Worked Solutions
Measurement & Geometry — Worked Solutions (Year 9 Maths)
Created with Intu AI
Studying this? See our Year 9 Maths course →
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.
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.