G16: Area & Volume
Know and apply formulae to calculate area of triangles, parallelograms, trapezia; volume of cuboids and prisms
📋 Key Concepts
Area is the amount of space inside a 2D shape, measured in square units (cm², m², etc.).
Volume is the amount of space inside a 3D shape, measured in cubic units (cm³, m³, etc.).
📝 Area of Rectangle
Area = length × width
Example 1
Find the area of a rectangle with length 8 cm and width 5 cm.
Solution:
Area = 8 × 5 = 40 cm²
📝 Area of Triangle
Area = ½ × base × height
Important: The height must be perpendicular to the base.
Example 2
Find the area of a triangle with base 10 cm and height 6 cm.
Solution:
Area = ½ × 10 × 6 = 30 cm²
Example 3
Find the area of a triangle with base 8 cm and perpendicular height 5 cm.
Solution:
Area = ½ × 8 × 5 = 20 cm²
📝 Area of Parallelogram
Area = base × height
Note: The height is the perpendicular height, not the slanted side.
Example 4
Find the area of a parallelogram with base 12 cm and height 7 cm.
Solution:
Area = 12 × 7 = 84 cm²
📝 Area of Trapezium
Area = ½(a + b) × height
Where a and b are the lengths of the parallel sides.
Example 5
Find the area of a trapezium with parallel sides 6 cm and 10 cm, and height 4 cm.
Solution:
Area = ½(6 + 10) × 4
= ½ × 16 × 4
= 32 cm²
📝 Volume of Cuboid
Volume = length × width × height
Example 6
Find the volume of a cuboid with dimensions 5 cm × 4 cm × 3 cm.
Solution:
Volume = 5 × 4 × 3 = 60 cm³
📝 Volume of Prism
Volume = area of cross-section × length
Prism: A 3D shape with the same cross-section throughout its length.
Example 7
A triangular prism has a cross-section with area 24 cm² and length 10 cm. Find the volume.
Solution:
Volume = 24 × 10 = 240 cm³
Example 8
Find the volume of a prism with triangular cross-section (base 6 cm, height 4 cm) and length 15 cm.
Solution:
Area of triangle = ½ × 6 × 4 = 12 cm²
Volume = 12 × 15 = 180 cm³
📝 Compound Shapes
Method: Split the shape into simpler shapes, find each area, then add them together.
Example 9
Find the area of an L-shape made from two rectangles: 8 cm × 5 cm and 4 cm × 3 cm.
Solution:
Area = (8 × 5) + (4 × 3) = 40 + 12 = 52 cm²
🧮 Worked Examples — Exam Style
Area formulas per shape; volumes = area of cross-section × depth. Compound shapes: split into known parts.
Find the volume of a triangular prism with cross-section area 24 cm² and length 15 cm.
V = area × length = 24 × 15 = 360 cm³
M1 for area × length, A1 for 360. For ANY prism: cross-section area × depth — the formula list collapses to one idea.
A trapezium has parallel sides 8 cm and 12 cm, height 5 cm. A prism built on it is 20 cm long. Find the volume.
Trapezium area = ½(8 + 12) × 5 = 50 cm² → V = 50 × 20 = 1000 cm³
M1 for the trapezium area, A1 for 1000. The classic two-step: 2-D area, then extend to 3-D.
Exam wording to expect: "find the volume/area of …", compound shapes, unit conversions (cm³ → litres).
❓ Practice Questions
Q1: Find the area of a triangle with base 14 cm and height 9 cm.
Q2: Find the area of a trapezium with parallel sides 5 cm and 9 cm, height 6 cm.
Q3: Find the volume of a cuboid 10 cm × 8 cm × 6 cm.
Q4: A parallelogram has area 56 cm² and base 8 cm. Find the height.
Q5: A prism has cross-sectional area 35 cm² and volume 210 cm³. Find its length.
Q6: A water tank is a cuboid 2 m × 1.5 m × 1.2 m. Find its capacity in litres
Q7: A cylinder has radius 5 cm and height 12 cm. Find its volume (1 d.p.)
✅ Answers
- 63 cm²
- 42 cm²
- 480 cm³
- 7 cm
- 6 cm
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. ½ × 14 × 9 = 63 cm²
Q2. ½(5 + 9) × 6 = ½ × 84 = 42 cm²
Q3. 10 × 8 × 6 = 480 cm³
Q4. 56 ÷ 8 = 7 cm
Q5. 210 ÷ 35 = 6 cm
Q6. V = 2 × 1.5 × 1.2 = 3.6 m³ = 3600 litres (1 m³ = 1000 L)
Q7. V = π × 25 × 12 = 942.5 cm³ (1 d.p.)
📌 Key Facts to Memorise
- Rectangle lw · Triangle ½bh · Parallelogram bh · Trapezium ½(a + b)h
- Cuboid lwh · Prism = cross-section area × length · Cylinder πr²h
- 1 litre = 1000 cm³ · 1 m³ = 1,000,000 cm³ = 1000 litres
- Compound shapes: split, compute, add (or subtract holes)
- Reverse: length = volume ÷ cross-section area
🗺️ In the Exam
Area and volume: 4–6 marks per paper, nearly always with a real object and a unit conversion. The cross-section idea converts every prism problem into one formula.
Links to: circles (G17), units (N13), similar shapes (G19).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Always include units in your answer (cm², m², cm³, etc.)
- For triangles, use perpendicular height, not the slanted side
- For trapezium, identify the two parallel sides first
- Check your answer is sensible - estimate roughly first
- For compound shapes, split into shapes you can calculate
- Remember: area = 2D, volume = 3D
🧠 Problem-Solving Strategies
Problem-Solving
Area of rectangle = l x w, triangle = 1/2 x b x h, parallelogram = b x h, trapezium = 1/2(a + b) x h, circle = pi x r squared. Surface area of cuboid = 2(lw + lh + wh). For composite shapes, split into simpler shapes. For surface area of prisms, find all face areas separately. When scaling, area scales as the square of the linear factor.
Multi-Step Problem
A trapezium has parallel sides 8 cm and 14 cm, height 5 cm. A triangle has the same area as the trapezium with base 11 cm. Find the height of the triangle.
Solution: Trapezium area = 1/2 x (8 + 14) x 5 = 1/2 x 22 x 5 = 55 cm squared. Triangle: 1/2 x 11 x h = 55. h = 55 x 2/11 = 10 cm.
⚠️ Common Errors
Watch Out!
1. Wrong: Calculating triangle area as base x height (forgetting the 1/2) Correct: Area of triangle = 1/2 x base x height. The 1/2 is essential. A triangle is half a parallelogram with the same base and height.
2. Wrong: Using the slant height instead of the perpendicular height for area Correct: Area formulas use the PERPENDICULAR height, not the slant height. The perpendicular height is the shortest distance from the base to the opposite vertex.
3. Wrong: For the trapezium, multiplying the sum of parallel sides by height without halving Correct: Trapezium area = 1/2 x (a + b) x h. You must halve the sum of the parallel sides before multiplying by height.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A garden is in the shape of a rectangle 12 m by 8 m with a semicircular patio of radius 4 m attached to one 8 m side. (a) Find the total area of the garden. (b) Turf costs £4.50 per m squared for the rectangular part and paving costs £12 per m squared for the patio. Find the total cost. (c) A circular pond of radius 1.5 m is added in the centre of the rectangular lawn. How much turf area is saved?
(a) Rectangle = 12 x 8 = 96 m squared. Semicircle = 1/2 x pi x 4 squared = 8pi = 25.13 m squared. Total = 121.1 m squared.
(b) Turf cost = 96 x 4.50 = £432. Paving cost = 25.13 x 12 = £301.59. Total = £733.59.
(c) Pond area = pi x 1.5 squared = 7.07 m squared. Turf saved = 7.07 m squared. Cost saved = 7.07 x 4.50 = £31.80.
Mark scheme: M1 rectangle and semicircle, A1 total area, M1 both costs, A1 total cost, M1 pond area, A1 turf saved
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Farmer Brown has 200 m of fencing to enclose a rectangular field next to a river (no fence needed on the river side).
(a) If the river forms one length, express the area in terms of the width w.
(b) What dimensions give the maximum area?
(c) The farmer says "A square field always gives the maximum area." Is this true when one side is a river? Explain.
Answers: (a) Three sides fenced: two widths and one length. 2w + l = 200, so l = 200 - 2w. Area = w x (200 - 2w) = 200w - 2w squared. (b) This is a quadratic with maximum at w = 200/4 = 50 m. Length = 200 - 100 = 100 m. Maximum area = 50 x 100 = 5000 m squared. (c) No — with a river, the optimal shape is 100 x 50 (a 2:1 rectangle), not a square. A square would give 66.7 x 66.7 = 4444 m squared, which is less. The river removes one side constraint, changing the optimal proportions.
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