — Faded example (creek)
Step 1: 5 measurements → 4 strips; h = 3 m. Step 2: d_f = 0.8; d_l = 1.1; middle = 1.6 + 2.4 + 1.9 = 5.9. Step 3: A ≈ (3/2) × (0.8 + 2 × 5.9 + 1.1) = 1.5 × (0.8 + 11.8 + 1.1) = 1.5 × 13.7 = 20.55. Conclusion: A ≈ 20.55 m².
Q4.1 — Single trapezium 30/50 by 20
A = ½(30 + 50) × 20 = ½ × 80 × 20 = 800 m².
Q4.2 — Strips from 4 measurements
3 strips.
Q4.3 — 4500 m² → hectares
4500 ÷ 10 000 = 0.45 ha.
Q4.4 — Substitute given values
h/2 = 3; bracket = 0 + 16 + 20 + 0 = 36; A = 3 × 36 = 108 m².
Q4.5 — Block 60/80 by 40 (one strip)
A ≈ (40/2)(60 + 80) = 20 × 140 = 2800 m².
Q4.6 — Widths 0, 9, 15, 11, 0 at h = 5
d_f = 0; d_l = 0; middle = 9 + 15 + 11 = 35. A ≈ (5/2)(0 + 70 + 0) = 2.5 × 70 = 175 m².
Q4.7 — Stream depths at h = 4
d_f = 0.5; d_l = 0.7; middle = 1.8 + 2.1 + 1.4 = 5.3. A ≈ (4/2)(0.5 + 10.6 + 0.7) = 2 × 11.8 = 23.6 m².
Q4.8 — Widths 0, 20, 32, 28, 17, 0 at h = 12
d_f = 0; d_l = 0; middle = 20 + 32 + 28 + 17 = 97. A ≈ (12/2)(0 + 194 + 0) = 6 × 194 = 1164 m².
Q4.9 — Paddock 130/110 by 90 (one strip), in hectares
A ≈ (90/2)(130 + 110) = 45 × 240 = 10 800 m² = 10 800 ÷ 10 000 = 1.08 ha.
Q4.10 — Widths 4, 10, 13, 11, 6 at h = 6
d_f = 4; d_l = 6; middle = 10 + 13 + 11 = 34. A ≈ (6/2)(4 + 68 + 6) = 3 × 78 = 234 m².
Q4.11 — Lake widths at h = 15
(a) d_f = 0; d_l = 0; middle = 18 + 26 + 30 + 24 + 16 = 114. A ≈ (15/2)(0 + 228 + 0) = 7.5 × 228 = 1710 m².
(b) 1710 ÷ 10 000 = 0.171 ha.
Q4.12 — Accuracy reasoning
Student B's estimate is more accurate. Narrower strips (smaller h) allow the trapezoidal approximation to follow a curving boundary more closely, reducing the systematic over- or under-estimation that occurs when a wide strip "cuts the corner" between two measurements. With twice as many measurements, B captures more of the boundary's true shape.