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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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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

Think First — your gut answer first

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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.

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.

$$A \approx \dfrac{h}{2}(d_f + 2d_m + d_l)$$

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.

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

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 ($d_1, d_2, d_3, d_4$).

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.

What to write in your book

One strip: $A \approx \frac{h}{2}(d_f + d_l)$ — this is just a trapezium with parallel sides $d_f$ and $d_l$.
Multiple strips: $A \approx \frac{h}{2}(d_f + 2d_m + d_l)$ where $d_m$ is the sum of all interior measurements.
Number of strips = number of measurements $-$ 1.
More strips = smaller $h$ = better approximation of the true boundary.

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.

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.

Identify: $d_f = 28$ m, $d_l = 34$ m, $h = 18$ m

Two parallel boundaries — one strip — two measurements.

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.

6 measurements → 5 strips; $h = 10$ m

Number of strips = number of measurements − 1.

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.

5 measurements → 4 strips; $h = 4$ m

Intervals are 4 m apart.

What to write in your book

Always count: strips = measurements − 1. Write out the list and circle first/last; box the middles.
Non-zero endpoints are common in river/channel problems — don't assume endpoints are 0.
Units matter: if widths are in metres, the area is in m². If converting to hectares, divide by 10 000.
To improve accuracy: use more strips (smaller $h$) so the straight edges better follow the curve.

Quick check: Widths at 5 m intervals are: 0, 8, 12, 6, 0 (metres). Which values go in the bracket as $d_f + 2d_m + d_l$?

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.

What to write in your book

Always label: circle first and last; add up the middle values; then apply the formula.
Check units: area in m²; convert to ha by dividing by 10 000.
More strips → smaller $h$ → trapeziums fit the boundary more closely → better estimate.

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 $d_m = 12 + 18 + 15 + 9 =$ .

Quick-fire practice · 5 calculations

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

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).

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.

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.

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.

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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Multiple choice

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

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)	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)

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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].

Boss battle · The Surveyor

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Five timed questions on the trapezoidal rule. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering trapezoidal rule questions. Pool: lesson 19.

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Module overview · Year 11 Maths Standard