Area, Perimeter, Surface Area and Volume Revision Notes | GCSE AQA Higher Further Maths

Worked example: Area of a Trapezium

A trapezium has parallel sides of length 5 cm and 9 cm, and a perpendicular height of 4 cm. Calculate its area.

1. $A = \frac{1}{2}(a + b)h$ — Identify this as a trapezium: two parallel sides and a perpendicular height are given. [M1 — correct formula]
2. $A = \frac{1}{2}(5 + 9) \times 4$ — $a = 5$ cm, $b = 9$ cm, $h = 4$ cm. Substitute carefully. [M1 — correct substitution]
3. $5 + 9 = 14$ — Work inside the bracket first.
4. $A = \frac{1}{2} \times 14 \times 4 = 7 \times 4 = 28$ — Halving 14 gives 7, then 7 × 4 = 28. [A1 — correct answer]
5. $A = 28 \text{ cm}^2$ — Area is always in square units. Missing units costs a mark on many mark schemes.

Answer: $A = 28 \text{ cm}^2$

Worked example: Volume of a Cylinder

A cylinder has radius 3 cm and height 10 cm. Calculate its volume. Give your answer in terms of $\pi$.

1. $V = \pi r^2 h$ — A cylinder is a circular prism. [M1 — correct formula]
2. $V = \pi \times 3^2 \times 10$ — $r = 3$ cm, $h = 10$ cm. Substitute before evaluating. [M1 — correct substitution]
3. $3^2 = 3 \times 3 = 9$ — Always expand the square explicitly to avoid errors.
4. $V = \pi \times 9 \times 10 = 90\pi$ — $9 \times 10 = 90$. Leaving the answer in terms of $\pi$ is exact and preferred unless a decimal is requested. [A1 — correct answer]
5. $V = 90\pi \text{ cm}^3$ — Volume is always in cubic units.

Answer: $V = 90\pi \text{ cm}^3 \approx 282.7 \text{ cm}^3$

Worked example: Total Surface Area of a Cylinder

Using the same cylinder (radius 3 cm, height 10 cm), calculate its total surface area. Give your answer to 1 decimal place.

1. $SA = 2\pi r h + 2\pi r^2$ — Curved surface + two circular ends. [M1 — correct formula structure]
2. $2\pi r h = 2 \times \pi \times 3 \times 10 = 60\pi$ — $2 \times 3 \times 10 = 60$. [M1 — correct substitution]
3. $2\pi r^2 = 2 \times \pi \times 3^2 = 2 \times \pi \times 9 = 18\pi$ — $3^2 = 3 \times 3 = 9$, then $2 \times 9 = 18$.
4. $SA = 60\pi + 18\pi = 78\pi$ — Combine like terms before converting to a decimal.
5. $SA = 78\pi \approx 78 \times 3.14159... \approx 245.0 \text{ cm}^2$ — Do not round until the final step. [A1 — correct answer to 1 d.p.]

Answer: $SA = 78\pi \approx 245.0 \text{ cm}^2$

Worked example: Volume of a Cone

A cone has base radius 4 cm and perpendicular height 9 cm. Calculate its volume. Give your answer to 3 significant figures.

1. $V = \frac{1}{3}\pi r^2 h$ — The cone formula is on the formula sheet. [M1 — correct formula]
2. $V = \frac{1}{3} \times \pi \times 4^2 \times 9$ — $r = 4$ cm, $h = 9$ cm. [M1 — correct substitution]
3. $4^2 = 4 \times 4 = 16$ — Expand the square explicitly.
4. $V = \frac{1}{3} \times \pi \times 16 \times 9 = \frac{1}{3} \times 144\pi = 48\pi$ — $16 \times 9 = 144$, then $144 \div 3 = 48$.
5. $V = 48\pi \approx 150.796... \approx 151 \text{ cm}^3$ — Round only at the final step. [A1 — correct answer to 3 s.f.]

Answer: $V = 48\pi \approx 151 \text{ cm}^3$ (to 3 s.f.)

Worked example: Volume of a Pyramid

A square-based pyramid has a base of side 6 cm and a perpendicular height of 8 cm. Calculate its volume.

1. $A_{\text{base}} = 6 \times 6 = 36 \text{ cm}^2$ — The base is a square, so area = side². [M1 — correct base area]
2. $V = \frac{1}{3} \times A_{\text{base}} \times h$ — This formula works for any pyramid. [M1 — correct formula]
3. $V = \frac{1}{3} \times 36 \times 8 = \frac{1}{3} \times 288 = 96$ — $36 \times 8 = 288$, then $288 \div 3 = 96$. [A1 — correct answer]
4. $V = 96 \text{ cm}^3$ — Volume is always in cubic units.

Answer: $V = 96 \text{ cm}^3$