Measurement — Worked Solutions (Preliminary Maths Standard)
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Worked examples for Preliminary Maths Standard measurement. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Maths Standard — Measurement. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Surface area of a closed cylinder
Question
A closed cylindrical can has a radius of $4$ cm and a height of $10$ cm. Calculate its total surface area, correct to the nearest square centimetre. Use $A = 2\pi r^2 + 2\pi r h$.
Solution
Substitute $r = 4$ and $h = 10$ into the surface-area formula.
$A = 2\pi (4)^2 + 2\pi (4)(10)$.
Two circular ends: $2\pi \times 16 = 32\pi$. The curved side: $2\pi \times 40 = 80\pi$.
$A = 32\pi + 80\pi = 112\pi \approx 351.86$, so $A \approx 352$ cm$^2$.
Round only at the very end. Rounding $\pi$ early drags the answer off by a square centimetre or two.
A closed cylinder has three pieces of surface: a top circle, a bottom circle, and the curved wall that wraps around. The formula bundles them together — $2\pi r^2$ is the two circular ends, and $2\pi r h$ is the wrapped-around side.
Putting in $r = 4$ and $h = 10$: $A = 2\pi (4)^2 + 2\pi (4)(10)$.
The two ends give $2\pi \times 16 = 32\pi$, and the side gives $2\pi \times 40 = 80\pi$.
Adding: $A = 112\pi \approx 351.86$ cm$^2$, which rounds to $352$ cm$^2$.
Picturing the can unrolled — two lids plus a rectangle that becomes the curved side — is what makes the formula make sense rather than something to memorise.
Substitute $r = 4$, $h = 10$.
- Ends: $2\pi r^2 = 2\pi (16) = 32\pi$
- Side: $2\pi r h = 2\pi (40) = 80\pi$
- $A = 112\pi \approx 351.86$
- $A \approx 352$ cm$^2$
Where the marks go
- 1 mark: Correct substitution into both terms of the formula
- 1 mark: Correct exact value $112\pi$ (or correct unrounded value)
- 1 mark: Correct answer rounded to the nearest cm$^2$ with units
Key idea
Total surface area of a closed cylinder = two circular ends ($2\pi r^2$) plus the curved side ($2\pi r h$); round only at the final step.
Example 2 — Right-angled trigonometry
Question
A ramp rises at an angle of $12^\circ$ to the horizontal. The ramp is $5$ m long along its sloped surface. Find the vertical height the ramp rises, correct to two decimal places.
Solution
The $5$ m sloped length is the hypotenuse, and the vertical height is opposite the $12^\circ$ angle. Opposite and hypotenuse → use sine.
$\sin 12^\circ = \dfrac{h}{5}$.
Rearrange: $h = 5 \sin 12^\circ$.
$h = 5 \times 0.20791\ldots = 1.0396\ldots \approx 1.04$ m.
Label which side is which before you pick the ratio — that one decision is where most marks are lost.
Let's set up the triangle. The ramp's sloped surface is the longest side, the hypotenuse, at $5$ m. The height we want goes straight up, which is the side opposite the $12^\circ$ angle.
When we have the opposite and the hypotenuse, the ratio that links them is sine: $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$.
So $\sin 12^\circ = \dfrac{h}{5}$, and to free $h$ we multiply both sides by $5$: $h = 5 \sin 12^\circ$.
Working it out: $h = 5 \times 0.2079 = 1.0396$, which rounds to $1.04$ m.
Choosing sine comes from naming the sides first — once you know it's opposite-and-hypotenuse, SOH tells you exactly which button to press.
Hypotenuse $= 5$, opposite $= h$, angle $= 12^\circ$. Use sine.
- $\sin 12^\circ = \dfrac{h}{5}$
- $h = 5 \sin 12^\circ$
- $h = 5 \times 0.20791 = 1.0396$
- $h \approx 1.04$ m
Where the marks go
- 1 mark: Identifies the correct ratio (sine) with opposite and hypotenuse
- 1 mark: Sets up and rearranges to $h = 5 \sin 12^\circ$
- 1 mark: Correct answer $h \approx 1.04$ m to two decimal places
Key idea
In a right-angled triangle, match the known and wanted sides to SOH CAH TOA — opposite with hypotenuse means sine.
Frequently asked questions
Step-by-step solutions to Measurement questions in Preliminary Maths Standard, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Standard, using the methods and notation expected in exams and assessments.