⭕ Area of a Circle

Cambridge Lower Secondary · Grade 7 · Geometry & Measure

Full Circle
A = πr²  ·  r = 6 cm
A = π × 36 ≈ 113.1 cm²
Given Diameter
d = 10 cm → r = 5 cm
A = π × 25 ≈ 78.5 cm²
Semicircle
r = 8 cm → A = πr² ÷ 2
A = π × 64 ÷ 2 ≈ 100.5 cm²

Watch a circle fill up! ⭕


What you'll learn:

  • The formula A = πr² and why it works
  • Finding area given the radius
  • Finding area given the diameter (halve it first!)
  • Working backwards: find r when you know A → r = √(A/π)
  • Area of a semicircle and quarter circle
  • Area of composite shapes involving circles
  • Comparing circle area to squares and rectangles

📖 Learn: Area of a Circle

Part 1: The Formula A = πr²

The area of a circle is the amount of space inside it. The formula uses π (pi) ≈ 3.14159… and the radius (r) — the distance from the centre to the edge.

📌 Formula: A = π × r²   (pi times radius squared)
The radius is half the diameter: r = d ÷ 2
Always square the radius first, then multiply by π.
💡 Use the π button on your calculator for full precision, or use π ≈ 3.142 when calculating by hand.

Part 2: Why A = πr² ? (The Sector Proof)

Imagine slicing a circle into many thin sectors (like pizza slices). Rearrange them alternating up/down — they form a rough rectangle!

Height of the "rectangle" = r (the radius)
Width of the "rectangle" = πr (half the circumference, because C = 2πr)
Area of the rectangle = height × width = r × πr = πr²
🍕 The more slices, the more rectangular it becomes. See the Visualizer tab for the animation!

Part 3: Area Given the Radius

When given the radius, substitute directly into A = πr².

Example: r = 5 cm  →  A = π × 5² = π × 25 ≈ 78.5 cm²
Example: r = 3.5 m  →  A = π × 3.5² = π × 12.25 ≈ 38.5 m²
Example: r = 12 mm  →  A = π × 144 ≈ 452.4 mm²
⚠️ The most common mistake is forgetting to square r before multiplying. r² ≠ 2r!

Part 4: Area Given the Diameter

When given the diameter, always halve it to find the radius first!

📌 r = d ÷ 2, then use A = πr²
Example: d = 14 cm → r = 7 cm → A = π × 49 ≈ 153.9 cm²
Example: d = 9 m → r = 4.5 m → A = π × 20.25 ≈ 63.6 m²
💡 Never use d directly in the formula! A = π(d/2)² is correct, but it's cleaner to halve first.

Part 5: Working Backwards — Find r Given Area

If you know the area and need to find the radius, rearrange the formula:

📌 A = πr²  →  r² = A ÷ π  →  r = √(A ÷ π)
Example: A = 200 cm² → r² = 200 ÷ π ≈ 63.66 → r = √63.66 ≈ 7.98 cm
Example: A = 50.3 m² → r² = 50.3 ÷ π ≈ 16.0 → r = √16 = 4.0 m
🔁 Steps: (1) Divide area by π. (2) Take the square root. That's your radius!

Part 5b: Semicircle and Quarter Circle

A semicircle is half a circle; a quarter circle is one quarter.

Semicircle area: A = πr² ÷ 2
Quarter circle area: A = πr² ÷ 4
Example (semi): r = 6 cm → A = π × 36 ÷ 2 ≈ 56.5 cm²
Example (quarter): r = 10 m → A = π × 100 ÷ 4 ≈ 78.5 m²
💡 Calculate the full circle area first, then divide by 2 or 4. Same radius, different fraction!

💡 Worked Examples

Example 1: Area Given Radius

Find the area of a circle with radius 7 cm. Give your answer to 1 decimal place.

Step 1: Write the formula: A = πr²
Step 2: Substitute: A = π × 7²
Step 3: Square: A = π × 49
Step 4: Multiply: A = 153.938… ≈ 153.9 cm²
📝 Always include the unit squared (cm², m², etc.). Area is always in square units.

Example 2: Area Given Diameter

A circular pond has a diameter of 18 m. Find its area to 1 d.p.

Step 1: Find radius: r = 18 ÷ 2 = 9 m
Step 2: Apply formula: A = π × 9²
Step 3: Square: A = π × 81
Step 4: Calculate: A = 254.469… ≈ 254.5 m²
⚠️ Common error: using d = 18 directly → π × 18² = 1017.9 (wrong — 4× too large!)

Example 3: Find Radius Given Area

A circle has area 314.2 cm². Find its radius to 1 d.p.

Step 1: Write rearranged formula: r = √(A ÷ π)
Step 2: Divide: 314.2 ÷ π = 314.2 ÷ 3.14159… ≈ 100.01
Step 3: Square root: r = √100.01 ≈ 10.0 cm
🔍 Check: π × 10² = 314.2 ✅

Example 4: Composite Shape

A rectangle is 10 cm × 6 cm. A semicircle of diameter 6 cm is added to one short end. Find the total area to 1 d.p.

Step 1: Rectangle area = 10 × 6 = 60 cm²
Step 2: Semicircle: d = 6 → r = 3 cm
Step 3: Semicircle area = π × 3² ÷ 2 = π × 9 ÷ 2 ≈ 14.1 cm²
Step 4: Total = 60 + 14.1 = 74.1 cm²
💡 For composite shapes: break into simpler parts, find each area, then add (or subtract if there's a hole).

⭕ Circle Area Visualizer

🍕 Sector Rearrangement — Why A = πr²

Watch the circle slice into sectors and rearrange into a rectangle. Height = r, Width = πr, so Area = πr²!


8

📐 Live Circle Calculator

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A = π × 6² = 113.1 cm²

🔷 Composite Shape Builder

Click shapes to add them to your composite figure (using r = 5 cm as the base unit). Click a placed shape to remove it.

⭕ Full Circle (r=5)
🌙 Semicircle (r=5)
🍕 Quarter (r=5)
🟦 5×5 Square
🟦 10×5 Rectangle
Click shapes above to add them here…
Total Area: 0.0 cm²

🤔 Is the Circle Bigger?

A circle and a square both have the same "size". The circle's radius = the square's side. Which has greater area?

7 cm

⭕ Circle (r = 7 cm)

🟦 Square (side = 7 cm)

Your prediction: which has greater area?

✏️ Exercise 1: Area Given Radius

Calculate the area of each circle. Give answers to 1 decimal place. Use A = πr²

✏️ Exercise 2: Area Given Diameter

Find the radius first (r = d ÷ 2), then calculate the area to 1 decimal place.

✏️ Exercise 3: Find the Radius

Use r = √(A ÷ π) to find the radius. Give answers to 1 decimal place.

🍕 Exercise 4: Semicircles & Quarter Circles

Calculate the shaded area to 1 decimal place. Remember: semi = πr²÷2, quarter = πr²÷4.

🔷 Exercise 5: Composite Shapes

Break each shape into parts, calculate each area, then combine. Give answers to 1 decimal place.

📝 Practice Questions

  1. Find the area of a circle with radius 4 cm. (1 d.p.)
  2. Find the area of a circle with radius 9 m. (1 d.p.)
  3. Find the area of a circle with radius 2.5 cm. (1 d.p.)
  4. Find the area of a circle with diameter 10 cm. (1 d.p.)
  5. Find the area of a circle with diameter 7 m. (1 d.p.)
  6. Find the area of a circle with diameter 13 cm. (1 d.p.)
  7. A circle has area 78.5 cm². Find its radius. (1 d.p.)
  8. A circle has area 200.96 cm². Find its radius. (1 d.p.)
  9. Find the area of a semicircle with radius 5 cm. (1 d.p.)
  10. Find the area of a semicircle with diameter 12 m. (1 d.p.)
  11. Find the area of a quarter circle with radius 8 cm. (1 d.p.)
  12. A rectangle (12 cm × 8 cm) has a semicircle of diameter 8 cm on one short end. Find the total area. (1 d.p.)
  13. A square of side 10 cm has a quarter circle of radius 10 cm cut from one corner. Find the remaining area. (1 d.p.)
  14. A circular pizza has diameter 30 cm. Find its area in cm². (1 d.p.)
  15. A circular flower bed has radius 3.5 m. Find the area in m². (1 d.p.)
  16. Two circles: Circle A has r = 6 cm, Circle B has r = 3 cm. By how much is Circle A's area greater than Circle B's? (1 d.p.)
  17. A circle and a square share the same "radius" — the circle has r = 5 cm and the square has side 5 cm. How much bigger is the circle's area than the square's area? (1 d.p.)
  18. A semicircle sits on top of a rectangle. The rectangle is 8 cm × 5 cm and the semicircle has diameter 8 cm. What is the total area? (1 d.p.)
  19. A dartboard has a bullseye circle of radius 6 mm inside a larger circle of radius 18 mm. Find the area of the ring between them. (1 d.p.)
  20. A running track has two semicircular ends each with radius 40 m and two straight sections 80 m long and 5 m wide. Find the area of the track surface only (not the inner field). (1 d.p.)
  1. A = π × 16 ≈ 50.3 cm²
  2. A = π × 81 ≈ 254.5 m²
  3. A = π × 6.25 ≈ 19.6 cm²
  4. r = 5 cm → A = π × 25 ≈ 78.5 cm²
  5. r = 3.5 m → A = π × 12.25 ≈ 38.5 m²
  6. r = 6.5 cm → A = π × 42.25 ≈ 132.7 cm²
  7. r = √(78.5 ÷ π) ≈ √25 = 5.0 cm
  8. r = √(200.96 ÷ π) ≈ √64 = 8.0 cm
  9. A = π × 25 ÷ 2 ≈ 39.3 cm²
  10. r = 6 m → A = π × 36 ÷ 2 ≈ 56.5 m²
  11. A = π × 64 ÷ 4 ≈ 50.3 cm²
  12. Rectangle: 96 cm². Semicircle (r=4): π×16÷2 ≈ 25.1. Total ≈ 121.1 cm²
  13. Square: 100 cm². Quarter circle (r=10): π×100÷4 ≈ 78.5. Remaining: 100 − 78.5 ≈ 21.5 cm²
  14. r = 15 cm → A = π × 225 ≈ 706.9 cm²
  15. A = π × 12.25 ≈ 38.5 m²
  16. A_A = π×36 ≈ 113.1; A_B = π×9 ≈ 28.3; Difference ≈ 84.8 cm²
  17. Circle: π×25 ≈ 78.5; Square: 25; Difference ≈ 53.5 cm²
  18. Rectangle: 40; Semicircle (r=4): ≈ 25.1; Total ≈ 65.1 cm²
  19. Large: π×324 ≈ 1017.9; Small: π×36 ≈ 113.1; Ring ≈ 904.8 mm²
  20. Two semicircle track rings (outer r=45, inner r=40). Ring area = π(45²−40²) = π×425 ≈ 1335.2. Two straight sections: 2×(80×5) = 800. Total ≈ 2135.2 m²

🏆 Challenge: Multi-Step Problems

  1. A garden contains a rectangular lawn (15 m × 10 m) with a circular pond of diameter 4 m cut into it. Find the remaining lawn area to 1 d.p.
  2. A semicircle is placed on top of a triangle. The triangle has base 10 cm and height 8 cm. The semicircle has diameter 10 cm. Find the total area to 1 d.p.
  3. Circle A has radius r. Circle B has radius 2r. Show that Circle B's area is exactly 4 times Circle A's area.
  4. A target consists of 3 concentric circles with radii 2 cm, 5 cm, and 9 cm. Find the area of each ring (between consecutive circles) to 1 d.p.
  5. A logo is made of a full circle of radius 8 cm with two quarter circles of radius 4 cm cut from opposite corners of an 8 cm × 8 cm square inside it. Find the logo's shaded area to 1 d.p. (Hint: find full circle area, subtract square, add back the two cut quarters' inverse.)
  6. A wheel has diameter 60 cm. It rolls 10 full rotations. The road surface it rolls over is a rectangle 10 cm wide and as long as the distance covered. Find the area of the road in m². (Hint: circumference × rotations = length.)
  7. A farmer has 100 m of fencing. He can make a circular pen or a square pen. Which encloses more area, and by how much? Give answers to 1 d.p.
  8. The ratio of two circles' radii is 3:5. The smaller circle has area 113.1 cm². Find the larger circle's area to 1 d.p. without measuring the larger radius.
  1. Lawn: 150 m². Pond (r=2): π×4 ≈ 12.6 m². Remaining ≈ 137.4 m²
  2. Triangle: ½×10×8 = 40 cm². Semicircle (r=5): π×25÷2 ≈ 39.3 cm². Total ≈ 79.3 cm²
  3. A_A = πr². A_B = π(2r)² = 4πr² = 4 × A_A ✅ Ratio is 4:1 (area scales as the square of the radius ratio)
  4. Bullseye: π×4 ≈ 12.6 cm². Ring 1 (5²−2²)×π = π×21 ≈ 65.97 → Ring 1 ≈ 65.97 − 12.6 → middle ring ≈ 65.97 cm²? Precisely: inner disc: 12.6, middle ring: π(25−4) = 65.97 ≈ 66.0 cm², outer ring: π(81−25) = π×56 ≈ 175.9 cm²
  5. Full circle (r=8): 201.1 cm². Square inside: 64 cm². Two quarter circles (r=4): 2×(π×16÷4) = 25.1 cm². Shaded = 201.1 − 64 + 25.1 ≈ 162.2 cm² (method depends on exact logo — accept equivalent working)
  6. C = π × 60 ≈ 188.5 cm = 1.885 m per rotation. Distance = 1.885 × 10 = 18.85 m. Road area = 18.85 × 0.10 ≈ 1.9 m²
  7. Square (side=25): area = 625 m². Circle (C=100 → r=100÷2π≈15.92): area = π×15.92²≈795.8 m². Circle is bigger by 795.8 − 625 ≈ 170.8 m²
  8. Ratio of radii = 3:5, so ratio of areas = 9:25. If small area = 113.1, large = 113.1 × 25/9 ≈ 314.2 cm²