Land blocks, harbours, and paddocks don't have neat rectangular edges. The trapezoidal rule lets you estimate the area of any irregular shape — provided you can measure a set of parallel widths at equal intervals.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A surveyor needs to estimate the area of a lake for a council report. The lake is roughly oval-shaped but with an irregular shoreline — definitely not a circle or ellipse. The surveyor can walk along one side and measure the width of the lake every 20 metres. How might you use those width measurements to estimate the area? What assumption would you be making?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Come back to this at the end of the lesson.
Wrong: A percentage over 100% is impossible.
Right: Percentages over 100% are valid and common in contexts like percentage increase, profit margins, and scale factors.
Understanding the Rule
Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.
The area of a single trapezium is $\frac{1}{2}(a+b) \times h$ where $a$ and $b$ are the parallel sides and $h$ is the perpendicular distance between them. The trapezoidal rule applies this repeatedly across an irregular region.
For one strip (2 measurements $d_1$, $d_2$, interval $h$): $$A \approx \frac{h}{2}(d_1 + d_2)$$
For two strips (3 measurements $d_1$, $d_2$, $d_3$, common interval $h$): $$A \approx \frac{h}{2}(d_1 + d_2) + \frac{h}{2}(d_2 + d_3) = \frac{h}{2}(d_1 + 2d_2 + d_3)$$
For $n$ strips ($n+1$ measurements), the pattern becomes: $$A \approx \frac{h}{2}(d_1 + 2d_2 + 2d_3 + \cdots + 2d_n + d_{n+1})$$
A block of land has a straight back boundary of 28 m and a straight front boundary of 34 m. The perpendicular depth between these boundaries is 18 m. Use the trapezoidal rule to estimate the area of the block.
A surveyor measures the widths of an irregular block of land at 10 m intervals from one end, obtaining the following measurements: 0, 14, 22, 19, 11, 0 (metres). Use the trapezoidal rule to estimate the area.
A cross-section of a river channel is measured at 4 m intervals. The depths (in metres) at each measurement point are: 1.2, 2.8, 3.5, 2.9, 1.6. Estimate the cross-sectional area of the river using the trapezoidal rule.
Section A — One Strip
Section B — Multiple Strips
Section C — Problem Solving
$A \approx \frac{30}{2}(45+62) = 15 \times 107 = \mathbf{1605 \text{ m}^2}$
$A \approx \frac{80}{2}(110+95) = 40 \times 205 = 8200 \text{ m}^2 = \mathbf{0.82 \text{ ha}}$
$h=8$; $d_f=0$, $d_l=0$; $d_m=12+18+15+9=54$; $A \approx \frac{8}{2}(0+108+0)=4 \times 108 = \mathbf{432 \text{ m}^2}$
$h=5$; $d_f=6$, $d_l=5$; $d_m=10+14+13+8=45$; $A \approx \frac{5}{2}(6+90+5)=2.5\times 101=\mathbf{252.5 \text{ m}^2}$
$h=2$; $d_f=0$, $d_l=0.6$; $d_m=1.4+2.2+1.8=5.4$; $A \approx \frac{2}{2}(0+10.8+0.6)=1\times 11.4=\mathbf{11.4 \text{ m}^2}$
$h=20$; $d_f=15$, $d_l=10$; $d_m=28+32+24=84$; $A \approx \frac{20}{2}(15+168+10)=10\times 193=1930 \text{ m}^2 = \mathbf{0.193 \text{ ha}}$
More strips means smaller intervals, so the trapezium approximation more closely follows the true curve of the boundary. More measurements → less "cutting of corners" → more accurate estimate.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Multiple Choice
1 Widths of an irregular block, measured at 6 m intervals, are: 0, 8, 14, 10, 0 (metres). Using the trapezoidal rule, the estimated area is:
? Regarding this topic, 1 Widths of an irregular block, measured at 6 m intervals, are: 0, 8, 14, 10, 0 (metres). Using the trapezoidal rule, the estimated area is:
2 A block of land has parallel boundaries of 40 m and 52 m separated by 25 m. The trapezoidal rule gives an area of:
? Regarding this topic, 2 A block of land has parallel boundaries of 40 m and 52 m separated by 25 m. The trapezoidal rule gives an area of:
3 Widths at 4 m intervals are: 3, 7, 9, 5 (metres). The trapezoidal rule estimate for the area is:
? Regarding this topic, 3 Widths at 4 m intervals are: 3, 7, 9, 5 (metres). The trapezoidal rule estimate for the area is:
Short Answer
SA 4 3 marks A surveyor records widths across an irregular block at 12 m intervals: 0, 18, 25, 20, 14, 0 (metres).
(a) How many strips does this represent? (1 mark)
(b) Estimate the area using the trapezoidal rule. (2 marks)
6 measurements → $\mathbf{5 \text{ strips}}$
$d_f=0$, $d_l=0$; $d_m=18+25+20+14=77$; $A \approx \frac{12}{2}(0+154+0)=6\times 154=\mathbf{924 \text{ m}^2}$
SA 5 4 marks Depths (in metres) across a harbour are measured at 15 m intervals: 2.0, 4.5, 6.2, 5.8, 3.1, 1.0. A planner wishes to estimate the cross-sectional area of the harbour.
(a) Apply the trapezoidal rule to estimate the cross-sectional area. (3 marks)
(b) The harbour is 200 m long. Estimate its volume in cubic metres, assuming the cross-section is uniform. (1 mark)
$h=15$; $d_f=2.0$, $d_l=1.0$; $d_m=4.5+6.2+5.8+3.1=19.6$; $A \approx \frac{15}{2}(2.0+39.2+1.0)=7.5\times 42.2=\mathbf{316.5 \text{ m}^2}$
$V = 316.5 \times 200 = \mathbf{63\,300 \text{ m}^3}$
SA 6 4 marks A rural block of land is surveyed along one boundary. Width measurements (in metres), taken at 25 m intervals, are recorded in the table below.
| Distance along boundary (m) | 0 | 25 | 50 | 75 | 100 |
|---|---|---|---|---|---|
| Width (m) | 30 | 44 | 52 | 41 | 28 |
(a) Use the trapezoidal rule to estimate the area of the block. (3 marks)
(b) Express this area in hectares. (1 mark)
$h=25$; $d_f=30$, $d_l=28$; $d_m=44+52+41=137$; $A \approx \frac{25}{2}(30+274+28)=12.5\times 332=\mathbf{4150 \text{ m}^2}$
$4150 \div 10\,000 = \mathbf{0.415 \text{ ha}}$
The Trapezoidal Rule