G17: Circles: Area & Circumference
Calculate circumference and area of circles; arcs and sectors; surface area and volume of spheres, pyramids, cones (Higher)
📋 Key Concepts
π (pi) ≈ 3.14159 or 22/7. In exams, use the π button on your calculator unless told otherwise.
📝 Circumference of a Circle
Circumference = 2πr or C = πd
Where r = radius, d = diameter
Example 1
Find the circumference of a circle with radius 7 cm.
Solution:
C = 2πr = 2 × π × 7 = 14π cm ≈ 43.98 cm
Example 2
Find the circumference of a circle with diameter 12 cm. Leave your answer in terms of π.
Solution:
C = πd = 12π cm
📝 Area of a Circle
Area = πr²
Example 3
Find the area of a circle with radius 5 cm.
Solution:
Area = πr² = π × 5² = 25π cm² ≈ 78.54 cm²
Example 4
Find the area of a circle with diameter 10 cm.
Solution:
Radius = 10 ÷ 2 = 5 cm
Area = π × 5² = 25π cm²
📝 Arc Length
Arc length = (θ/360) × 2πr
Where θ is the angle at the centre in degrees.
Example 5
Find the length of an arc with radius 8 cm and angle 45°.
Solution:
Arc length = (45/360) × 2π × 8
= (1/8) × 16π
= 2π cm ≈ 6.28 cm
📝 Area of a Sector
Area of sector = (θ/360) × πr²
Example 6
Find the area of a sector with radius 10 cm and angle 60°.
Solution:
Area = (60/360) × π × 10²
= (1/6) × 100π
= 50π/3 cm² ≈ 52.36 cm²
📝 Higher: Volume and Surface Area of a Sphere
Volume = (4/3)πr³
Surface area = 4πr²
Surface area = 4πr²
Example 7
Find the volume and surface area of a sphere with radius 6 cm.
Solution:
Volume = (4/3)π × 6³ = (4/3)π × 216 = 288π cm³
Surface area = 4π × 6² = 144π cm²
📝 Higher: Volume and Surface Area of a Cone
Volume = (1/3)πr²h
Surface area = πrl + πr²
Surface area = πrl + πr²
Where r = radius, h = height, l = slant height
Note: l² = r² + h² (by Pythagoras)
Example 8
Find the volume of a cone with radius 5 cm and height 12 cm.
Solution:
Volume = (1/3)π × 5² × 12
= (1/3)π × 300
= 100π cm³
📝 Higher: Volume of a Pyramid
Volume = (1/3) × base area × height
Example 9
Find the volume of a square-based pyramid with base 8 cm × 8 cm and height 12 cm.
Solution:
Volume = (1/3) × 8 × 8 × 12
= (1/3) × 768
= 256 cm³
🧮 Worked Examples — Exam Style
Circles: C = πd = 2πr, A = πr². The diameter-to-radius slip is the most common mark lost.
A wheel has radius 35 cm. How far does it travel in 10 rotations?
Circumference = 2π × 35 = 70π cm ≈ 219.9 cm
10 rotations: 219.9 × 10 ≈ 2199 cm ≈ 22.0 m
M1 for the circumference, A1 for the distance. One rotation = one circumference — the standard context.
The area of a circle is 50 cm². Find its radius (2 d.p.).
πr² = 50 → r² = 50/π = 15.9155 → r = 3.99 cm
M1 for r² = A/π, A1 for 3.99. Reverse circle questions need the calculator square root.
Exam wording to expect: "in terms of π", "to 1 decimal place", semicircles and compound shapes.
❓ Practice Questions
Q1: Find the circumference of a circle with radius 14 cm.
Q2: Find the area of a circle with diameter 20 cm. Leave in terms of π.
Q3: Find the arc length with radius 9 cm and angle 120°.
Q4: Find the area of a sector with radius 6 cm and angle 90°.
Q5: (Higher) Find the volume of a sphere with radius 3 cm.
Q6: A circular pond has diameter 6 m. Find its area (1 d.p.)
Q7: A semicircular window has radius 40 cm. Find its area (1 d.p.)
✅ Answers
- 28π cm ≈ 87.96 cm
- Radius = 10 cm, Area = 100π cm²
- (120/360) × 2π × 9 = 6π cm ≈ 18.85 cm
- (90/360) × π × 36 = 9π cm²
- (4/3)π × 27 = 36π cm³
📋 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. 2π × 14 = 28π ≈ 87.96 cm
Q2. r = 10 → 100π cm²
Q3. (120/360) × 2π × 9 = (1/3) × 18π = 6π ≈ 18.85 cm
Q4. (90/360) × π × 36 = 9π cm²
Q5. (4/3)π × 27 = 36π cm³
Q6. r = 3 → A = π × 9 = 28.3 m² (1 d.p.)
Q7. Half of π × 40² = π × 1600/2 = 2513 cm² (1 d.p.)
📌 Key Facts to Memorise
- C = 2πr = πd · A = πr² — check radius vs diameter FIRST
- Semicircle area = πr²/2; perimeter = πr + 2r
- Reverse: r = √(A/π); d = C/π
- "In terms of π" = leave π in the answer (exact)
- Sphere volume (4/3)πr³, cylinder V = πr²h — Higher only
🗺️ In the Exam
Circles: 3–5 marks per paper, usually embedded in compound shapes or real objects (ponds, wheels, pizzas). The diameter/radius switch is the targeted trap.
Links to: arcs and sectors (G18), area/volume (G16), perimeter work (G).
🧭 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
- Check if the question gives radius or diameter
- Leave answers in terms of π unless asked for a decimal
- For arc/sector, remember to divide by 360
- Higher students: learn all the formulae
- Check units are consistent
- For cones, the slant height is not the same as the height
🧠 Problem-Solving Strategies
Problem-Solving
Circumference = 2 x pi x r = pi x d. Area = pi x r squared. Use pi = 3.14 or 3.142, or leave in terms of pi. When finding the radius from the circumference, divide by 2pi. When finding the radius from the area, divide by pi then square root. Always check whether you need radius or diameter.
Multi-Step Problem
A circular running track has an inner circumference of 100 m. The track is 3 m wide. Find the area of the track surface.
Solution: Inner radius: 2 x pi x r = 100, r = 100/(2pi) = 15.92 m. Outer radius = 15.92 + 3 = 18.92 m. Track area = pi x 18.92 squared - pi x 15.92 squared = pi(18.92 squared - 15.92 squared) = pi(357.97 - 253.45) = pi x 104.52 = 328.3 m squared.
⚠️ Common Errors
Watch Out!
1. Wrong: Using the diameter instead of the radius in the area formula: pi x d squared Correct: Area = pi x r squared. If you have the diameter, first halve it to get the radius, then square.
2. Wrong: Confusing circumference and area formulas Correct: Circumference = 2 x pi x r (linear, gives a length). Area = pi x r squared (gives a squared unit). Check the units to verify: circumference is in cm, area in cm squared.
3. Wrong: Squaring pi along with the radius: pi squared x r squared Correct: Area = pi x r squared. Pi is a constant multiplier — only the radius gets squared.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A garden has a circular pond of radius 2 m surrounded by a path of width 1.5 m. (a) Find the area of the path. (b) The path is paved with slabs costing £18 per m squared. Find the total cost. (c) A second circular pond has the same area as the path. Find its radius.
(a) Inner radius = 2 m. Outer radius = 2 + 1.5 = 3.5 m. Path area = pi x 3.5 squared - pi x 2 squared = pi(12.25 - 4) = 8.25pi = 25.9 m squared.
(b) Cost = 25.9 x 18 = £466.20.
(c) Area of second pond = 8.25pi. pi x r squared = 8.25pi. r squared = 8.25. r = 2.87 m (2 d.p.).
Mark scheme: M1 both areas, A1 path area 8.25pi, M1 cost, A1 £466.20, M1 equating areas, A1 r = 2.87
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A pizza has radius 15 cm. A smaller pizza has radius 10 cm.
(a) How many times larger is the area of the big pizza than the small one?
(b) The big pizza costs £9 and the small costs £5. Which gives better value per cm squared?
(c) A student says "The big pizza is 50% wider so it has 50% more area." Explain the error.
Answers: (a) Area ratio = 15 squared / 10 squared = 225/100 = 2.25. The big pizza is 2.25 times larger (125% more area). (b) Big: 9/(pi x 225) = £0.01273/cm squared. Small: 5/(pi x 100) = £0.01592/cm squared. The big pizza gives better value. (c) Area scales with the square of the radius. A 50% increase in radius gives (1.5) squared = 2.25 times the area, which is a 125% increase, not 50%. The student confuses linear and area scaling.
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