Worked Solutions
Measurement — Worked Solutions (Preliminary Maths Standard)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our Preliminary Maths Standard course →
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.
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.