Circle Area Problems Worksheet | Printable Maths Worksheets - Free Printable

Let’s solve each problem one by one. We’ll use the formula for the area of a circle:
Area = π × r², where r is the radius.

For parts of circles (like semicircles, quarters, etc.), we multiply the full circle area by the fraction that’s shown.

We’ll round all answers to 2 decimal places as instructed.

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Problem 1: Semicircle with diameter 8 cm

- Radius = 8 ÷ 2 = 4 cm
- Full circle area = π × 4² = π × 16 ≈ 50.265
- Semicircle = half of that → 50.265 ÷ 2 ≈ 25.13 cm²

✔ Check: ½ × π × 4² = 8π ≈ 25.13 — correct.

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Problem 2: Quarter circle with radius 4.7 mm

- Radius = 4.7 mm
- Full circle area = π × (4.7)² = π × 22.09 ≈ 69.397
- Quarter circle = ¼ of that → 69.397 ÷ 4 ≈ 17.35 mm²

✔ Check: ¼ × π × 4.7² = (π × 22.09)/4 ≈ 17.35 — correct.

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Problem 3: Three-quarters circle with radius 11 m

- Radius = 11 m
- Full circle area = π × 11² = π × 121 ≈ 380.133
- Three-quarters = ¾ × 380.133 ≈ 285.10 m²

✔ Check: 0.75 × π × 121 = 90.75π ≈ 285.10 — correct.

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Problem 4: Two quarter-circles making a semicircle? Wait — look at diagram.

Actually, it shows two separate red quarter-circles, but together they form a semicircle. The total height is 16 mm — which is the diameter of the full circle these come from.

So:

- Diameter = 16 mm → Radius = 8 mm
- Each piece is a quarter-circle → two pieces = half a circle
- Area = ½ × π × 8² = ½ × π × 64 = 32π ≈ 100.53 mm²

✔ Check: 2 × (¼ × π × 8²) = ½ × π × 64 = 100.53 — correct.

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Problem 5: Two sectors — looks like a “bowtie” shape. Each sector has radius 12 cm and angle? From diagram, each seems to be a quarter-circle? Actually, looking closely — the dashed line suggests symmetry. It appears to be two quarter-circles, so again, total = half a circle.

Wait — actually, in many such diagrams, if it's symmetric and labeled with radius 12 cm, and looks like two 90° sectors, then yes — total area = ½ × π × r².

But let’s confirm: if each sector is 90°, then two make 180° → semicircle.

→ Radius = 12 cm

→ Area = ½ × π × 12² = ½ × π × 144 = 72π ≈ 226.19 cm²

✔ Check: 2 × (¼ × π × 144) = ½ × 144π = 72π ≈ 226.19 — correct.

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Problem 6: Yellow shape — looks like three-quarters of a circle? But wait — the vertical arrow says 4 m — that’s the radius, because it goes from center to edge.

Looking at the shape: it’s missing one quarter — so it’s three-quarters of a circle.

→ Radius = 4 m

→ Full circle = π × 4² = 16π ≈ 50.265

→ Three-quarters = 0.75 × 50.265 ≈ 37.70 m²

✔ Check: ¾ × π × 16 = 12π ≈ 37.70 — correct.

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Problem 7: Square with side 12 cm, white circle inside — shaded area is square minus circle.

- Square area = 12 × 12 = 144 cm²
- Circle diameter = 12 cm → radius = 6 cm
- Circle area = π × 6² = 36π ≈ 113.097
- Shaded area = 144 - 113.097 ≈ 30.90 cm²

✔ Check: 144 - 36π ≈ 144 - 113.097 = 30.903 → rounded to 30.90 — correct.

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Problem 8: Shape looks like a quarter-circle cut out of a square? Wait — no. Actually, it’s a quarter-circle sector — gray area is the curved part. The height is 110 km — that’s the radius.

It’s a quarter-circle! Because it’s bounded by two straight lines (radii) and an arc — forming 90°.

→ Radius = 110 km

→ Area = ¼ × π × (110)² = ¼ × π × 12100 = 3025π ≈ 9503.32 km²

✔ Check: 0.25 × π × 12100 = 3025π ≈ 9503.317 → rounded to 9503.32 — correct.

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Problem 9: Square with diagonal 19 mm — shaded area is square minus white quarter-circle? Wait — look carefully.

The white region is a quarter-circle whose radius equals the side of the square? No — the diagonal is given as 19 mm.

Actually, the dashed line is the diagonal of the square, and also the radius of the quarter-circle? That doesn’t make sense geometrically.

Wait — re-examining: the white region is a quarter-circle drawn from one corner, sweeping across the square. The dashed line is the diagonal of the square — labeled 19 mm.

In a square, diagonal = side × √2 → so side = diagonal / √2 = 19 / √2

But the quarter-circle — its radius must be equal to the side of the square, because it starts at one corner and curves to adjacent corners.

So:

→ Diagonal = 19 mm → Side = 19 / √2 ≈ 13.435 mm

→ Area of square = side² = (19/√2)² = 361 / 2 = 180.5 mm²

→ Area of quarter-circle = ¼ × π × (side)² = ¼ × π × 180.5 ≈ 0.25 × 3.1416 × 180.5 ≈ 141.76 mm²

Wait — that can’t be right because the quarter-circle should fit inside the square — but if radius = side, then quarter-circle area is ¼ π s², and square is s² — so shaded area = s² - ¼ π s² = s²(1 - π/4)

But here, we’re told the diagonal is 19 mm.

So:

s = 19 / √2

s² = (19)^2 / 2 = 361 / 2 = 180.5 mm² ← area of square

Quarter-circle area = ¼ × π × s² = ¼ × π × 180.5 ≈ 0.25 × 3.1415926535 × 180.5 ≈ let’s compute:

First, π × 180.5 ≈ 3.1415926535 × 180.5 ≈ 567.057

Then divide by 4 → 567.057 ÷ 4 ≈ 141.764

Shaded area = square - quarter-circle = 180.5 - 141.764 ≈ 38.74 mm²

✔ Alternative way: shaded = s² (1 - π/4) = 180.5 × (1 - 0.785398) ≈ 180.5 × 0.214602 ≈ 38.74 — same.

Correct.

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Now compiling all final answers:

1) 25.13
2) 17.35
3) 285.10
4) 100.53
5) 226.19
6) 37.70
7) 30.90
8) 9503.32
9) 38.74

All units included as per problem.

Final Answer:
1) 25.13 cm²
2) 17.35 mm²
3) 285.10 m²
4) 100.53 mm²
5) 226.19 cm²
6) 37.70 m²
7) 30.90 cm²
8) 9503.32 km²
9) 38.74 mm²