Estimating areas using the trapezoidal rule
We have been looking at how to calculate the area of different shapes such as rectangles, triangles and parallelograms. In this section we investigate a method that will allow us to calculate the area of irregular shapes, such as the block of land below which is bounded by a river on one side. We will use a method of approximation called the trapezoidal rule.
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Aerial view of paddock |
The trapezoidal rule is named because it uses trapeziums to approximate areas. Trapeziums are useful because they can have a slanted side which gives them more flexibility and a better fit, compared to rectangles or other shapes, when trying to match irregular areas.
For one application of the trapezoidal rule, the area, $A$A, is given by:
$A\approx\frac{h}{2}(d_f+d_l)$A≈h2(df+dl)
Where:
| $d_f$df | $=$= | first measurement |
| $d_l$dl | $=$= | last measurement |
| $h$h | $=$= | distance between successive measurements |
Proof:
The area of a trapezium is:
$Area=\frac{h}{2}(a+b)$Area=h2(a+b) 
Remember $h$h is perpendicular to the parallel sides $a$a and $b$b.
In the case of the irregular area, $a=d_f$a=df and $b=d_l$b=dl with $h$h the distance between the sides
Hence:
$Area\approx\frac{h}{2}(d_f+d_l)$Area≈h2(df+dl)
Worked example
Example 1
For the block of land shown, use two applications of the trapezoidal rule to approximate the area.
Think: When we are asked to use two applications of the trapezoidal rule, it means that we should break the area up into two equal height trapeziums as shown below.
Do: Using the trapezoidal rule:
| Area | $\approx$≈ | Area of trapezium 1 $+$+ Area of trapezium 2 |
For trapezium 1:
| $d_f$df | $=$= | $210$210 m | |
| $d_l$dl | $=$= | $258$258 m | |
| $h$h | $=$= | $220$220 m |
For trapezium 2:
| $d_f$df | $=$= | $258$258 m | |
| $d_l$dl | $=$= | $320$320 m | |
| $h$h | $=$= | $220$220 m |
Therefore:
| Area | $\approx$≈ | $\frac{220}{2}(210+258)+\frac{220}{2}(258+320)$2202(210+258)+2202(258+320) | |
| Area | $\approx$≈ | $115060$115060m2 |
The trapezoidal rule extended
Notice, in the previous example, that the middle distance of $258$258 m was repeated in both calculations for trapezium $1$1 and $2$2.
| Area | $\approx$≈ | $\frac{220}{2}(210+258)+\frac{220}{2}(258+320)$2202(210+258)+2202(258+320) |
It is possible to develop a rule where we can find the area of multiple trapeziums in one go. Let's look at another example to see if we can identify the pattern.
Worked example
Example 2
To estimate the area below, we can use three trapeziums to find a fairly accurate approximation:
Working from left to right with the three trapeziums with $h=\frac{1020}{3}=340$h=10203=340 m, our previous rule gives us:
| Area | $\approx$≈ | $\frac{340}{2}(410+310)+\frac{340}{2}(310+450)+\frac{340}{2}(450+360)$3402(410+310)+3402(310+450)+3402(450+360) |
Noticing that all the distances are multiplied by $\frac{340}{2}$3402 we can take that out as a factor:
| Area | $\approx$≈ | $\frac{340}{2}(410+310+310+450+450+360)$3402(410+310+310+450+450+360) |
And then noticing that all of the 'middle' distances occur twice we can write:
| Area | $\approx$≈ | $\frac{340}{2}(410+2\times(310+450)+360)$3402(410+2×(310+450)+360) | |
| Area | $\approx$≈ | $389300$389300m2 |
For multiple applications of the trapezoidal rule:
$Area\approx\frac{h}{2}(d_f+2\times$Area≈h2(df+2×(The middle distances)$+d_l)$+dl)
Where:
| $d_f$df | $=$= | first measurement |
| $d_l$dl | $=$= | last measurement |
| The middle distances | $=$= | all distances excluding the first and the last |
| $h$h | $=$= | distance between successive measurements |
Be careful to notice that $d_f$df and $d_l$dl are now the first and last distances of the whole area, not of each trapezium.
Practice questions
Question 1
The following piece of land has straight boundaries on the east, west and south borders and follows a creek at the north.
The land has been divided into two sections to allow us to use the trapezoidal rule to approximate the area of the land.
Note: the diagram is not to scale.
Approximate Area $1$1 using one application of the trapezoidal rule.
Approximate Area $2$2 using one application of the trapezoidal rule.
Hence, find the approximate area of the piece of land using your answers to parts (a) and (b).
Is the actual area of the land greater than or less than this calculated area?
Greater than
ALess than
B
Question 2
A surveyor provided the following diagram with measurements for a property she was mapping out.
Find the approximate total area of the property by using three applications of the trapezoidal rule.
Give your answer in square metres.
The average weekly rainfall is $34$34 mm. Calculate the total volume of water that falls on the land in cubic metres.
Round your answer to two decimal places.
Estimating volumes
We can go even further and use the trapezoidal rule to find volumes. A common application of this is calculating rainfall, which is often reported as a height in millimetres over a certain area. Once we calculate the area using trapezoidal rule, written as $A$A, and the given rainfall written as $h$h, then we calculate the volume of rainfall over a given area using $V=Ah$V=Ah. In other words, the volume equals the area of the land multiplied by the rainfall. This calculation tells us how much water would be sitting above the land if the rain did not run off or pass through the ground.
Worked example
Example 3
We will revisit our first area example and calculate the number of litres of water produced by a rainfall of $80$80 mm.
With this rainfall we can consider the volume of water to be a prism with the area of the end equal to what we calculated in example 1. This was an area $=115060$=115060 m2
When finding the volume of rainfall, we want to make sure the units for the height of the rainfall match the units of the area. Here, we are given the average rainfall of this month last year as $80$80 mm, but the area of the paddock is measured in $m^2$m2. So, we can calculate the volume of water on the paddock by using the formula:
| Volume | $=$= | $Ah$Ah | (Formula for the volume of rainfall) |
| $h$h | $=$= | $\frac{80}{1000}$801000 | (Converting the rainfall to metres) |
| $=$= | $0.08$0.08 m | (Simplifying the expression) | |
| Volume | $=$= | $115060\times0.08$115060×0.08 | (Substituting the values into formula) |
| $=$= | $9204.8$9204.8 m3 | (Simplifying the expression) |
That is a lot of water! Of course most of the water is soaked up by the ground, but knowing this amount precisely is very important for agriculture, as well as disaster response in the case of a flood or drought.
Practice question
Question 3
The following piece of land has straight boundaries on the east, west and south borders and follows a creek at the north.
The land has been divided into two sections so we can use the trapezoidal rule to approximate the area.
Find the approximate area of the piece of land by using two applications of the trapezoidal rule.
Give your answer in square metres.
$35.2$35.2 mm of rain fell during a heavy storm. Find the volume of water that lands on this property in cubic metres.
Round your answer to the nearest cubic metre.