Measurement & Geometry — Worked Solutions (Year 9 Maths)
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Worked examples for Year 9 Maths Measurement & Geometry. 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 Year 9 Maths — Measurement & Geometry. 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 and volume
Question
A solid cylinder has radius $5\text{ cm}$ and height $12\text{ cm}$. Find its volume and its total surface area. Give each answer to the nearest whole number, using $\pi = 3.14$.
Solution
Two separate formulas — volume $V = \pi r^2 h$ and total surface area $A = 2\pi r^2 + 2\pi r h$.
Volume: $V = 3.14 \times 5^2 \times 12 = 3.14 \times 25 \times 12 = 942\text{ cm}^3$.
Surface area: the two circular ends are $2\pi r^2 = 2 \times 3.14 \times 25 = 157$, and the curved side is $2\pi r h = 2 \times 3.14 \times 5 \times 12 = 376.8$. Add them: $157 + 376.8 = 533.8 \approx 534\text{ cm}^2$.
Don't mix up the units — volume is cubic, area is square. Losing the unit loses a mark.
A cylinder is just a circle that's been pushed up into a tube, so its volume is the area of the circular base times the height: $V = \pi r^2 h$.
$V = 3.14 \times 5^2 \times 12$. Working inside out, $5^2 = 25$, then $3.14 \times 25 = 78.5$, and $78.5 \times 12 = 942\text{ cm}^3$.
The surface area is made of three pieces: the top circle, the bottom circle, and the curved part. The two circles together are $2\pi r^2 = 2 \times 3.14 \times 25 = 157$. If you imagine unrolling the curved side, it becomes a rectangle whose width is the circle's circumference, giving $2\pi r h = 2 \times 3.14 \times 5 \times 12 = 376.8$.
Adding the pieces: $157 + 376.8 = 533.8 \approx 534\text{ cm}^2$. Picturing the curved side as an unrolled rectangle is the trick that makes the formula make sense.
Volume: $V = \pi r^2 h$.
- $V = 3.14 \times 25 \times 12 = 942\text{ cm}^3$
Surface area: $A = 2\pi r^2 + 2\pi r h$.
- Ends: $2 \times 3.14 \times 25 = 157$
- Curved: $2 \times 3.14 \times 5 \times 12 = 376.8$
- Total: $533.8 \approx 534\text{ cm}^2$
Where the marks go
- 1 mark: Correct volume formula and substitution $\pi r^2 h$
- 1 mark: Correct volume $942\text{ cm}^3$
- 1 mark: Correct surface area expression $2\pi r^2 + 2\pi r h$ with substitution
- 1 mark: Correct surface area $534\text{ cm}^2$ with units
Key idea
A cylinder's volume is base area times height ($\pi r^2 h$); its surface area is the two ends ($2\pi r^2$) plus the curved side ($2\pi r h$).
Example 2 — Right-angled trigonometry
Question
In a right-angled triangle, the angle at $A$ is $35^\circ$ and the side opposite this angle is $8\text{ cm}$. Find the length of the hypotenuse, correct to one decimal place.
Solution
Label the sides relative to the $35^\circ$ angle: opposite $= 8$, and we want the hypotenuse. Opposite and hypotenuse means sine (SOH).
$\sin 35^\circ = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{8}{h}$.
Rearrange: $h = \dfrac{8}{\sin 35^\circ} = \dfrac{8}{0.5736} = 13.9\text{ cm}$.
Pick the ratio from the two sides involved — don't reach for the calculator until you've named opp, adj or hyp.
The first job in any trig question is to label the sides from the angle's point of view. The $8\text{ cm}$ side is across from the $35^\circ$ angle, so it's the opposite, and the side we want is the hypotenuse (the longest side, across from the right angle).
Opposite and hypotenuse together point us to sine, because SOH tells us $\sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$. So $\sin 35^\circ = \dfrac{8}{h}$.
The unknown $h$ is on the bottom, so we swap it with $\sin 35^\circ$: $h = \dfrac{8}{\sin 35^\circ}$. Since $\sin 35^\circ \approx 0.5736$, we get $h = \dfrac{8}{0.5736} \approx 13.9\text{ cm}$.
Naming the sides before choosing the ratio is what stops the classic mix-up between sine and cosine.
Sides from $35^\circ$: opp $= 8$, want hyp → sine.
- $\sin 35^\circ = \frac{8}{h}$
- $h = \frac{8}{\sin 35^\circ} = \frac{8}{0.5736}$
- $h \approx 13.9\text{ cm}$
Where the marks go
- 1 mark: Identifies sine as the correct ratio (opposite and hypotenuse)
- 1 mark: Correct equation $\sin 35^\circ = \frac{8}{h}$ and rearrangement
- 1 mark: Correct hypotenuse $13.9\text{ cm}$ to one decimal place
Key idea
Label the sides relative to the given angle, then pick the ratio (SOH-CAH-TOA) that uses the two sides involved; here opposite and hypotenuse means sine.
Frequently asked questions
Step-by-step solutions to Measurement & Geometry questions in Year 9 Maths, 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 Year 9 Maths, using the methods and notation expected in exams and assessments.