Math worksheet for calculating area and perimeter of rectangles with various dimensions.

To solve the problem, we need to calculate the area and perimeter for each rectangle or square provided in the image. Let's go through each one step by step.

Formulas:

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

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

Solutions:

#### Rectangle 1:
- Dimensions: 9 ft (length), 6 ft (width)
- Area:
\[
\text{Area} = 9 \, \text{ft} \times 6 \, \text{ft} = 54 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (9 \, \text{ft} + 6 \, \text{ft}) = 2 \times 15 \, \text{ft} = 30 \, \text{ft}
\]

#### Rectangle 2:
- Dimensions: 3 ft (length), 11 ft (width)
- Area:
\[
\text{Area} = 3 \, \text{ft} \times 11 \, \text{ft} = 33 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (3 \, \text{ft} + 11 \, \text{ft}) = 2 \times 14 \, \text{ft} = 28 \, \text{ft}
\]

#### Square 3:
- Dimensions: 13 cm (side length)
- Area:
\[
\text{Area} = 13 \, \text{cm} \times 13 \, \text{cm} = 169 \, \text{cm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 13 \, \text{cm} = 52 \, \text{cm}
\]

#### Rectangle 4:
- Dimensions: 22 in (length), 19 in (width)
- Area:
\[
\text{Area} = 22 \, \text{in} \times 19 \, \text{in} = 418 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (22 \, \text{in} + 19 \, \text{in}) = 2 \times 41 \, \text{in} = 82 \, \text{in}
\]

#### Rectangle 5:
- Dimensions: 25 m (length), 16 m (width)
- Area:
\[
\text{Area} = 25 \, \text{m} \times 16 \, \text{m} = 400 \, \text{m}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (25 \, \text{m} + 16 \, \text{m}) = 2 \times 41 \, \text{m} = 82 \, \text{m}
\]

#### Rectangle 6:
- Dimensions: 10 yd (length), 2 yd (width)
- Area:
\[
\text{Area} = 10 \, \text{yd} \times 2 \, \text{yd} = 20 \, \text{yd}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (10 \, \text{yd} + 2 \, \text{yd}) = 2 \times 12 \, \text{yd} = 24 \, \text{yd}
\]

#### Rectangle 7:
- Dimensions: 6 mm (length), 1 mm (width)
- Area:
\[
\text{Area} = 6 \, \text{mm} \times 1 \, \text{mm} = 6 \, \text{mm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (6 \, \text{mm} + 1 \, \text{mm}) = 2 \times 7 \, \text{mm} = 14 \, \text{mm}
\]

#### Square 8:
- Dimensions: 21 km (side length)
- Area:
\[
\text{Area} = 21 \, \text{km} \times 21 \, \text{km} = 441 \, \text{km}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 21 \, \text{km} = 84 \, \text{km}
\]

Final Answers:

1. Rectangle 1:
- Area: \( 54 \, \text{ft}^2 \)
- Perimeter: \( 30 \, \text{ft} \)

2. Rectangle 2:
- Area: \( 33 \, \text{ft}^2 \)
- Perimeter: \( 28 \, \text{ft} \)

3. Square 3:
- Area: \( 169 \, \text{cm}^2 \)
- Perimeter: \( 52 \, \text{cm} \)

4. Rectangle 4:
- Area: \( 418 \, \text{in}^2 \)
- Perimeter: \( 82 \, \text{in} \)

5. Rectangle 5:
- Area: \( 400 \, \text{m}^2 \)
- Perimeter: \( 82 \, \text{m} \)

6. Rectangle 6:
- Area: \( 20 \, \text{yd}^2 \)
- Perimeter: \( 24 \, \text{yd} \)

7. Rectangle 7:
- Area: \( 6 \, \text{mm}^2 \)
- Perimeter: \( 14 \, \text{mm} \)

8. Square 8:
- Area: \( 441 \, \text{km}^2 \)
- Perimeter: \( 84 \, \text{km} \)

Boxed Final Answer:

\[
\boxed{
\begin{array}{ll}
\text{Rectangle 1:} & \text{Area} = 54 \, \text{ft}^2, \, \text{Perimeter} = 30 \, \text{ft} \\
\text{Rectangle 2:} & \text{Area} = 33 \, \text{ft}^2, \, \text{Perimeter} = 28 \, \text{ft} \\
\text{Square 3:} & \text{Area} = 169 \, \text{cm}^2, \, \text{Perimeter} = 52 \, \text{cm} \\
\text{Rectangle 4:} & \text{Area} = 418 \, \text{in}^2, \, \text{Perimeter} = 82 \, \text{in} \\
\text{Rectangle 5:} & \text{Area} = 400 \, \text{m}^2, \, \text{Perimeter} = 82 \, \text{m} \\
\text{Rectangle 6:} & \text{Area} = 20 \, \text{yd}^2, \, \text{Perimeter} = 24 \, \text{yd} \\
\text{Rectangle 7:} & \text{Area} = 6 \, \text{mm}^2, \, \text{Perimeter} = 14 \, \text{mm} \\
\text{Square 8:} & \text{Area} = 441 \, \text{km}^2, \, \text{Perimeter} = 84 \, \text{km} \\
\end{array}
}
\]