Measurement & Geometry — Worked Solutions (Year 8 Maths)
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Worked examples for Year 8 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 8 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 — Circles and volume
Question
A cylindrical water tank has a circular base of radius $1.5$ m and a height of $2$ m.
(a) Find the circumference of the base, correct to one decimal place.
(b) Find the volume of the tank, correct to one decimal place. (Use $\pi \approx 3.14$.)
Solution
(a) Circumference is $C = 2\pi r = 2 \times 3.14 \times 1.5 = 9.42$, so $C \approx 9.4$ m.
(b) Volume of a cylinder is the base area times the height: $V = \pi r^2 h$.
Base area $= 3.14 \times 1.5^2 = 3.14 \times 2.25 = 7.065$ m². Multiply by height $2$: $V = 14.13$ m³, so $V \approx 14.1$ m³.
Square the radius before multiplying — a common slip is doing $(\pi r)^2$. And keep full figures until the final rounding.
(a) The circumference is the distance around the circle, and the formula is $C = 2\pi r$. With $r = 1.5$: $C = 2 \times 3.14 \times 1.5 = 9.42$, which rounds to $9.4$ m.
(b) A cylinder is just a circle pushed straight up, so its volume is the area of the circular base multiplied by the height. The base area is $\pi r^2 = 3.14 \times 1.5^2$. Remember $1.5^2 = 2.25$, so the area is $3.14 \times 2.25 = 7.065$ m².
Now stack it up to the height of $2$ m: $V = 7.065 \times 2 = 14.13$ m³, which rounds to $14.1$ m³.
Thinking of volume as "base area times height" works for any prism or cylinder — the only new part for a circle is using $\pi r^2$ for that base.
$r = 1.5$ m, $h = 2$ m, $\pi \approx 3.14$.
- (a) $C = 2\pi r = 2 \times 3.14 \times 1.5 = 9.42 \approx 9.4$ m
- Base area $= \pi r^2 = 3.14 \times 2.25 = 7.065$ m²
- (b) $V = \pi r^2 h = 7.065 \times 2 = 14.13 \approx 14.1$ m³
Answers: $9.4$ m; $14.1$ m³.
Where the marks go
- 1 mark: Correct circumference $C \approx 9.4$ m
- 1 mark: Correct base area $\pi r^2 = 7.065$ m²
- 1 mark: Correct volume $V \approx 14.1$ m³
Key idea
Circumference uses $C = 2\pi r$; the volume of a cylinder is the base area $\pi r^2$ times the height.
Example 2 — Congruence and Pythagoras
Question
Two right-angled triangles, $\triangle ABC$ and $\triangle PQR$, are congruent. In $\triangle ABC$ the right angle is at $B$, with $AB = 6$ cm and $BC = 8$ cm.
(a) Find the length of the hypotenuse $AC$.
(b) State the length of the longest side of $\triangle PQR$, giving a reason.
Solution
(a) The hypotenuse is opposite the right angle, so $AC^2 = AB^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100$. Therefore $AC = \sqrt{100} = 10$ cm.
(b) Congruent triangles are identical in size and shape, so matching sides are equal. The longest side of $\triangle PQR$ equals the longest side of $\triangle ABC$, which is $AC = 10$ cm.
Use Pythagoras only for the hypotenuse here, and remember congruence means the triangles are equal, not just similar — so the side lengths carry straight across.
(a) Because the right angle is at $B$, the side $AC$ across from it is the hypotenuse — the longest side. Pythagoras' theorem says the square of the hypotenuse equals the sum of the squares of the other two sides: $AC^2 = 6^2 + 8^2 = 36 + 64 = 100$. Taking the square root, $AC = 10$ cm.
(b) Congruent triangles are exact copies of one another — same angles and same side lengths — so each side of $\triangle PQR$ matches a side of $\triangle ABC$. The longest side of $\triangle ABC$ is the hypotenuse $AC = 10$ cm, so the longest side of $\triangle PQR$ must also be $10$ cm.
The key idea is that congruence is stronger than just "same shape": it guarantees the measurements are identical, which is why we can copy the $10$ cm straight over.
Right angle at $B$, so $AC$ is the hypotenuse.
- (a) $AC^2 = 6^2 + 8^2 = 36 + 64 = 100$
- $AC = \sqrt{100} = 10$ cm
- (b) Congruent ⇒ matching sides equal
- Longest side of $\triangle PQR = AC = 10$ cm
Answers: $10$ cm; $10$ cm (congruent triangles have equal corresponding sides).
Where the marks go
- 1 mark: Sets up Pythagoras as $AC^2 = 6^2 + 8^2$
- 1 mark: Correct hypotenuse $AC = 10$ cm
- 1 mark: States $10$ cm with a reason based on congruence
Key idea
Pythagoras gives the hypotenuse from the two shorter sides; congruent triangles have equal corresponding sides.
Frequently asked questions
Step-by-step solutions to Measurement & Geometry questions in Year 8 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 8 Maths, using the methods and notation expected in exams and assessments.