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G17: Circles: Area & Circumference

Foundation Higher AQAEdexcelOCREduqasCCEA

Calculate circumference and area of circles; arcs and sectors; surface area and volume of spheres, pyramids, cones (Higher)

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📋 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
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

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

❓ 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.

✅ Answers

  1. 28π cm ≈ 87.96 cm
  2. Radius = 10 cm, Area = 100π cm²
  3. (120/360) × 2π × 9 = 6π cm ≈ 18.85 cm
  4. (90/360) × π × 36 = 9π cm²
  5. (4/3)π × 27 = 36π cm³

🎯 Exam Tips

🧠 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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