Worked Solutions
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.
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.