Calculate the approximate area of an irregularly shaped region using the trapezoidal rule, A = h/2 (d_f + d_l), and apply the rule repeatedly to find the approximate area between an irregular boundary and a straight line, where d_f and d_l are the first and last measurements and h is the strip width

A full-mark response identifies that one application means a single strip whose width is the whole frontage, so $h = 45$, $d_f = 16$ and $d_l = 22$, and writes the rule $A \approx \frac{h}{2}(d_f + d_l)$ before substituting (the formula line carries the method mark even if the arithmetic slips). Substituting gives $A \approx \frac{45}{2}(16 + 22) = 22.5 \times 38 = 855$ m$^2$. Markers award the final mark for the correct value with the unit m$^2$ and the word "approximately" or the $\approx$ sign, since the trapezoidal rule only estimates. A common one-mark loss is halving $45$ to make two strips; with a single application the whole frontage is the strip width.

Markers first look for the strip width, $h = \frac{80}{4} = 20$ m, then for the correct repeated rule with the ends counted once and the interior offsets doubled: $A \approx \frac{h}{2}(d_f + 2(d_1 + d_2 + d_3) + d_l)$. Substituting the offsets $0, 11, 17, 13, 0$ gives $A \approx \frac{20}{2}(0 + 2(11 + 17 + 13) + 0) = 10 \times (2 \times 41) = 10 \times 82 = 820$ m$^2$. One mark is for $h$, one for the correctly structured formula (interior doubled), and one for the answer $820$ m$^2$. The most penalised error is doubling every offset or doubling the zero ends; only the interior offsets are doubled.

For part (a), markers expect $h = \frac{90}{5} = 18$ m, then $A \approx \frac{18}{2}(0 + 2(14 + 20 + 18 + 12) + 0)$. The interior sum is $64$, so $A \approx 9 \times (2 \times 64) = 9 \times 128 = 1152$ m$^2$ - this earns three of the four marks (one for $h$, one for the structured substitution, one for the value with units). Part (b) earns the final mark for a clear statement that more strips means a smaller $h$, so the straight chords across the tops follow the curved boundary more closely and the estimate is generally more accurate. A response that just says "yes" without linking narrower strips to a closer fit does not earn the reasoning mark.

Identify the values. With one application the strip width $h$ is the whole length of the straight fence, and $d_f$ and $d_l$ are the two end measurements:

Substitute into the rule.

Evaluate. Work out the bracket first, then multiply:

so the area of the block is approximately $840$ m$^2$. (One application treats the whole block as a single trapezium, which is why $h$ is the full $40$ m here, not a smaller strip.)

Set up the single application. The straight edge is the strip width, and the two widths are the end offsets:

Apply the rule.

so the garden bed is approximately $480$ m$^2$. (The rule is just the area of a trapezium, the average of the two parallel sides, $\tfrac{12 + 20}{2} = 16$, times the width $30$, which also gives $480$.)

Recognise this is two strips. Three offsets $20$ m apart make two equal strips, so the strip width is $h = 20$ and the offsets in order are $8$, $14$ and $10$.

Apply the rule to each strip and add. Use the repeated form, in which the two end offsets are counted once and the interior offset is doubled:

Evaluate. Inside the bracket, $8 + 28 + 10 = 46$, so

so the area is approximately $460$ m$^2$. (Check by adding two trapezia: $\frac{20}{2}(8 + 14) + \frac{20}{2}(14 + 10) = 220 + 240 = 460$, which agrees.)

Find the strip width. A baseline of $80$ m split into four equal strips gives

List the offsets and apply the repeated rule. The five offsets are $0, 12, 16, 9, 0$. The ends ($0$ and $0$) are counted once, the three interior offsets ($12$, $16$, $9$) are doubled:

Evaluate. The interior sum is $12 + 16 + 9 = 37$, so

so the field is approximately $740$ m$^2$. (The end offsets are zero because the boundary meets the baseline there, which is common when a curved edge starts and finishes on the survey line.)

Part (a) - strip width and number of strips. The offsets are $10$ m apart, so $h = 10$ m. There are seven offsets, which means six gaps, so there are six strips. (A quick check: six strips of $10$ m span the full $60$ m fence.)

Part (b) - apply the repeated rule. The seven offsets are $0, 6, 11, 13, 10, 5, 0$. The two ends ($0$ and $0$) count once; the five interior offsets ($6, 11, 13, 10, 5$) are doubled:

Substitute. The interior sum is $6 + 11 + 13 + 10 + 5 = 45$, so

so the paddock is approximately $450$ m$^2$. (The most common slip here is to use $h = 60$, the whole fence, instead of $h = 10$, the gap between offsets. With several strips, $h$ is always the spacing between adjacent offsets.)

Part (a) - find the strip width. A baseline of $100$ m in five equal strips gives

Apply the repeated rule. There are six offsets, so the first ($12$) and last ($8$) are counted once and the four interior offsets ($18, 22, 19, 14$) are doubled:

Evaluate. The interior sum is $18 + 22 + 19 + 14 = 73$, so the bracket is $12 + 2(73) + 8 = 12 + 146 + 8 = 166$, and

so the surface area is approximately $1660$ m$^2$.

Part (b) - convert to hectares. One hectare is $10\,000$ m$^2$, so divide:

so the reservoir covers about $0.166$ ha. (Note the end offsets are not zero here, because both banks are open water at the ends of the baseline, so $12$ and $8$ each belong to a single strip and are not doubled.)

Part (a) - set up the repeated rule. The widths are measured every $3$ m, so $h = 3$ m. The six offsets are $0, 4, 7, 6, 4, 0$. The ends ($0$ and $0$) count once and the four interior widths ($4, 7, 6, 4$) are doubled:

Evaluate. The interior sum is $4 + 7 + 6 + 4 = 21$, so

so the lawn is approximately $63$ m$^2$.

Part (b) - cost the turf. At $8.50 per square metre,

so the turf costs about $535.50. (Because the trapezoidal rule only estimates the area, the landscaper would round up and buy a little extra; the estimate is the basis for the order, not the exact final bill.)