Geometry & MeasuresExam Tips
Exam Tips
Part of Circumference & Area of Circles · GCSE GCSE Mathematics revision
This exam tips covers Exam Tips within Circumference & Area of Circles for GCSE Mathematics. Revise Circumference & Area of Circles in Geometry & Measures for GCSE Mathematics with 11 exam-style questions and 7 flashcards. This topic appears regularly enough that it should still be part of a steady revision cycle. It is section 9 of 10 in this topic. Treat this as a marking guide for what examiners are looking for, not just a fact list.
Topic position
Section 9 of 10
Practice
11 questions
Recall
7 flashcards
Exam Tips
- If given diameter, HALVE IT to get radius before using A = πr²
- Leave answers in terms of π if asked (e.g., 25π cm²)
- Use the π button on your calculator, not 3.14
- Check your units - circumference uses cm, area uses cm²
Practice Questions
Q1 Find the circumference of a circle with diameter 10 cm (1 dp) [2 marks]
[1] C = πd = π × 10
[1] C = 31.4 cm
Q2 Find the area of a circle with radius 5 cm (1 dp) [2 marks]
[1] A = πr² = π × 5² = π × 25
[1] A = 78.5 cm²
Q3 Find the area of a circle with diameter 8 cm (1 dp) [3 marks]
[1] Radius = 8 ÷ 2 = 4 cm
[1] A = πr² = π × 4² = π × 16
[1] A = 50.3 cm²
Q4 A circular pond has radius 6 m. Find the distance around the pond (1 dp) [2 marks]
[1] C = 2πr = 2 × π × 6
[1] C = 37.7 m
Q5 Calculate the circumference of a circle with diameter 12 cm. Give your answer to 1 decimal place. [2 marks]
[1] C = πd = π × 12
[1] = 37.7 cm (1 dp)
Examiner says: Use the π button on your calculator, not 3.14 or 3.142 - it's more accurate!
Q6 Find the area of a circle with radius 6 cm. Give your answer to 1 decimal place. [2 marks]
[1] A = πr² = π × 6² = π × 36
[1] = 113.1 cm² (1 dp)
Q7 A circular table has diameter 120 cm. Calculate its area in m². Give your answer to 2 decimal places. [3 marks]
[1] Radius = 120 ÷ 2 = 60 cm = 0.6 m
[1] A = πr² = π × 0.6² = π × 0.36
[1] = 1.13 m² (2 dp)
Examiner says: Don't forget to convert to metres first! 120 cm = 1.2 m, so radius = 0.6 m.
Q8 A bicycle wheel has diameter 70 cm. How far does the bicycle travel in one complete revolution of the wheel? Give your answer to the nearest cm. [3 marks]
[1] Distance = circumference of wheel
[1] C = πd = π × 70
[1] = 220 cm (nearest cm)
Examiner says: One revolution = one complete turn = circumference distance traveled.
Q9 A circle has radius 9 cm. Find its circumference. Give your answer to 1 decimal place. [3 marks]
[1] C = 2πr (using radius formula)
[1] C = 2 × π × 9 = 18π
[1] = 56.5 cm (1 dp)
Q10 A circular lawn has diameter 8 m. Grass seed costs £2.50 per m². What is the total cost to seed the lawn? Give your answer to the nearest penny. [4 marks]
[1] Radius = 8 ÷ 2 = 4 m
[1] Area = πr² = π × 4² = 16π ≈ 50.27 m²
[1] Cost = 50.27 × £2.50
[1] = £125.66
Q11 Find the area of a circle with radius 7 cm. Leave your answer in terms of π. [2 marks]
[1] A = πr² = π × 7²
[1] = 49π cm²
Examiner says: When asked to leave in terms of π, DON'T calculate - just leave the π symbol in your answer!
Q12 The circumference of a circle is 50 cm. Find its radius. Give your answer to 2 decimal places. [4 marks]
[1] C = 2πr, so 50 = 2πr
[1] r = 50 ÷ (2π)
[1] r = 50 ÷ 6.283...
[1] = 7.96 cm (2 dp)
Q13 A circular garden has area 100 m². Calculate its diameter. Give your answer to 1 decimal place. [4 marks]
[1] A = πr², so 100 = πr²
[1] r² = 100 ÷ π = 31.831...
[1] r = √31.831 = 5.642 m
[1] Diameter = 2 × 5.642 = 11.3 m (1 dp)
Q14 A semicircular window has diameter 80 cm. Find: (a) the perimeter, (b) the area. Give answers to 1 decimal place. [5 marks]
[1] (a) Perimeter = semicircle arc + diameter
[1] Arc = ½ × πd = ½ × π × 80 = 40π ≈ 125.7 cm
[1] Perimeter = 125.7 + 80 = 205.7 cm
[1] (b) Area = ½ × πr² = ½ × π × 40² = 800π
[1] = 2513.3 cm² (1 dp)
Q15 A circular ring (annulus) has outer radius 10 cm and inner radius 7 cm. Find the area of the ring. Give your answer to 1 decimal place. [5 marks]
[1] Outer circle area = π × 10² = 100π cm²
[1] Inner circle area = π × 7² = 49π cm²
[1] Ring area = 100π - 49π = 51π
[2] = 160.2 cm² (1 dp)
Examiner says: Annulus (ring) area = large circle - small circle. You can factor: π(R² - r²).
Q16 A car wheel has diameter 55 cm. The car travels 3 km. How many complete revolutions does the wheel make? [6 marks]
[1] Circumference = πd = π × 55 ≈ 172.79 cm
[1] Distance traveled = 3 km = 3000 m = 300,000 cm
[2] Number of revolutions = 300,000 ÷ 172.79
[1] = 1736.04...
[1] = 1736 complete revolutions
Q17 The radius of a circle is increased by 20%. Find the percentage increase in: (a) circumference, (b) area. [6 marks]
[1] Let original radius = r. New radius = 1.2r
[1] (a) Original C = 2πr, new C = 2π(1.2r) = 2.4πr
[1] Increase = (2.4πr - 2πr)/(2πr) × 100 = 20%
[1] (b) Original A = πr², new A = π(1.2r)² = 1.44πr²
[1] Increase = (1.44πr² - πr²)/(πr²) × 100
[1] = 0.44 × 100 = 44%
Examiner says: Circumference is linear (scales by 1.2), but area is squared (scales by 1.2² = 1.44)!
Q18 A running track consists of two straight sections of 100 m joined by two semicircular ends with diameter 60 m. Find the total distance around the track. [6 marks]
[1] Two straight sections = 2 × 100 = 200 m
[2] Two semicircular ends = one complete circle with diameter 60 m
[1] Circular part = πd = π × 60 ≈ 188.5 m
[1] Total distance = 200 + 188.5
[1] = 388.5 m (or 389 m to nearest metre)
Q19 A square has side length 10 cm. A circle is inscribed inside it (touching all four sides). Find the area of the shaded region between the square and circle. Give your answer to 2 decimal places. [7 marks]
[1] Square area = 10² = 100 cm²
[2] Circle diameter = square side = 10 cm, so radius = 5 cm
[2] Circle area = πr² = π × 5² = 25π ≈ 78.54 cm²
[1] Shaded area = 100 - 78.54
[1] = 21.46 cm² (2 dp)
Q20 The area of a circle is numerically equal to its circumference. Find the radius. [7 marks]
[2] Area = Circumference, so πr² = 2πr
[2] Dividing both sides by πr: r = 2
[2] Check: A = π(2)² = 4π, C = 2π(2) = 4π ✓
[1] Radius = 2 units
Examiner says: "Numerically equal" means the numbers are the same, ignoring units (cm² vs cm).
Q21 A circular pizza has diameter 30 cm. One slice is a sector with angle 60° at the centre. Find the area of the slice. Give your answer to 1 decimal place. [8 marks]
[1] Radius = 30 ÷ 2 = 15 cm
[2] Full circle area = πr² = π × 15² = 225π cm²
[2] Sector angle is 60° out of 360°, so fraction = 60/360 = 1/6
[2] Sector area = 1/6 × 225π = 37.5π
[1] = 117.8 cm² (1 dp)
Examiner says: Sector area = (angle/360) × πr² - remember the fraction of the full circle!
Q22 A piece of wire 100 cm long is bent into a circle. Find: (a) the radius, (b) the area enclosed. Give answers to 2 decimal places. [8 marks]
[2] (a) Wire length = circumference, so 2πr = 100
[2] r = 100 ÷ (2π) = 100 ÷ 6.283... = 15.92 cm (2 dp)
[2] (b) A = πr² = π × 15.92²
[2] = π × 253.45 = 796.18 cm² (2 dp)
Q23 Four identical circles of radius 5 cm are placed so that each touches two others. The centres form a square. Find the area of the region inside the square but outside the circles. Give your answer to 1 decimal place. [9 marks]
[2] Distance between adjacent centres = 2 × radius = 10 cm
[1] Square side = 10 cm, so square area = 10² = 100 cm²
[2] Each circle contributes a quarter-circle inside the square
[2] Total circle area inside = 4 × (¼πr²) = πr² = π × 5² = 25π ≈ 78.54 cm²
[1] Shaded area = 100 - 78.54
[1] = 21.5 cm² (1 dp)
Q24 Two circles have areas in the ratio 4:9. The smaller circle has radius 6 cm. Find: (a) the radius of the larger circle, (b) the ratio of their circumferences. [9 marks]
[2] (a) Area ratio = 4:9, so radius ratio = √4 : √9 = 2:3
[2] If smaller radius = 6, then 2 parts = 6, so 1 part = 3
[1] Larger radius = 3 parts = 3 × 3 = 9 cm
[2] (b) Circumference ratio = radius ratio (since C = 2πr)
[2] Ratio = 2:3 (or 6:9)
Examiner says: Area scales as radius², so radius scales as √(area ratio).
Q25 A circle has radius 8 cm. A chord is 6 cm from the centre. Find the area of the minor segment (region between chord and arc). Give your answer to 2 decimal places. [Use: area of segment = area of sector - area of triangle] [10 marks]
[2] Triangle: half-chord = √(8² - 6²) = √28 = 5.29 cm (Pythagoras)
[2] Triangle area = ½ × base × height = ½ × (2×5.29) × 6 = 31.75 cm²
[2] Angle: cos(θ/2) = 6/8 = 0.75, so θ/2 = 41.41°, θ = 82.82°
[2] Sector area = (82.82/360) × π × 8² = 46.18 cm²
[1] Segment area = 46.18 - 31.75
[1] = 14.43 cm² (2 dp)
Q26 A sector of a circle has radius 12 cm and arc length 15 cm. Find: (a) the angle of the sector in degrees, (b) the area of the sector. Give answers to 1 decimal place. [10 marks]
[2] (a) Arc length = (θ/360) × 2πr, so 15 = (θ/360) × 2π × 12
[2] 15 = (θ/360) × 24π
[2] θ = (15 × 360)/(24π) = 71.6°
[1] Angle = 71.6° (1 dp)
[2] (b) Sector area = ½ × r × arc = ½ × 12 × 15
[1] = 90.0 cm²
Examiner says: Alternative sector area formula: ½rL where L is arc length - often faster!
Q27 A Norman window consists of a rectangle topped by a semicircle. The perimeter of the window is 6 m and the width is x m. Show that the area is A = 3x - x²(1 + π/4), and find the value of x that maximises the area. [11 marks]
[2] Let height of rectangle = h. Perimeter = 2h + x + πx/2 = 6
[2] So 2h = 6 - x - πx/2, h = 3 - x/2 - πx/4
[2] Area = rectangle + semicircle = xh + ½π(x/2)²
[2] A = x(3 - x/2 - πx/4) + πx²/8 = 3x - x²/2 - πx²/4 + πx²/8
[1] A = 3x - x²/2 - πx²/8 = 3x - x²(1/2 + π/8) = 3x - x²(4/8 + π/8) = 3x - x²(4+π)/8
[1] Note: Question uses π/4 factor, alternative form. Maximum: dA/dx = 0
[1] x ≈ 1.68 m (calculus solution)
Q28 Prove that if a circle of radius r is inscribed in a square, the ratio of the circle's area to the square's area is π:4. [11 marks]
[2] Circle inscribed means it touches all four sides
[2] Square side = circle diameter = 2r
[2] Square area = (2r)² = 4r²
[2] Circle area = πr²
[2] Ratio = πr² : 4r² = π : 4 ✓
[1] This ratio is constant for ANY circle inscribed in a square!
Q29 A sphere has surface area 314 cm². Find: (a) its radius, (b) its volume. Use formulas: Surface area = 4πr², Volume = (4/3)πr³. Give answers to 1 decimal place. [12 marks]
[2] (a) Surface area = 4πr², so 314 = 4πr²
[2] r² = 314 ÷ (4π) = 314 ÷ 12.566 = 24.99
[2] r = √24.99 = 5.0 cm (1 dp)
[2] (b) V = (4/3)πr³ = (4/3)π × 5³
[2] = (4/3)π × 125 = 500π/3
[2] = 523.6 cm³ (1 dp)
Examiner says: This extends circle formulas to 3D - spheres use similar ideas but with cubed/squared relationships!
Q30 An infinite sequence of circles: first circle has radius 1, second has radius 1/2, third has radius 1/4, and so on (each radius is half the previous). Find the sum of: (a) all the circumferences, (b) all the areas. [12 marks]
[2] (a) Circumferences: 2π(1) + 2π(1/2) + 2π(1/4) + ...
[2] = 2π(1 + 1/2 + 1/4 + ...) - geometric series with a=1, r=1/2
[2] Sum = 2π × [1/(1-1/2)] = 2π × 2 = 4π (or 12.57 units)
[2] (b) Areas: π(1)² + π(1/2)² + π(1/4)² + ...
[2] = π(1 + 1/4 + 1/16 + ...) - geometric series with a=1, r=1/4
[2] Sum = π × [1/(1-1/4)] = π × (4/3) = 4π/3 (or 4.19 square units)
Examiner says: This is A-level maths! Infinite geometric series: S = a/(1-r) when |r| < 1.
Keep building this topic
Read this section alongside the surrounding pages in Circumference & Area of Circles. That gives you the full topic sequence instead of a single isolated revision point.
Practice Questions for Circumference & Area of Circles
A circle has radius r. Which pair of formulas gives the correct circumference C and area A?
Explain why the circumference of a circle is measured in cm (or other length units) but the area is measured in cm² (square units).
Quick Recall Flashcards
Circumference
C = πd or C = 2πr
Area of Circle
A = πr²
11 questions on Circumference & Area of Circles — practise free
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