Perimeter of Squares Worksheets - Free Printable

To solve the problem of finding the area and perimeter of each shape in the provided worksheet, we need to use the following formulas:

Formulas:

1. Area of a Square:
\[
\text{Area} = \text{side} \times \text{side} = s^2
\]

2. Perimeter of a Square:
\[
\text{Perimeter} = 4 \times \text{side}
\]

3. Area of a Rectangle:
\[
\text{Area} = \text{length} \times \text{width}
\]

4. Perimeter of a Rectangle:
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width})
\]

Now, let's solve each problem step by step.

---

Problem 1: Square with side = 7 cm

- Area:
\[
\text{Area} = 7 \times 7 = 49 \, \text{cm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 7 = 28 \, \text{cm}
\]

Problem 2: Rectangle with length = 8 cm, width = 2 cm

- Area:
\[
\text{Area} = 8 \times 2 = 16 \, \text{cm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (8 + 2) = 2 \times 10 = 20 \, \text{cm}
\]

Problem 3: Square with side = 9 in

- Area:
\[
\text{Area} = 9 \times 9 = 81 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 9 = 36 \, \text{in}
\]

Problem 4: Square with side = 5 ft

- Area:
\[
\text{Area} = 5 \times 5 = 25 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 5 = 20 \, \text{ft}
\]

Problem 5: Rectangle with length = 6 in, width = 4 in

- Area:
\[
\text{Area} = 6 \times 4 = 24 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (6 + 4) = 2 \times 10 = 20 \, \text{in}
\]

Problem 6: Square with side = 3 cm

- Area:
\[
\text{Area} = 3 \times 3 = 9 \, \text{cm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 3 = 12 \, \text{cm}
\]

Problem 7: Square with side = 8 in

- Area:
\[
\text{Area} = 8 \times 8 = 64 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 8 = 32 \, \text{in}
\]

Problem 8: Rectangle with length = 11 ft, width = 6 ft

- Area:
\[
\text{Area} = 11 \times 6 = 66 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (11 + 6) = 2 \times 17 = 34 \, \text{ft}
\]

Problem 9: Square with side = 7 cm

- Area:
\[
\text{Area} = 7 \times 7 = 49 \, \text{cm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 7 = 28 \, \text{cm}
\]

Problem 10: Rectangle with length = 12 m, width = 2 m

- Area:
\[
\text{Area} = 12 \times 2 = 24 \, \text{m}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (12 + 2) = 2 \times 14 = 28 \, \text{m}
\]

Problem 11: Rectangle with length = 11 cm, width = 4 cm

- Area:
\[
\text{Area} = 11 \times 4 = 44 \, \text{cm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (11 + 4) = 2 \times 15 = 30 \, \text{cm}
\]

Problem 12: Rectangle with length = 7 in, width = 3 in

- Area:
\[
\text{Area} = 7 \times 3 = 21 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (7 + 3) = 2 \times 10 = 20 \, \text{in}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & \text{Area} = 49 \, \text{cm}^2, \, \text{Perimeter} = 28 \, \text{cm} \\
2. & \text{Area} = 16 \, \text{cm}^2, \, \text{Perimeter} = 20 \, \text{cm} \\
3. & \text{Area} = 81 \, \text{in}^2, \, \text{Perimeter} = 36 \, \text{in} \\
4. & \text{Area} = 25 \, \text{ft}^2, \, \text{Perimeter} = 20 \, \text{ft} \\
5. & \text{Area} = 24 \, \text{in}^2, \, \text{Perimeter} = 20 \, \text{in} \\
6. & \text{Area} = 9 \, \text{cm}^2, \, \text{Perimeter} = 12 \, \text{cm} \\
7. & \text{Area} = 64 \, \text{in}^2, \, \text{Perimeter} = 32 \, \text{in} \\
8. & \text{Area} = 66 \, \text{ft}^2, \, \text{Perimeter} = 34 \, \text{ft} \\
9. & \text{Area} = 49 \, \text{cm}^2, \, \text{Perimeter} = 28 \, \text{cm} \\
10. & \text{Area} = 24 \, \text{m}^2, \, \text{Perimeter} = 28 \, \text{m} \\
11. & \text{Area} = 44 \, \text{cm}^2, \, \text{Perimeter} = 30 \, \text{cm} \\
12. & \text{Area} = 21 \, \text{in}^2, \, \text{Perimeter} = 20 \, \text{in} \\
\end{array}
}
\]