Your weak spots

Insights load after your first practice round.

MST-12-S2-05 ~35 min ⚡ +90 XP available

The Trapezoidal Rule and Volume of Rainfall

Real blocks of land have curved edges and no area formula. The trapezoidal rule replaces the curve with straight lines you can measure, and then rainfall turns that area into litres.

Today's hook, A roof measures $12\text{ m}$ by $8\text{ m}$ and $25\text{ mm}$ of rain falls on it. Is that enough water to fill a bathtub, a rainwater tank, or a swimming pool?

0/5QUESTS

Your guided path0 of 7 done · keep going

Orient and prepare

Capture your estimate and preview the key measurement language.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any question in this focus area.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any question

Think first, your gut answer

+5 XP warm-up

A roof measures $12\text{ m}$ by $8\text{ m}$. Overnight, $25\text{ mm}$ of rain falls.

Without calculating write down how many litres you think landed on that roof, and say whether $25\text{ mm}$ sounds like a lot of water to you.

auto-saved

What you will be able to do

+5 XP to read

Use the trapezoidal rule to estimate the area of a shape with one curved or irregular side
Apply the rule with one application, using two offsets
Apply the rule with two applications, using three equally spaced offsets
Explain why the answer is an estimate and how to make it more accurate
Turn a rainfall depth in millimetres into a volume of water in cubic metres and litres

+ +

Preview the vocabulary

Review the essential terms before using them in the worked methods.

Key terms

+5 XP to read

Offset: A measured width across the shape, taken at right angles to a straight baseline. Like this: standing on a fence line and measuring $12\text{ m}$ across to the creek gives an offset of $12\text{ m}$.
Strip width, h: The distance along the baseline between two neighbouring offsets, always equal. Like this: if the block is $40\text{ m}$ long and you take offsets at each end and in the middle, then $h = 20\text{ m}$.
One application: Using the rule once, with exactly two offsets and one strip. Like this: offsets of $12\text{ m}$ and $16\text{ m}$ that are $8\text{ m}$ apart give an area of $112\text{ m}^2$.
Volume of rainfall: The water landing on a surface, found by multiplying the area by the rainfall depth in metres. Like this: $10\text{ mm}$ of rain on $100\text{ m}^2$ is $100 \times 0.01 = 1\text{ m}^3$, which is $1000$ litres.

Use the trapezoidal rule

Apply the rule carefully to one measured cross-section.

The trapezoidal rule, one application

+10 XP

Real paddocks and blocks of land are bounded by creeks and roads that curve. There is no area formula for those. So we pretend the curve is a straight line between two measurements, which turns the shape into a trapezium we can handle.

Measure across the shape at two places, $h$ apart. Those two measurements are the parallel sides of a trapezium. Average them, then multiply by the distance between them. That average is why the formula has a half in it.

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

Here $h$ is the strip width, $d_f$ is the first offset and $d_l$ is the last offset. The formula is on the HSC reference sheet, so the marks come from choosing the right numbers, not from remembering it.

Worked example 1, one application

$h = 8\text{ m}$, $\;d_f = 12\text{ m}$, $\;d_l = 16\text{ m}$

Two offsets, $8\text{ m}$ apart, so one strip and one application.

$A \approx \dfrac{8}{2}(12 + 16)$

Substitute into the rule. Halve the strip width first, it keeps the numbers small.

$A \approx 4 \times 28$

The bracket is the sum of the two offsets, not their average.

$A \approx 112\text{ m}^2$

Check: $112 \div 4 = 28$, which is $12 + 16$.

Why it is only an estimate. The rule joins the tops of the two offsets with a straight line. If the real boundary bulges outwards the rule under-counts, and if it curves inwards the rule over-counts. Write "approximately" in your answer.

In your book. Copy the formula, then draw one trapezium and label $h$, $d_f$ and $d_l$ on your own diagram. Write one line saying why the half is there.

Quick check. Offsets of $10\text{ m}$ and $14\text{ m}$ are $5\text{ m}$ apart. Estimate the area using one application.

Work with several offsets

Organise repeated measurements and combine trapezoidal estimates.

Two applications, three offsets

+10 XP

One straight line across a long, wavy boundary is a poor guess. Take a third measurement in the middle and you get two trapeziums instead of one, and the estimate improves.

You can just do the rule twice and add. But the middle offset belongs to both strips, so it gets counted twice, and collecting the terms gives a single formula worth knowing.

Adding the two strips: $\frac{h}{2}(d_0 + d_1) + \frac{h}{2}(d_1 + d_2)$. The $d_1$ appears in both brackets, so it becomes $2d_1$ when you collect them.

$A \approx \dfrac{h}{2}\left(d_0 + 2d_1 + d_2\right)$

Worked example 2, two applications

Offsets $10\text{ m}$, $14\text{ m}$, $12\text{ m}$, with $h = 6\text{ m}$

Three equally spaced offsets, so two strips.

$A \approx \dfrac{6}{2}\left(10 + 2 \times 14 + 12\right)$

Only the middle offset is doubled. The two ends are not.

$A \approx 3 \times (10 + 28 + 12) = 3 \times 50$

Work the bracket fully before multiplying.

$A \approx 150\text{ m}^2$

Check as two separate strips: $3(10+14) = 72$ and $3(14+12) = 78$, and $72 + 78 = 150$.

Find $h$ before you start. Questions often give the total length instead of the strip width. If a block is $40\text{ m}$ long with three offsets, the strips are $h = 40 \div 2 = 20\text{ m}$, not $40\text{ m}$.

Worked example 3, working out h first

A paddock $40\text{ m}$ long, offsets $25\text{ m}$, $32\text{ m}$, $28\text{ m}$

Three offsets means two strips across the $40\text{ m}$.

$h = 40 \div 2 = 20\text{ m}$

Divide the total length by the number of strips, not the number of offsets.

$A \approx \dfrac{20}{2}\left(25 + 2 \times 32 + 28\right) = 10 \times 117$

$25 + 64 + 28 = 117$.

$A \approx 1170\text{ m}^2$

Check: $10(25+32) = 570$ and $10(32+28) = 600$, and $570 + 600 = 1170$.

In your book. Write the two-application formula and circle the $2$ in front of the middle offset. Beside it write "ends once, middle twice".

Did you get this? True or false: taking more offsets, so that the strips are narrower, generally makes the trapezoidal estimate more accurate.

Connect rainfall depth and volume

Convert rainfall depth into a volume over a known area.

Volume of rainfall

+10 XP

Rainfall is reported as a depth. Saying $25\text{ mm}$ fell means that if none of it drained away, it would sit $25\text{ mm}$ deep everywhere. So the water is just a very flat box: area multiplied by depth.

The one trap is units. The area is in square metres but the depth is in millimetres, and you cannot multiply those. Convert the depth to metres first, using $1\text{ mm} = 0.001\text{ m}$.

$V = A \times d \qquad 1\text{ m}^3 = 1000\text{ L}$

Worked example 4, the hook roof

Roof $12\text{ m}$ by $8\text{ m}$, rainfall $25\text{ mm}$

Find the area first, then deal with the depth.

$A = 12 \times 8 = 96\text{ m}^2$

A plain rectangle here, but it could come from the trapezoidal rule.

$d = 25\text{ mm} = 0.025\text{ m}$

Divide by $1000$. This is the step that costs the marks.

$V = 96 \times 0.025 = 2.4\text{ m}^3 = 2400\text{ L}$

Check: $2400 \div 1000 = 2.4$, and $2.4 \div 96 = 0.025\text{ m}$.

Worked example 5, joining the two ideas

The paddock from worked example 3, area $1170\text{ m}^2$

An estimated area feeds straight into a volume.

Rainfall $18\text{ mm} = 0.018\text{ m}$

Convert before multiplying, every time.

$V \approx 1170 \times 0.018 = 21.06\text{ m}^3$

Still approximate, because the area was.

$V \approx 21\,060\text{ L}$

Check: $21.06 \div 1170 = 0.018$.

Millimetres to metres

Divide by $1000$. Writing $25\text{ mm}$ as $0.25\text{ m}$ is the single most common error in this topic.

Cubic metres to litres

Multiply by $1000$. One cubic metre of water is $1000$ litres, and it has a mass of one tonne.

Keep the word approximately

If the area came from the trapezoidal rule, the volume is an estimate too. Say so.

In your book. Write the chain: area, then depth in metres, then volume, then litres. Put the hook roof beside it and finish the calculation.

Fill the gaps. A roof measures $10\text{ m}$ by $6\text{ m}$ and $20\text{ mm}$ of rain falls. Area $=$ m$^2$, depth $=$ m, volume $=$ m$^3$, which is litres.

Show what you can do

Answer the exam-style questions, then compare your working with the model answers.

Quick-check drill

work before revealing answers

Choose an option for each fixed drill question, then reveal the concept-labelled explanations below.

Drill 1. Offsets of $8\text{ m}$ and $14\text{ m}$ are $10\text{ m}$ apart. Use one application of the trapezoidal rule to estimate the area.

$110\text{ m}^2$
$220\text{ m}^2$
$55\text{ m}^2$
$22\text{ m}^2$

Drill 2. Three offsets of $6\text{ m}$, $9\text{ m}$ and $7\text{ m}$ are each $4\text{ m}$ apart. Estimate the area.

$44\text{ m}^2$
$62\text{ m}^2$
$124\text{ m}^2$
$31\text{ m}^2$

Drill 3. A roof measures $15\text{ m}$ by $10\text{ m}$ and $20\text{ mm}$ of rain falls on it. What volume of water lands on the roof?

$3000\text{ L}$
$300\text{ L}$
$30\,000\text{ L}$
$3\text{ L}$

Drill 4. Which expression is the trapezoidal rule for one application?

$\dfrac{h}{2}(d_f + d_l)$
$h(d_f + d_l)$
$\dfrac{h}{2}(d_f \times d_l)$
$\dfrac{h}{2}(d_f - d_l)$

Drill 5. A rainfall of $25\text{ mm}$ is written in metres as:

$0.025\text{ m}$
$0.25\text{ m}$
$2.5\text{ m}$
$0.0025\text{ m}$

Short answer

ApplyBand 43 marks

SA 1. A block of land has one straight side and one creek boundary. Two offsets are measured, $9\text{ m}$ and $15\text{ m}$, and they are $12\text{ m}$ apart. Use one application of the trapezoidal rule to estimate the area of the block, and explain why your answer is an estimate. (3 marks)

auto-saved

ApplyBand 54 marks

SA 2. A paddock is $24\text{ m}$ long. Three equally spaced offsets are measured across it: $20\text{ m}$, $26\text{ m}$ and $22\text{ m}$. (a) Estimate the area of the paddock. (2 marks) (b) $30\text{ mm}$ of rain falls on the paddock. Find the volume of water in litres. (2 marks)

auto-saved

AnalyseBand 54 marks

SA 3. A roof measures $18\text{ m}$ by $9\text{ m}$ and drains into an empty $5000\text{ L}$ tank. A storm drops $40\text{ mm}$ of rain. (a) Find the volume of water that lands on the roof, in litres. (3 marks) (b) Does the tank overflow, and if so by how much? (1 mark)

auto-saved

📖 Comprehensive answers (click to reveal)

One trapezoidal application. $A \approx \tfrac{10}{2}(8 + 14) = 5 \times 22 = 110\text{ m}^2$. Option B forgets to halve the strip width, and option D leaves out $h$ altogether.

Several trapezoidal strips. $A \approx \tfrac{4}{2}(6 + 2 \times 9 + 7) = 2 \times 31 = 62\text{ m}^2$. Option A comes from adding the three offsets without doubling the middle one.

Rainfall volume. $A = 150\text{ m}^2$ and $d = 20\text{ mm} = 0.020\text{ m}$, so $V = 150 \times 0.02 = 3\text{ m}^3 = 3000\text{ L}$. Option C treats $20\text{ mm}$ as $0.2\text{ m}$.

Structure of the trapezoidal rule. The rule averages the two offsets, which is the sum halved, and then multiplies by the distance between them. Option B is the sum without the average, giving double the area.

Millimetres to metres. $1\text{ mm} = 0.001\text{ m}$, so $25\text{ mm} = 25 \div 1000 = 0.025\text{ m}$. Option B divides by $100$ instead of $1000$, which is the error that turns a full tank into an empty one.

SA 1 (3 marks): $A \approx \tfrac{12}{2}(9 + 15)$ [1] $= 6 \times 24 = 144\text{ m}^2$ [1]. It is an estimate because the rule replaces the curved creek boundary with a straight line between the two offsets, so any bulge outwards is not counted and any curve inwards is over-counted [1].

SA 2 (4 marks): (a) Three offsets means two strips, so $h = 24 \div 2 = 12\text{ m}$ [1]. $A \approx \tfrac{12}{2}(20 + 2 \times 26 + 22) = 6 \times 94 = 564\text{ m}^2$ [1]. Check as two strips: $6(20+26) = 276$ and $6(26+22) = 288$, and $276 + 288 = 564$. (b) $30\text{ mm} = 0.030\text{ m}$ [1], so $V \approx 564 \times 0.03 = 16.92\text{ m}^3 = 16\,920\text{ L}$ [1].

SA 3 (4 marks): (a) $A = 18 \times 9 = 162\text{ m}^2$ [1]. $40\text{ mm} = 0.040\text{ m}$ [1]. $V = 162 \times 0.04 = 6.48\text{ m}^3 = 6480\text{ L}$ [1]. (b) Yes. $6480\text{ L}$ is more than the $5000\text{ L}$ the tank holds, so it overflows by $6480 - 5000 = 1480\text{ L}$ [1].

Retrieve, reflect and finish

Revisit your opening idea, then use the topic challenge and mark the lesson complete.

Revisit your thinking

+5 XP

Go back to what you wrote in section 01. The roof is $96\text{ m}^2$, the rain is $0.025\text{ m}$ deep, so $2.4\text{ m}^3$ landed, which is $2400$ litres. That is well past a bathtub and nowhere near a swimming pool, and it is why one storm can fill a rainwater tank.

auto-saved

Take the full module quiz

quiz

A full module quiz covering every lesson in this focus area, not just this one.

Start the module quiz →

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 3 · Scale drawings Lesson 5 · Capture-recapture →

Focus area overview · Year 12 · Maths Standard