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Module 2 · MS-M1 · L19 of 22 ~50 min ⚡ +95 XP available

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.

Today's hook, A surveyor needs to estimate the area of a lake for a council report. The lake has an irregular shoreline. They can measure the width every 20 metres along one side. How do you turn a list of widths into an area?
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The formulas you need to own

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
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Think First, your gut answer first
+5 XP warm-up

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.

Without calculating how might you use those width measurements to estimate the area? What assumption would you be making?

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Come back to this at the end of the lesson.

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The formulas you need to own
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The trapezoidal rule comes in two forms. Lock both in, everything else in this lesson is just applying them.

One strip (2 measurements): area of a single trapezium, use when you have just a front and back boundary.

Multiple strips ($n+1$ measurements): the first and last measurements appear once; all interior measurements are doubled.

Comparison of the one-strip and multiple-strip trapezoidal rule formulas with a shared variable legend.
$$A \approx \dfrac{h}{2}(d_f + 2d_m + d_l)$$
Five equally spaced measurements divide the region into four strips whose areas are summed.
One strip: $A \approx \frac{h}{2}(d_f + d_l)$
Just the area of a single trapezium: $\tfrac{1}{2}(a+b) \times h$ where $a$ and $b$ are the parallel sides.
Multiple strips: $A \approx \frac{h}{2}(d_f + 2d_m + d_l)$
$d_m$ is the sum of all interior (middle) measurements. First and last appear once; all others are doubled.
Strips vs measurements
$n$ strips require $n+1$ measurements. e.g. 4 strips → 5 measurements.
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What you'll master
Know

Key facts

  • 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

Concepts

  • 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

Skills

  • 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 ruleA 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.
Strips vs measurementsIf there are $n$ strips, there are $n+1$ measurements. e.g. 3 strips require 4 measurements ($d_1, d_2, d_3, d_4$).
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From trapezium to trapezoidal rule

The area of a single trapezium is where and are the parallel sides and is the perpendicular distance between them.

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From trapezium to trapezoidal rule
core concept

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.

$$\text{One strip: } A \approx \frac{h}{2}(d_1 + d_2)$$
$$\text{Two strips: } A \approx \frac{h}{2}(d_1 + 2d_2 + d_3)$$
$$\text{$n$ strips: } A \approx \frac{h}{2}(d_1 + 2d_2 + 2d_3 + \cdots + 2d_n + d_{n+1})$$
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.

Trapezoidal rule: A ≈ (h/2)(d_first + d_last + 2×(all middle heights)) where h is the uniform interval width. Adding more strips improves the approximation.

Pause, copy the trapezoidal rule A ≈ (h/2)(d₀ + dₙ + 2(d₁ + d₂ + ⋯ + dₙ₋₁)), where h is the strip width, d₀ and dₙ are the first and last heights, and all middle heights are doubled, into your book.

Did you get this? True or false: in the trapezoidal rule, the first and last measurements are each multiplied by 2, while all interior measurements appear only once.

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · SINGLE STRIP

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.

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Identify: $d_f = 28$ m, $d_l = 34$ m, $h = 18$ m
Two parallel boundaries, one strip, two measurements.
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PROBLEM 2 · MULTIPLE STRIPS, IRREGULAR LAND BLOCK

A surveyor measures the widths of an irregular block of land at 10 m intervals from one end, obtaining: 0, 14, 22, 19, 11, 0 (metres). Use the trapezoidal rule to estimate the area.

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6 measurements → 5 strips; $h = 10$ m
Number of strips = number of measurements − 1.
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PROBLEM 3 · MULTIPLE STRIPS, NON-ZERO ENDPOINTS

A cross-section of a river channel is measured at 4 m intervals. The depths (in metres) at each point are: 1.2, 2.8, 3.5, 2.9, 1.6. Estimate the cross-sectional area using the trapezoidal rule.

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5 measurements → 4 strips; $h = 4$ m
Intervals are 4 m apart.
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Quick check: Widths at 5 m intervals are: 0, 8, 12, 6, 0 (metres). Which values go in the bracket as $d_f + 2d_m + d_l$?

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Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Trap 01
Doubling the first or last measurement
The first and last measurements each appear once in the bracket. Only the interior measurements are multiplied by 2. Always label first, last, and middle before substituting.
Trap 02
Miscounting strips and measurements
6 measurements means 5 strips, not 6. Always verify: strips = measurements − 1. Check this before writing down $h$.
Trap 03
Forgetting to add $d_m$ values before doubling
$2d_m$ means $2 \times (\text{sum of all middle values})$, not $2 \times$ each separately then listed. Sum the middles first, then multiply by 2.

Fill the gap: Measurements at 8 m intervals are: 0, 12, 18, 15, 9, 0. There are measurements, so there are strips. The sum of the interior measurements is $d_m = 12 + 18 + 15 + 9 =$ .

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Quick-fire practice · 5 calculations

Quick-fire practice · 5 calculations

Quick-fire practice · 5 calculations

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A trapezoidal land block has parallel sides of 45 m and 62 m, separated by a perpendicular distance of 30 m. Estimate the area.

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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).

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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.

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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.

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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 in hectares.

Two truths, one lie: Three of these statements are correct; one is wrong. Which is the lie?

Match it: Measurements: 3, 7, 9, 5 (metres) at 4 m intervals. Match each part of the formula to the correct value.

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Revisit your thinking

Look back at what you wrote in the Think First section.

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Revisit your thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 43 marks

Q1. 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)

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ApplyBand 44 marks

Q2. 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) 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)

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ApplyBand 54 marks

Q3. 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)

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📖 Comprehensive answers (click to reveal)

Drill 1: $A \approx \frac{30}{2}(45+62) = 15 \times 107 = \mathbf{1605 \text{ m}^2}$

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

Drill 3: $h=8$; $d_f=0$, $d_l=0$; $d_m=12+18+15+9=54$; $A \approx \frac{8}{2}(0+108+0)=4 \times 108 = \mathbf{432 \text{ m}^2}$

Drill 4: $h=2$; $d_f=0$, $d_l=0.6$; $d_m=1.4+2.2+1.8=5.4$; $A \approx \frac{2}{2}(0+10.8+0.6)=1\times 11.4=\mathbf{11.4 \text{ m}^2}$

Drill 5: $h=20$; $d_f=15$, $d_l=10$; $d_m=28+32+24=84$; $A \approx \frac{20}{2}(15+168+10)=10\times 193=1930 \text{ m}^2 = \mathbf{0.193 \text{ ha}}$

Q1 (3 marks): (a) 6 measurements → $\mathbf{5 \text{ strips}}$ [1]. (b) $d_f=0$, $d_l=0$; $d_m=18+25+20+14=77$; $A \approx \frac{12}{2}(0+154+0)=6\times 154=\mathbf{924 \text{ m}^2}$ [2].

Q2 (4 marks): (a) $h=15$; $d_f=2.0$, $d_l=1.0$; $d_m=4.5+6.2+5.8+3.1=19.6$; $A \approx \frac{15}{2}(2.0+39.2+1.0)=7.5\times 42.2=\mathbf{316.5 \text{ m}^2}$ [3]. (b) $V = 316.5 \times 200 = \mathbf{63\,300 \text{ m}^3}$ [1].

Q3 (4 marks): (a) $h=25$; $d_f=30$, $d_l=28$; $d_m=44+52+41=137$; $A \approx \frac{25}{2}(30+274+28)=12.5\times 332=\mathbf{4150 \text{ m}^2}$ [3]. (b) $4150 \div 10\,000 = \mathbf{0.415 \text{ ha}}$ [1].