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The Trapezoidal Rule

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.

50–55 min MS-M1 3 MC 3 SA Lesson 19 of 22 Free
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Choose how you work: type answers on screen, or work in your book.

Think First

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.

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Come back to this at the end of the lesson.

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Trapezoidal Rule — This Lesson

$A \approx \dfrac{h}{2}(d_f + d_l)$   (one strip)
h — the perpendicular distance between the two parallel sides (the "step" interval) $d_f$ — the length of the first parallel side $d_l$ — the length of the last parallel side This is just the area of one trapezium: $\frac{1}{2}(a+b) \times h$
$A \approx \dfrac{h}{2}(d_f + 2d_m + d_l)$   (multiple strips)
h — the common interval between measurements $d_f$ — first measurement (one end) $d_l$ — last measurement (other end) $d_m$ — sum of all middle (interior) measurements The first and last measurements appear once; all interior measurements appear twice
TRAPEZOIDAL RULE — 4 STRIPS (5 MEASUREMENTS) d₁=0 d₂ d₃ d₄ d₅ d₆=0 h A ≈ h/2 × (d₁ + 2d₂ + 2d₃ + 2d₄ + 2d₅ + d₆) first and last once; all middle values doubled strip 1 strip 2 strip 3 strip 4
Interactive

Trapezoidal Rule Explorer — adjust the widths and strip interval to see the calculation update live

5 m
0 m
7 m
13 m
9 m
0 m

🧠 Know

  • The trapezoidal rule formula: $A \approx \frac{h}{2}(d_f + 2d_m + d_l)$
  • $h$ is the interval between measurements; $d_f$ and $d_l$ appear once; all interior $d$ values appear twice
  • The rule gives an approximation, not an exact area

💡 Understand

  • Why the trapezoidal rule is needed — real-world shapes are rarely geometric
  • How each strip is treated as a trapezium, and summing the strips gives the total
  • How increasing the number of strips improves accuracy

✅ Can Do

  • Apply the trapezoidal rule with 2, 3, 4 or more measurements
  • Identify which values are "first", "last", and "middle" in a data set
  • Solve problems involving irregular land blocks and cross-sections
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Key Terms

Trapezoidal rule A numerical method for estimating the area of an irregular shape by dividing it into strips, each approximated as a trapezium
Interval (h) The constant perpendicular distance between successive parallel measurements; must be equal across all strips
Offsets (d values) The measured parallel widths (or depths) at regular intervals across the shape; often called "offsets" in surveying contexts
Number of strips vs measurements If there are $n$ strips, there are $n+1$ measurements. e.g. 3 strips require 4 measurements ($d_1, d_2, d_3, d_4$)

Misconceptions to Fix

✗

Wrong: The trapezoidal rule gives an exact answer for the area under any curve

✗

Right: The trapezoidal rule gives an approximation. It overestimates for concave-up curves and underestimates for concave-down curves. More trapezoids improve the approximation.

Understanding the Rule

Key Point

Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.

Key Terms
FormulaA rule showing the relationship between variables using symbols.
SubstitutionReplacing variables with their known values in an equation.
Unit ConversionChanging a measurement from one unit to another.
CapacityThe amount of liquid a container can hold, measured in litres or millilitres.
PerimeterThe total distance around the outside of a shape.
AreaThe amount of space inside a two-dimensional shape.

From Trapezium to Trapezoidal Rule

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: $$A \approx \frac{h}{2}(d_1 + 2d_2 + 2d_3 + \cdots + 2d_n + d_{n+1})$$

Must do — identify first, last, and middle: Before applying the formula, label which measurement is first ($d_f$), which is last ($d_l$), and which are middle ($d_m$). Only the first and last are multiplied by 1; all others are multiplied by 2.
Common error: Doubling the first or last measurement, or not doubling an interior one. Always write out the bracket carefully: $d_f + 2d_2 + 2d_3 + \ldots + 2d_{n} + d_l$ and check the count.
Worked Example 1 Single Strip

Problem

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.

Solution

1 Identify: $d_f = 28$ m, $d_l = 34$ m, $h = 18$ m Two parallel boundaries → one strip → two measurements
Worked Example 2 Multiple Strips — Irregular Land Block

Problem

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.

Solution

1 6 measurements → 5 strips; $h = 10$ m Number of strips = number of measurements − 1
Worked Example 3 Multiple Strips — Non-Zero Endpoints

Problem

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.

Solution

1 5 measurements → 4 strips; $h = 4$ m Intervals are 4 m apart

Practice Questions

Section A — One Strip

  1. A trapezoidal land block has parallel sides of 45 m and 62 m, separated by a perpendicular distance of 30 m. Estimate the area.
  2. A paddock has two parallel fence lines of length 110 m and 95 m, 80 m apart. Estimate the area in hectares (1 ha = 10 000 m²).

Section B — Multiple Strips

  1. Measurements of a block's width, taken every 8 m, are: 0, 12, 18, 15, 9, 0 (metres). Use the trapezoidal rule to estimate the area.
  2. A surveyor records widths at 5 m intervals across an irregular field: 6, 10, 14, 13, 8, 5 (metres). Estimate the area.
  3. Cross-sectional depths of a drainage channel, measured at 2 m intervals, are: 0, 1.4, 2.2, 1.8, 0.6 (metres). Estimate the cross-sectional area.

Section C — Problem Solving

  1. A block of land is measured at 20 m intervals. The widths are 15 m, 28 m, 32 m, 24 m, and 10 m. Estimate the total area and express it in hectares.
  2. Two surveyors use the trapezoidal rule to estimate the same irregular area. Surveyor A uses 4 measurements at 10 m intervals; Surveyor B uses 7 measurements at 5 m intervals. Explain why Surveyor B's estimate is likely to be more accurate.

Q1

$A \approx \frac{30}{2}(45+62) = 15 \times 107 = \mathbf{1605 \text{ m}^2}$

Q2

$A \approx \frac{80}{2}(110+95) = 40 \times 205 = 8200 \text{ m}^2 = \mathbf{0.82 \text{ ha}}$

Q3

$h=8$; $d_f=0$, $d_l=0$; $d_m=12+18+15+9=54$; $A \approx \frac{8}{2}(0+108+0)=4 \times 108 = \mathbf{432 \text{ m}^2}$

Q4

$h=5$; $d_f=6$, $d_l=5$; $d_m=10+14+13+8=45$; $A \approx \frac{5}{2}(6+90+5)=2.5\times 101=\mathbf{252.5 \text{ m}^2}$

Q5

$h=2$; $d_f=0$, $d_l=0.6$; $d_m=1.4+2.2+1.8=5.4$; $A \approx \frac{2}{2}(0+10.8+0.6)=1\times 11.4=\mathbf{11.4 \text{ m}^2}$

Q6

$h=20$; $d_f=15$, $d_l=10$; $d_m=28+32+24=84$; $A \approx \frac{20}{2}(15+168+10)=10\times 193=1930 \text{ m}^2 = \mathbf{0.193 \text{ ha}}$

Q7

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.

Revisit Your Initial Thinking

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:

A   192 m²
B   96 m²
C   192 m²
D   384 m²
A — $h=6$; $d_f=0$, $d_l=0$; $d_m=8+14+10=32$; $A \approx \frac{6}{2}(0+64+0)=3\times 64=192 \text{ m}^2$.

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:

A   1150 m²
B   575 m²
C   2300 m²
D   1300 m²
A — $A \approx \frac{25}{2}(40+52) = 12.5 \times 92 = 1150 \text{ m}^2$.

3 Widths at 4 m intervals are: 3, 7, 9, 5 (metres). The trapezoidal rule estimate for the area is:

A   64 m²
B   80 m²
C   96 m²
D   48 m²
C — $h=4$; $d_f=3$, $d_l=5$; $d_m=7+9=16$; $A \approx \frac{4}{2}(3+32+5)=2\times 40=80$ — wait: $3+32+5=40$; $2\times 40=80$... rechecking: $d_m=16$, so $2d_m=32$; bracket $=3+32+5=40$; $A=2\times 40=80$ m² → answer is B 80 m². Option A results from not doubling the middle values.

Short Answer

01

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)

Work in your book
Saved

(a)

6 measurements → $\mathbf{5 \text{ strips}}$

(b)

$d_f=0$, $d_l=0$; $d_m=18+25+20+14=77$; $A \approx \frac{12}{2}(0+154+0)=6\times 154=\mathbf{924 \text{ m}^2}$

02

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)

Work in your book
Saved

(a)

$h=15$; $d_f=2.0$, $d_l=1.0$; $d_m=4.5+6.2+5.8+3.1=19.6$; $A \approx \frac{15}{2}(2.0+39.2+1.0)=7.5\times 42.2=\mathbf{316.5 \text{ m}^2}$

(b)

$V = 316.5 \times 200 = \mathbf{63\,300 \text{ m}^3}$

03

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)0255075100
Width (m)3044524128

(a) Use the trapezoidal rule to estimate the area of the block.  (3 marks)

(b) Express this area in hectares.  (1 mark)

Work in your book
Saved

(a)

$h=25$; $d_f=30$, $d_l=28$; $d_m=44+52+41=137$; $A \approx \frac{25}{2}(30+274+28)=12.5\times 332=\mathbf{4150 \text{ m}^2}$

(b)

$4150 \div 10\,000 = \mathbf{0.415 \text{ ha}}$

Consolidation Game

The Trapezoidal Rule