Find the area and perimeter of the following rectangular shapes.

To solve the problem of finding the area and perimeter of the given rectangular shapes, we will use the following formulas:

Formulas:

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

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

Step-by-Step Solution:

#### Rectangle 1:
- Dimensions: Length = 31 yd, Width = 6 yd
- Area:
\[
\text{Area} = 31 \, \text{yd} \times 6 \, \text{yd} = 186 \, \text{yd}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (31 \, \text{yd} + 6 \, \text{yd}) = 2 \times 37 \, \text{yd} = 74 \, \text{yd}
\]

#### Rectangle 2:
- Dimensions: Length = 22 in, Width = 19 in
- 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 3:
- Dimensions: Length = 24 in, Width = 23 in
- Area:
\[
\text{Area} = 24 \, \text{in} \times 23 \, \text{in} = 552 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (24 \, \text{in} + 23 \, \text{in}) = 2 \times 47 \, \text{in} = 94 \, \text{in}
\]

#### Rectangle 4:
- Dimensions: Length = 24 ft, Width = 17 ft
- Area:
\[
\text{Area} = 24 \, \text{ft} \times 17 \, \text{ft} = 408 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (24 \, \text{ft} + 17 \, \text{ft}) = 2 \times 41 \, \text{ft} = 82 \, \text{ft}
\]

#### Rectangle 5:
- Dimensions: Length = 30 ft, Width = 16 ft
- Area:
\[
\text{Area} = 30 \, \text{ft} \times 16 \, \text{ft} = 480 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (30 \, \text{ft} + 16 \, \text{ft}) = 2 \times 46 \, \text{ft} = 92 \, \text{ft}
\]

#### Rectangle 6:
- Dimensions: Length = 25 yd, Width = 13 yd
- Area:
\[
\text{Area} = 25 \, \text{yd} \times 13 \, \text{yd} = 325 \, \text{yd}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (25 \, \text{yd} + 13 \, \text{yd}) = 2 \times 38 \, \text{yd} = 76 \, \text{yd}
\]

#### Rectangle 7:
- Dimensions: Length = 31 yd, Width = 12 yd
- Area:
\[
\text{Area} = 31 \, \text{yd} \times 12 \, \text{yd} = 372 \, \text{yd}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (31 \, \text{yd} + 12 \, \text{yd}) = 2 \times 43 \, \text{yd} = 86 \, \text{yd}
\]

#### Rectangle 8:
- Dimensions: Length = 18 ft, Width = 12 ft
- Area:
\[
\text{Area} = 18 \, \text{ft} \times 12 \, \text{ft} = 216 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (18 \, \text{ft} + 12 \, \text{ft}) = 2 \times 30 \, \text{ft} = 60 \, \text{ft}
\]

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Rectangle 1:} & \text{Area} = 186 \, \text{yd}^2, \, \text{Perimeter} = 74 \, \text{yd} \\
\text{Rectangle 2:} & \text{Area} = 418 \, \text{in}^2, \, \text{Perimeter} = 82 \, \text{in} \\
\text{Rectangle 3:} & \text{Area} = 552 \, \text{in}^2, \, \text{Perimeter} = 94 \, \text{in} \\
\text{Rectangle 4:} & \text{Area} = 408 \, \text{ft}^2, \, \text{Perimeter} = 82 \, \text{ft} \\
\text{Rectangle 5:} & \text{Area} = 480 \, \text{ft}^2, \, \text{Perimeter} = 92 \, \text{ft} \\
\text{Rectangle 6:} & \text{Area} = 325 \, \text{yd}^2, \, \text{Perimeter} = 76 \, \text{yd} \\
\text{Rectangle 7:} & \text{Area} = 372 \, \text{yd}^2, \, \text{Perimeter} = 86 \, \text{yd} \\
\text{Rectangle 8:} & \text{Area} = 216 \, \text{ft}^2, \, \text{Perimeter} = 60 \, \text{ft} \\
\end{array}
}
\]