Mathematics Standard • Year 11 • Module 2 • Lesson 19
The Trapezoidal Rule — Problem Set
Apply the trapezoidal rule to authentic survey, environmental and engineering scenarios — paddocks, lakes, channels and roads.
Problem 1 — Estimating a paddock area
A surveyor pegs out an irregular paddock and records the width at 25 m intervals across one boundary. The widths are: 0, 38, 55, 62, 48, 30, 0 (metres).
Set up: What are we solving for?
(i) How many strips does this represent, and what is h? 1 mark
(ii) Estimate the area using the trapezoidal rule, in m². 2 marks
(iii) Express the area in hectares, to 2 decimal places. 1 mark
Stuck? Identify first/last (both 0 here) and the sum of the 5 interior values, then substitute.Problem 2 — Cross-sectional area of a river
A hydrologist measures the depth of a river at 2 m intervals across its width. The depths (in metres) at each measurement point are: 0.4, 1.6, 2.8, 3.1, 2.2, 1.0, 0.3.
Set up: What are we solving for?
(i) Identify d_f, d_l and the sum of the interior depths. 1 mark
(ii) Estimate the cross-sectional area of the river, in m². 2 marks
(iii) The river flows at an average speed of 0.7 m/s. Estimate the volumetric flow rate (volume per second) in m³/s. 2 marks
Stuck on (iii)? Flow rate = cross-sectional area × speed. Units: m² × m/s = m³/s.Problem 3 — Lake area, two surveyors
Two surveyors estimate the area of the same small lake. They each measure the width of the lake at regular intervals along one straight reference line.
Surveyor A (4 measurements at 20 m intervals): 0, 32, 36, 0 (m).
Surveyor B (7 measurements at 10 m intervals): 0, 18, 28, 36, 30, 17, 0 (m).
Set up: What are we solving for?
(i) Estimate the area using Surveyor A's data. 2 marks
(ii) Estimate the area using Surveyor B's data. 2 marks
(iii) By how many square metres do the two estimates differ, and which is likely to be more accurate? Explain in one sentence. 2 marks
Stuck? More measurements = narrower strips = better tracking of the curving boundary.Problem 4 — Earthworks volume for a road cutting
Engineers need to estimate the volume of soil to be removed for a 80 m long straight road cutting. Cross-sectional measurements (depth below ground level) are taken every 4 m across the cutting, giving the cross-sectional area at one position. The depths (m) are: 0, 1.5, 2.4, 2.8, 2.1, 1.2, 0.
Set up: What are we solving for?
(i) Estimate the cross-sectional area, in m². 2 marks
(ii) Assuming the cross-section is approximately constant along the 80 m length, estimate the volume of soil to be removed, in m³. 2 marks
(iii) A truck holds 8 m³ of soil. How many full truckloads must be removed? 2 marks
Stuck on (iii)? Divide the volume by 8 and round UP to make sure all soil is removed (you can't leave a fraction behind).Problem 5 — Working backwards from a known area
A council reports the area of a small grass reserve as exactly 1500 m². A surveyor takes widths at h = 10 m intervals across the reserve: 0, w_1, w_2, w_3, 0 (m), where w_1 and w_3 are equal. After the survey, the trapezoidal rule estimate of 1500 m² is obtained with w_2 = 40 m.
Set up: What are we solving for?
(i) How many strips are there, and what is the value of (h/2)? 1 mark
(ii) Write the trapezoidal rule equation that the surveyor's data must satisfy, in terms of w_1 (use w_3 = w_1). 2 marks
(iii) Solve to find w_1, the common width at the two outer interior points. 3 marks
Stuck? Substitute everything you know into A ≈ (h/2)(d_f + 2d_m + d_l) = 1500 and solve for the unknown.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Paddock area
Set up. 7 measurements at h = 25 m → 6 strips; substitute into A ≈ (h/2)(d_f + 2d_m + d_l), then convert to hectares.
(i) 6 strips; h = 25 m.
(ii) d_f = 0; d_l = 0; middle = 38 + 55 + 62 + 48 + 30 = 233. A ≈ (25/2)(0 + 466 + 0) = 12.5 × 466 = 5825 m².
(iii) 5825 ÷ 10 000 = 0.58 ha (to 2 d.p.).
Problem 2 — River cross-section
Set up. 7 depths at h = 2 m → 6 strips; calculate the cross-section, then multiply by speed for flow rate.
(i) d_f = 0.4; d_l = 0.3; interior sum = 1.6 + 2.8 + 3.1 + 2.2 + 1.0 = 10.7.
(ii) A ≈ (2/2)(0.4 + 2 × 10.7 + 0.3) = 1 × (0.4 + 21.4 + 0.3) = 22.1 m².
(iii) Flow rate = 22.1 × 0.7 = 15.47 m³/s.
Problem 3 — Two surveyors
Set up. Apply the formula to each data set, compare numerically, then comment on accuracy.
(i) Surveyor A: 4 measurements at h = 20 m → 3 strips; d_f = 0, d_l = 0, middle = 32 + 36 = 68. A ≈ (20/2)(0 + 136 + 0) = 10 × 136 = 1360 m².
(ii) Surveyor B: 7 measurements at h = 10 m → 6 strips; d_f = 0, d_l = 0, middle = 18 + 28 + 36 + 30 + 17 = 129. A ≈ (10/2)(0 + 258 + 0) = 5 × 258 = 1290 m².
(iii) Difference = 1360 − 1290 = 70 m². Surveyor B's estimate is likely more accurate because the narrower strips (h = 10 m vs 20 m) follow the lake's true curved boundary more closely.
Problem 4 — Road cutting
Set up. Cross-sectional area via trapezoidal rule × constant 80 m length = volume; then truckloads via division, rounded UP.
(i) 7 depths at h = 4 m → 6 strips; d_f = 0, d_l = 0, middle = 1.5 + 2.4 + 2.8 + 2.1 + 1.2 = 10. A ≈ (4/2)(0 + 20 + 0) = 2 × 20 = 40 m².
(ii) Volume = 40 × 80 = 3200 m³.
(iii) 3200 ÷ 8 = 400. 400 truckloads. (If the division wasn't exact — e.g. 400.3 — you'd round UP to 401 because no truckload can carry a partial 0.3 load and still remove all soil.)
Problem 5 — Solving for an unknown width
Set up. Substitute known values into the formula with the unknown w_1 (= w_3), then solve.
(i) 5 measurements → 4 strips; h/2 = 10/2 = 5.
(ii) A ≈ (10/2)(0 + 2w_1 + 2 × 40 + 2w_1 + 0) = 5 × (4w_1 + 80) = 1500, i.e. 5(4w_1 + 80) = 1500.
(iii) 4w_1 + 80 = 300; 4w_1 = 220; w_1 = 55 m. (Both outer interior widths equal 55 m.)