Measurement & Geometry — Worked Solutions (Year 10 Maths)
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Worked examples for Year 10 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 10 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.
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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 — Trigonometry with bearings
Question
A ship sails $12$ km from port $P$ on a bearing of $050^\circ$ to point $Q$, then $9$ km on a bearing of $140^\circ$ to point $R$. Show that angle $PQR = 90^\circ$, then find the distance $PR$, correct to one decimal place.
Solution
First the angle. The bearing of $Q$ from $P$ is $050^\circ$, so the back bearing of $P$ from $Q$ is $050^\circ + 180^\circ = 230^\circ$. The next leg leaves $Q$ on $140^\circ$.
Angle $PQR = 230^\circ - 140^\circ = 90^\circ$. So triangle $PQR$ is right-angled at $Q$.
Because it's right-angled, use Pythagoras: $PR^2 = 12^2 + 9^2 = 144 + 81 = 225$.
$PR = \sqrt{225} = 15$ km exactly, so $PR = 15.0$ km. Don't reach for the sine rule when the angle is a right angle — Pythagoras is faster and exact.
Bearings are measured clockwise from north, so let's track the directions carefully.
The ship travels from $P$ to $Q$ on $050^\circ$. Looking back the other way, from $Q$ towards $P$, we turn through $180^\circ$, giving a back bearing of $230^\circ$. The second leg, from $Q$ to $R$, is on $140^\circ$.
The angle inside the triangle at $Q$ is the difference between these two directions out of $Q$: $230^\circ - 140^\circ = 90^\circ$. That confirms angle $PQR = 90^\circ$.
Now, why does that help? A right angle means we can use Pythagoras' theorem with $PR$ as the hypotenuse: $PR^2 = 12^2 + 9^2 = 144 + 81 = 225$.
Taking the square root, $PR = 15$ km, so to one decimal place $PR = 15.0$ km.
Angle at $Q$:
- Back bearing of $P$ from $Q$: $050^\circ + 180^\circ = 230^\circ$
- Leg to $R$ on $140^\circ$
- $\angle PQR = 230^\circ - 140^\circ = 90^\circ$
Right-angled → Pythagoras:
- $PR^2 = 12^2 + 9^2 = 225$
- $PR = 15.0$ km
Where the marks go
- 1 mark: Finds the back bearing $230^\circ$ for $P$ from $Q$
- 1 mark: Shows angle $PQR = 90^\circ$
- 1 mark: Applies Pythagoras: $PR^2 = 12^2 + 9^2$
- 1 mark: Correct distance $PR = 15.0$ km
Key idea
Convert bearings into the interior angle using back bearings ($+180^\circ$); a right angle lets you use Pythagoras instead of the sine rule.
Example 2 — Surface area of a composite solid
Question
A solid is made of a cylinder of radius $5$ cm and height $12$ cm, topped by a hemisphere of the same radius. Find the total surface area, correct to the nearest square centimetre. Use $\pi = 3.1416$.
Solution
Break the surface into pieces: the circular base, the curved side of the cylinder, and the curved top of the hemisphere. The flat top of the cylinder is covered by the hemisphere, so it doesn't count.
Base (circle): $\pi r^2 = \pi (5)^2 = 25\pi$.
Cylinder side: $2\pi r h = 2\pi (5)(12) = 120\pi$.
Hemisphere curved surface: half a sphere $= \tfrac{1}{2}(4\pi r^2) = 2\pi (5)^2 = 50\pi$.
Total $= 25\pi + 120\pi + 50\pi = 195\pi \approx 195 \times 3.1416 = 612.6$, so $613$ cm$^2$. The trap is including the join face — once you cap it with the hemisphere, that disc is internal.
With a composite solid the smart move is to picture every surface you'd actually paint, and leave out any face that's hidden where the two shapes meet.
Here we have three exposed surfaces. The bottom is a full circle: $\pi r^2 = 25\pi$. The side of the cylinder is its curved surface: $2\pi r h = 2\pi (5)(12) = 120\pi$.
For the dome on top, a whole sphere has surface $4\pi r^2$, so a hemisphere's curved part is half of that: $2\pi r^2 = 2\pi (25) = 50\pi$.
Why don't we add the cylinder's flat top? Because the hemisphere sits exactly on it — that disc is sealed inside and never exposed.
Adding up: $25\pi + 120\pi + 50\pi = 195\pi$. Using $\pi = 3.1416$, that's $612.6$, which rounds to $613$ cm$^2$.
Exposed surfaces only (cylinder top is hidden).
- Base circle: $\pi r^2 = 25\pi$
- Cylinder curved side: $2\pi r h = 120\pi$
- Hemisphere curved: $2\pi r^2 = 50\pi$
- Total: $195\pi \approx 612.6$
$\approx 613$ cm$^2$.
Where the marks go
- 1 mark: Correct base area $25\pi$
- 1 mark: Correct cylinder curved surface $120\pi$
- 1 mark: Correct hemisphere curved surface $50\pi$ (and excludes the hidden top)
- 1 mark: Sums and rounds to $613$ cm$^2$
Key idea
For composite solids add only the exposed surfaces — the face where two shapes join is internal, so leave it out.
Frequently asked questions
Step-by-step solutions to Measurement & Geometry questions in Year 10 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 10 Maths, using the methods and notation expected in exams and assessments.