Area and perimeter worksheets (rectangles and squares) - Free Printable
Educational worksheet: Area and perimeter worksheets (rectangles and squares). Download and print for classroom or home learning activities.
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Step-by-step solution for: Area and perimeter worksheets (rectangles and squares)
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Show Answer Key & Explanations
Step-by-step solution for: Area and perimeter worksheets (rectangles and squares)
To solve the problems related to Area and Perimeter, we need to use the appropriate formulas for each shape. Let's go through each problem step by step.
---
1. Rectangle/Square:
- Area = Length × Width
- Perimeter = 2 × (Length + Width)
2. Triangle:
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} \)
- Perimeter = Sum of all sides
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- Area = Length × Width = \( 20 \, \text{m} \times 4 \, \text{m} = 80 \, \text{m}^2 \)
- Perimeter = 2 × (Length + Width) = \( 2 \times (20 \, \text{m} + 4 \, \text{m}) = 2 \times 24 \, \text{m} = 48 \, \text{m} \)
Answer:
- Area: \( 80 \, \text{m}^2 \)
- Perimeter: \( 48 \, \text{m} \)
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- Area = Side × Side = \( 10 \, \text{m} \times 10 \, \text{m} = 100 \, \text{m}^2 \)
- Perimeter = 4 × Side = \( 4 \times 10 \, \text{m} = 40 \, \text{m} \)
Answer:
- Area: \( 100 \, \text{m}^2 \)
- Perimeter: \( 40 \, \text{m} \)
---
- Area = Length × Width = \( 9 \, \text{m} \times 3 \, \text{m} = 27 \, \text{m}^2 \)
- Perimeter = 2 × (Length + Width) = \( 2 \times (9 \, \text{m} + 3 \, \text{m}) = 2 \times 12 \, \text{m} = 24 \, \text{m} \)
Answer:
- Area: \( 27 \, \text{m}^2 \)
- Perimeter: \( 24 \, \text{m} \)
---
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 7 \, \text{m} \times 8 \, \text{m} = \frac{1}{2} \times 56 \, \text{m}^2 = 28 \, \text{m}^2 \)
- Perimeter: Not enough information to calculate the perimeter (we only have the base and height, not the lengths of the other two sides).
Answer:
- Area: \( 28 \, \text{m}^2 \)
- Perimeter: Cannot be determined with the given information.
---
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10 \, \text{m} \times 12 \, \text{m} = \frac{1}{2} \times 120 \, \text{m}^2 = 60 \, \text{m}^2 \)
- Perimeter: Not enough information to calculate the perimeter (we only have the base and height, not the lengths of the other two sides).
Answer:
- Area: \( 60 \, \text{m}^2 \)
- Perimeter: Cannot be determined with the given information.
---
- Area = Side × Side = \( 12 \, \text{m} \times 12 \, \text{m} = 144 \, \text{m}^2 \)
- Perimeter = 4 × Side = \( 4 \times 12 \, \text{m} = 48 \, \text{m} \)
Answer:
- Area: \( 144 \, \text{m}^2 \)
- Perimeter: \( 48 \, \text{m} \)
---
- Area = Side × Side = \( 4 \, \text{m} \times 4 \, \text{m} = 16 \, \text{m}^2 \)
- Perimeter = 4 × Side = \( 4 \times 4 \, \text{m} = 16 \, \text{m} \)
Answer:
- Area: \( 16 \, \text{m}^2 \)
- Perimeter: \( 16 \, \text{m} \)
---
- Area = Length × Width = \( 11 \, \text{m} \times 3 \, \text{m} = 33 \, \text{m}^2 \)
- Perimeter = 2 × (Length + Width) = \( 2 \times (11 \, \text{m} + 3 \, \text{m}) = 2 \times 14 \, \text{m} = 28 \, \text{m} \)
Answer:
- Area: \( 33 \, \text{m}^2 \)
- Perimeter: \( 28 \, \text{m} \)
---
1. Area: \( 80 \, \text{m}^2 \), Perimeter: \( 48 \, \text{m} \)
2. Area: \( 100 \, \text{m}^2 \), Perimeter: \( 40 \, \text{m} \)
3. Area: \( 27 \, \text{m}^2 \), Perimeter: \( 24 \, \text{m} \)
4. Area: \( 28 \, \text{m}^2 \), Perimeter: Cannot be determined
5. Area: \( 60 \, \text{m}^2 \), Perimeter: Cannot be determined
6. Area: \( 144 \, \text{m}^2 \), Perimeter: \( 48 \, \text{m} \)
7. Area: \( 16 \, \text{m}^2 \), Perimeter: \( 16 \, \text{m} \)
8. Area: \( 33 \, \text{m}^2 \), Perimeter: \( 28 \, \text{m} \)
\boxed{
\begin{array}{ll}
1. & \text{Area: } 80 \, \text{m}^2, \text{ Perimeter: } 48 \, \text{m} \\
2. & \text{Area: } 100 \, \text{m}^2, \text{ Perimeter: } 40 \, \text{m} \\
3. & \text{Area: } 27 \, \text{m}^2, \text{ Perimeter: } 24 \, \text{m} \\
4. & \text{Area: } 28 \, \text{m}^2, \text{ Perimeter: Cannot be determined} \\
5. & \text{Area: } 60 \, \text{m}^2, \text{ Perimeter: Cannot be determined} \\
6. & \text{Area: } 144 \, \text{m}^2, \text{ Perimeter: } 48 \, \text{m} \\
7. & \text{Area: } 16 \, \text{m}^2, \text{ Perimeter: } 16 \, \text{m} \\
8. & \text{Area: } 33 \, \text{m}^2, \text{ Perimeter: } 28 \, \text{m} \\
\end{array}
}
---
Formulas Needed:
1. Rectangle/Square:
- Area = Length × Width
- Perimeter = 2 × (Length + Width)
2. Triangle:
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} \)
- Perimeter = Sum of all sides
---
Problem 1: Rectangle (20m × 4m)
- Area = Length × Width = \( 20 \, \text{m} \times 4 \, \text{m} = 80 \, \text{m}^2 \)
- Perimeter = 2 × (Length + Width) = \( 2 \times (20 \, \text{m} + 4 \, \text{m}) = 2 \times 24 \, \text{m} = 48 \, \text{m} \)
Answer:
- Area: \( 80 \, \text{m}^2 \)
- Perimeter: \( 48 \, \text{m} \)
---
Problem 2: Square (10m × 10m)
- Area = Side × Side = \( 10 \, \text{m} \times 10 \, \text{m} = 100 \, \text{m}^2 \)
- Perimeter = 4 × Side = \( 4 \times 10 \, \text{m} = 40 \, \text{m} \)
Answer:
- Area: \( 100 \, \text{m}^2 \)
- Perimeter: \( 40 \, \text{m} \)
---
Problem 3: Rectangle (9m × 3m)
- Area = Length × Width = \( 9 \, \text{m} \times 3 \, \text{m} = 27 \, \text{m}^2 \)
- Perimeter = 2 × (Length + Width) = \( 2 \times (9 \, \text{m} + 3 \, \text{m}) = 2 \times 12 \, \text{m} = 24 \, \text{m} \)
Answer:
- Area: \( 27 \, \text{m}^2 \)
- Perimeter: \( 24 \, \text{m} \)
---
Problem 4: Triangle (Base = 7m, Height = 8m)
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 7 \, \text{m} \times 8 \, \text{m} = \frac{1}{2} \times 56 \, \text{m}^2 = 28 \, \text{m}^2 \)
- Perimeter: Not enough information to calculate the perimeter (we only have the base and height, not the lengths of the other two sides).
Answer:
- Area: \( 28 \, \text{m}^2 \)
- Perimeter: Cannot be determined with the given information.
---
Problem 5: Triangle (Base = 10m, Height = 12m)
- Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10 \, \text{m} \times 12 \, \text{m} = \frac{1}{2} \times 120 \, \text{m}^2 = 60 \, \text{m}^2 \)
- Perimeter: Not enough information to calculate the perimeter (we only have the base and height, not the lengths of the other two sides).
Answer:
- Area: \( 60 \, \text{m}^2 \)
- Perimeter: Cannot be determined with the given information.
---
Problem 6: Square (12m × 12m)
- Area = Side × Side = \( 12 \, \text{m} \times 12 \, \text{m} = 144 \, \text{m}^2 \)
- Perimeter = 4 × Side = \( 4 \times 12 \, \text{m} = 48 \, \text{m} \)
Answer:
- Area: \( 144 \, \text{m}^2 \)
- Perimeter: \( 48 \, \text{m} \)
---
Problem 7: Square (4m × 4m)
- Area = Side × Side = \( 4 \, \text{m} \times 4 \, \text{m} = 16 \, \text{m}^2 \)
- Perimeter = 4 × Side = \( 4 \times 4 \, \text{m} = 16 \, \text{m} \)
Answer:
- Area: \( 16 \, \text{m}^2 \)
- Perimeter: \( 16 \, \text{m} \)
---
Problem 8: Rectangle (11m × 3m)
- Area = Length × Width = \( 11 \, \text{m} \times 3 \, \text{m} = 33 \, \text{m}^2 \)
- Perimeter = 2 × (Length + Width) = \( 2 \times (11 \, \text{m} + 3 \, \text{m}) = 2 \times 14 \, \text{m} = 28 \, \text{m} \)
Answer:
- Area: \( 33 \, \text{m}^2 \)
- Perimeter: \( 28 \, \text{m} \)
---
Final Answers:
1. Area: \( 80 \, \text{m}^2 \), Perimeter: \( 48 \, \text{m} \)
2. Area: \( 100 \, \text{m}^2 \), Perimeter: \( 40 \, \text{m} \)
3. Area: \( 27 \, \text{m}^2 \), Perimeter: \( 24 \, \text{m} \)
4. Area: \( 28 \, \text{m}^2 \), Perimeter: Cannot be determined
5. Area: \( 60 \, \text{m}^2 \), Perimeter: Cannot be determined
6. Area: \( 144 \, \text{m}^2 \), Perimeter: \( 48 \, \text{m} \)
7. Area: \( 16 \, \text{m}^2 \), Perimeter: \( 16 \, \text{m} \)
8. Area: \( 33 \, \text{m}^2 \), Perimeter: \( 28 \, \text{m} \)
\boxed{
\begin{array}{ll}
1. & \text{Area: } 80 \, \text{m}^2, \text{ Perimeter: } 48 \, \text{m} \\
2. & \text{Area: } 100 \, \text{m}^2, \text{ Perimeter: } 40 \, \text{m} \\
3. & \text{Area: } 27 \, \text{m}^2, \text{ Perimeter: } 24 \, \text{m} \\
4. & \text{Area: } 28 \, \text{m}^2, \text{ Perimeter: Cannot be determined} \\
5. & \text{Area: } 60 \, \text{m}^2, \text{ Perimeter: Cannot be determined} \\
6. & \text{Area: } 144 \, \text{m}^2, \text{ Perimeter: } 48 \, \text{m} \\
7. & \text{Area: } 16 \, \text{m}^2, \text{ Perimeter: } 16 \, \text{m} \\
8. & \text{Area: } 33 \, \text{m}^2, \text{ Perimeter: } 28 \, \text{m} \\
\end{array}
}
Parent Tip: Review the logic above to help your child master the concept of area and perimeter worksheets grade 3.
