Circle math worksheet with formulas and problems for calculating circumference, area, diameter, and radius.
A worksheet titled "Circles" that teaches how to calculate circumference and area using formulas, with examples and nine practice problems involving circles and a semicircle.
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Show Answer Key & Explanations
Step-by-step solution for: Circumference and Are A of A Circle | PDF | Area | Quantity
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Show Answer Key & Explanations
Step-by-step solution for: Circumference and Are A of A Circle | PDF | Area | Quantity
Let’s solve each problem step by step. We’ll use:
- Circumference = π × diameter (or 2 × π × radius)
- Area = π × radius²
- For semi-circle perimeter: half the circumference + diameter
- Use π ≈ 3.14 for calculations unless told otherwise.
---
Problem 1:
Circle with diameter = 6 cm → radius = 3 cm
Circumference = π × 6 = 3.14 × 6 = 18.84 cm
Area = π × 3² = 3.14 × 9 = 28.26 cm²
---
Problem 2:
Diameter = 9 cm → radius = 4.5 cm
Circumference = π × 9 = 3.14 × 9 = 28.26 cm
Area = π × (4.5)² = 3.14 × 20.25 = 63.585 cm² → round to 1 d.p.? Let’s check instructions — examples used 1 d.p., so we’ll do same.
→ 63.6 cm²
Wait — let me recalculate area carefully:
4.5 × 4.5 = 20.25
20.25 × 3.14 =
20 × 3.14 = 62.8
0.25 × 3.14 = 0.785
Total = 62.8 + 0.785 = 63.585 → yes, 63.6 cm² (to 1 d.p.)
---
Problem 3:
Radius = 4 cm → diameter = 8 cm
Circumference = π × 8 = 3.14 × 8 = 25.12 cm
Area = π × 4² = 3.14 × 16 = 50.24 cm²
---
Problem 4:
Radius = 5 cm → diameter = 10 cm
Circumference = π × 10 = 3.14 × 10 = 31.4 cm
Area = π × 5² = 3.14 × 25 = 78.5 cm²
---
Problem 5:
Circumference = 25.1 cm
We know: C = π × d → d = C ÷ π
d = 25.1 ÷ 3.14 = ?
Let’s calculate:
3.14 × 8 = 25.12 → very close to 25.1
So d ≈ 8.0 cm (since 25.1 ÷ 3.14 = 7.993... → rounds to 8.0)
Check: 3.14 × 8 = 25.12 → which is 0.02 over, but given input was 25.1 (probably rounded), so answer is 8.0 cm
---
Problem 6:
Circumference = 37.7 cm
C = 2 × π × r → r = C ÷ (2π)
r = 37.7 ÷ (2 × 3.14) = 37.7 ÷ 6.28
Calculate:
6.28 × 6 = 37.68 → very close to 37.7
So r ≈ 6.0 cm
Check: 2 × 3.14 × 6 = 37.68 → rounds to 37.7 → correct.
---
Problem 7:
Area = 153.9 cm²
A = π × r² → r² = A ÷ π = 153.9 ÷ 3.14
Calculate:
3.14 × 49 = 153.86 → very close to 153.9
So r² ≈ 49 → r = √49 = 7.0 cm
Check: 3.14 × 49 = 153.86 → rounds to 153.9 → perfect.
---
Problem 8:
Area = 113.1 cm²
A = π × r² → r² = 113.1 ÷ 3.14
Calculate:
3.14 × 36 = 113.04 → very close to 113.1
So r² ≈ 36 → r = 6 cm → diameter = 2 × 6 = 12.0 cm
Check: 3.14 × 36 = 113.04 → rounds to 113.1 → good.
---
Problem 9:
Semi-circle with diameter = 10 cm → radius = 5 cm
Perimeter of semi-circle = half circumference + diameter
Half circumference = (π × d) ÷ 2 = (3.14 × 10) ÷ 2 = 31.4 ÷ 2 = 15.7 cm
Add diameter: 15.7 + 10 = 25.7 cm
Area of semi-circle = half of full circle area
Full area = π × r² = 3.14 × 25 = 78.5 cm²
Half = 78.5 ÷ 2 = 39.25 cm² → to 1 d.p.? Examples used 1 d.p., so 39.3 cm²? Wait — 39.25 is exactly halfway. Usually round up → 39.3 cm²
But let’s check example: they did 28.3 from 28.26? No — in example: π×9=28.27… they wrote 28.3 → so yes, round to 1 d.p.
So area = 39.25 → 39.3 cm²
Wait — actually, 78.5 ÷ 2 = 39.25 → if we’re being precise, and using π=3.14, then it’s exact. But since the question says “calculate”, and examples rounded to 1 d.p., we should too.
Alternatively, maybe keep as 39.25? But example showed rounding. Let’s follow example style.
In example: π×9 = 28.274... → they wrote 28.3 → so round to 1 decimal place.
So 39.25 → round to 1 d.p. → look at second decimal: 5 → round up → 39.3 cm²
Perimeter: 15.7 + 10 = 25.7 → already 1 d.p.
---
Now compile all answers clearly.
Final Answer:
1. Circumference: 18.8 cm, Area: 28.3 cm²
*(Wait — earlier I had 18.84 and 28.26 — need to round to 1 d.p. like examples)*
Hold on! In the example, they calculated π×4=12.56 → wrote 12.6 cm (to 1 d.p.)
And π×9=28.27 → wrote 28.3 cm² (to 1 d.p.)
So we must round ALL answers to 1 decimal place.
Let me redo all with proper rounding:
---
Problem 1:
d=6 → r=3
C = π×6 = 18.849... → 18.8 cm (wait, 18.849 rounds to 18.8? No — 18.849 to 1 d.p.: look at hundredths digit is 4 <5 → so 18.8? But 18.849 is closer to 18.8 than 18.9? Actually no — standard rounding: 18.849 → tenths place is 8, hundredths is 4 → less than 5 → so 18.8? But wait, 18.849 is actually 18.8 when rounded to 1 d.p.? Let me think:
Actually, 18.849:
- The first decimal is 8 (tenths)
- Second decimal is 4 (hundredths) → since 4<5, we leave tenths as is → 18.8 cm
But in example: π×4=12.566... → they wrote 12.6 → because 12.566 → hundredths is 6≥5 → round up tenths from 5 to 6 → 12.6
Similarly, here 18.849 → hundredths is 4 → so 18.8? But that seems low. Let me calculate numerically:
π ≈ 3.1416
3.1416 × 6 = 18.8496 → to 1 decimal place: look at second decimal (4) → less than 5 → so 18.8? But 18.8496 is closer to 18.8 than 18.9? Actually, 18.85 would be midpoint. Since 18.8496 < 18.85, it rounds down to 18.8.
But let's check what the example did:
Example: diameter 4cm → C=π×4=12.566... → they wrote 12.6 → which is correct rounding (since 12.566 → hundredths=6≥5 → round up 5 to 6 → 12.6)
For our case: 18.8496 → hundredths=4 → so should be 18.8? But I feel like maybe they expect 18.8 or 18.9? Let me see common practice.
Actually, many textbooks use π=3.14 for such problems.
If we use π=3.14:
C = 3.14 × 6 = 18.84 → to 1 d.p.: 18.8 (since 4<5)
Area = 3.14 × 9 = 28.26 → to 1 d.p.: 28.3 (since 6≥5)
Yes, that matches example style.
So let's standardize: use π=3.14 for all calculations, then round final answer to 1 decimal place.
Redo all with π=3.14:
---
Problem 1:
d=6, r=3
C = 3.14 × 6 = 18.84 → 18.8 cm
A = 3.14 × 9 = 28.26 → 28.3 cm²
Problem 2:
d=9, r=4.5
C = 3.14 × 9 = 28.26 → 28.3 cm
A = 3.14 × (4.5)^2 = 3.14 × 20.25 = let's compute:
20.25 × 3 = 60.75
20.25 × 0.14 = 2.835
Total = 60.75 + 2.835 = 63.585 → 63.6 cm²
Problem 3:
r=4, d=8
C = 3.14 × 8 = 25.12 → 25.1 cm
A = 3.14 × 16 = 50.24 → 50.2 cm²
Problem 4:
r=5, d=10
C = 3.14 × 10 = 31.4 → 31.4 cm
A = 3.14 × 25 = 78.5 → 78.5 cm² (already 1 d.p.)
Problem 5:
C=25.1 = πd → d=25.1 / 3.14 = ?
25.1 ÷ 3.14 = 7.9936... → 8.0 cm (to 1 d.p.)
Problem 6:
C=37.7 = 2πr → r=37.7/(2×3.14)=37.7/6.28=5.999... → 6.0 cm
Problem 7:
A=153.9 = πr² → r²=153.9/3.14=49.0127... → r=√49.0127≈7.0009 → 7.0 cm
Problem 8:
A=113.1 = πr² → r²=113.1/3.14=36.019... → r=6.0016 → d=12.003 → 12.0 cm
Problem 9:
Semi-circle, d=10, r=5
Perimeter = (πd)/2 + d = (3.14×10)/2 + 10 = 31.4/2 + 10 = 15.7 + 10 = 25.7 cm
Area = (πr²)/2 = (3.14×25)/2 = 78.5/2 = 39.25 → 39.3 cm² (round to 1 d.p.)
All set.
Final Answer:
1. Circumference: 18.8 cm, Area: 28.3 cm²
2. Circumference: 28.3 cm, Area: 63.6 cm²
3. Circumference: 25.1 cm, Area: 50.2 cm²
4. Circumference: 31.4 cm, Area: 78.5 cm²
5. Diameter: 8.0 cm
6. Radius: 6.0 cm
7. Radius: 7.0 cm
8. Diameter: 12.0 cm
9. Perimeter: 25.7 cm, Area: 39.3 cm²
- Circumference = π × diameter (or 2 × π × radius)
- Area = π × radius²
- For semi-circle perimeter: half the circumference + diameter
- Use π ≈ 3.14 for calculations unless told otherwise.
---
Problem 1:
Circle with diameter = 6 cm → radius = 3 cm
Circumference = π × 6 = 3.14 × 6 = 18.84 cm
Area = π × 3² = 3.14 × 9 = 28.26 cm²
---
Problem 2:
Diameter = 9 cm → radius = 4.5 cm
Circumference = π × 9 = 3.14 × 9 = 28.26 cm
Area = π × (4.5)² = 3.14 × 20.25 = 63.585 cm² → round to 1 d.p.? Let’s check instructions — examples used 1 d.p., so we’ll do same.
→ 63.6 cm²
Wait — let me recalculate area carefully:
4.5 × 4.5 = 20.25
20.25 × 3.14 =
20 × 3.14 = 62.8
0.25 × 3.14 = 0.785
Total = 62.8 + 0.785 = 63.585 → yes, 63.6 cm² (to 1 d.p.)
---
Problem 3:
Radius = 4 cm → diameter = 8 cm
Circumference = π × 8 = 3.14 × 8 = 25.12 cm
Area = π × 4² = 3.14 × 16 = 50.24 cm²
---
Problem 4:
Radius = 5 cm → diameter = 10 cm
Circumference = π × 10 = 3.14 × 10 = 31.4 cm
Area = π × 5² = 3.14 × 25 = 78.5 cm²
---
Problem 5:
Circumference = 25.1 cm
We know: C = π × d → d = C ÷ π
d = 25.1 ÷ 3.14 = ?
Let’s calculate:
3.14 × 8 = 25.12 → very close to 25.1
So d ≈ 8.0 cm (since 25.1 ÷ 3.14 = 7.993... → rounds to 8.0)
Check: 3.14 × 8 = 25.12 → which is 0.02 over, but given input was 25.1 (probably rounded), so answer is 8.0 cm
---
Problem 6:
Circumference = 37.7 cm
C = 2 × π × r → r = C ÷ (2π)
r = 37.7 ÷ (2 × 3.14) = 37.7 ÷ 6.28
Calculate:
6.28 × 6 = 37.68 → very close to 37.7
So r ≈ 6.0 cm
Check: 2 × 3.14 × 6 = 37.68 → rounds to 37.7 → correct.
---
Problem 7:
Area = 153.9 cm²
A = π × r² → r² = A ÷ π = 153.9 ÷ 3.14
Calculate:
3.14 × 49 = 153.86 → very close to 153.9
So r² ≈ 49 → r = √49 = 7.0 cm
Check: 3.14 × 49 = 153.86 → rounds to 153.9 → perfect.
---
Problem 8:
Area = 113.1 cm²
A = π × r² → r² = 113.1 ÷ 3.14
Calculate:
3.14 × 36 = 113.04 → very close to 113.1
So r² ≈ 36 → r = 6 cm → diameter = 2 × 6 = 12.0 cm
Check: 3.14 × 36 = 113.04 → rounds to 113.1 → good.
---
Problem 9:
Semi-circle with diameter = 10 cm → radius = 5 cm
Perimeter of semi-circle = half circumference + diameter
Half circumference = (π × d) ÷ 2 = (3.14 × 10) ÷ 2 = 31.4 ÷ 2 = 15.7 cm
Add diameter: 15.7 + 10 = 25.7 cm
Area of semi-circle = half of full circle area
Full area = π × r² = 3.14 × 25 = 78.5 cm²
Half = 78.5 ÷ 2 = 39.25 cm² → to 1 d.p.? Examples used 1 d.p., so 39.3 cm²? Wait — 39.25 is exactly halfway. Usually round up → 39.3 cm²
But let’s check example: they did 28.3 from 28.26? No — in example: π×9=28.27… they wrote 28.3 → so yes, round to 1 d.p.
So area = 39.25 → 39.3 cm²
Wait — actually, 78.5 ÷ 2 = 39.25 → if we’re being precise, and using π=3.14, then it’s exact. But since the question says “calculate”, and examples rounded to 1 d.p., we should too.
Alternatively, maybe keep as 39.25? But example showed rounding. Let’s follow example style.
In example: π×9 = 28.274... → they wrote 28.3 → so round to 1 decimal place.
So 39.25 → round to 1 d.p. → look at second decimal: 5 → round up → 39.3 cm²
Perimeter: 15.7 + 10 = 25.7 → already 1 d.p.
---
Now compile all answers clearly.
Final Answer:
1. Circumference: 18.8 cm, Area: 28.3 cm²
*(Wait — earlier I had 18.84 and 28.26 — need to round to 1 d.p. like examples)*
Hold on! In the example, they calculated π×4=12.56 → wrote 12.6 cm (to 1 d.p.)
And π×9=28.27 → wrote 28.3 cm² (to 1 d.p.)
So we must round ALL answers to 1 decimal place.
Let me redo all with proper rounding:
---
Problem 1:
d=6 → r=3
C = π×6 = 18.849... → 18.8 cm (wait, 18.849 rounds to 18.8? No — 18.849 to 1 d.p.: look at hundredths digit is 4 <5 → so 18.8? But 18.849 is closer to 18.8 than 18.9? Actually no — standard rounding: 18.849 → tenths place is 8, hundredths is 4 → less than 5 → so 18.8? But wait, 18.849 is actually 18.8 when rounded to 1 d.p.? Let me think:
Actually, 18.849:
- The first decimal is 8 (tenths)
- Second decimal is 4 (hundredths) → since 4<5, we leave tenths as is → 18.8 cm
But in example: π×4=12.566... → they wrote 12.6 → because 12.566 → hundredths is 6≥5 → round up tenths from 5 to 6 → 12.6
Similarly, here 18.849 → hundredths is 4 → so 18.8? But that seems low. Let me calculate numerically:
π ≈ 3.1416
3.1416 × 6 = 18.8496 → to 1 decimal place: look at second decimal (4) → less than 5 → so 18.8? But 18.8496 is closer to 18.8 than 18.9? Actually, 18.85 would be midpoint. Since 18.8496 < 18.85, it rounds down to 18.8.
But let's check what the example did:
Example: diameter 4cm → C=π×4=12.566... → they wrote 12.6 → which is correct rounding (since 12.566 → hundredths=6≥5 → round up 5 to 6 → 12.6)
For our case: 18.8496 → hundredths=4 → so should be 18.8? But I feel like maybe they expect 18.8 or 18.9? Let me see common practice.
Actually, many textbooks use π=3.14 for such problems.
If we use π=3.14:
C = 3.14 × 6 = 18.84 → to 1 d.p.: 18.8 (since 4<5)
Area = 3.14 × 9 = 28.26 → to 1 d.p.: 28.3 (since 6≥5)
Yes, that matches example style.
So let's standardize: use π=3.14 for all calculations, then round final answer to 1 decimal place.
Redo all with π=3.14:
---
Problem 1:
d=6, r=3
C = 3.14 × 6 = 18.84 → 18.8 cm
A = 3.14 × 9 = 28.26 → 28.3 cm²
Problem 2:
d=9, r=4.5
C = 3.14 × 9 = 28.26 → 28.3 cm
A = 3.14 × (4.5)^2 = 3.14 × 20.25 = let's compute:
20.25 × 3 = 60.75
20.25 × 0.14 = 2.835
Total = 60.75 + 2.835 = 63.585 → 63.6 cm²
Problem 3:
r=4, d=8
C = 3.14 × 8 = 25.12 → 25.1 cm
A = 3.14 × 16 = 50.24 → 50.2 cm²
Problem 4:
r=5, d=10
C = 3.14 × 10 = 31.4 → 31.4 cm
A = 3.14 × 25 = 78.5 → 78.5 cm² (already 1 d.p.)
Problem 5:
C=25.1 = πd → d=25.1 / 3.14 = ?
25.1 ÷ 3.14 = 7.9936... → 8.0 cm (to 1 d.p.)
Problem 6:
C=37.7 = 2πr → r=37.7/(2×3.14)=37.7/6.28=5.999... → 6.0 cm
Problem 7:
A=153.9 = πr² → r²=153.9/3.14=49.0127... → r=√49.0127≈7.0009 → 7.0 cm
Problem 8:
A=113.1 = πr² → r²=113.1/3.14=36.019... → r=6.0016 → d=12.003 → 12.0 cm
Problem 9:
Semi-circle, d=10, r=5
Perimeter = (πd)/2 + d = (3.14×10)/2 + 10 = 31.4/2 + 10 = 15.7 + 10 = 25.7 cm
Area = (πr²)/2 = (3.14×25)/2 = 78.5/2 = 39.25 → 39.3 cm² (round to 1 d.p.)
All set.
Final Answer:
1. Circumference: 18.8 cm, Area: 28.3 cm²
2. Circumference: 28.3 cm, Area: 63.6 cm²
3. Circumference: 25.1 cm, Area: 50.2 cm²
4. Circumference: 31.4 cm, Area: 78.5 cm²
5. Diameter: 8.0 cm
6. Radius: 6.0 cm
7. Radius: 7.0 cm
8. Diameter: 12.0 cm
9. Perimeter: 25.7 cm, Area: 39.3 cm²
Parent Tip: Review the logic above to help your child master the concept of circumference and area worksheet 1.