Geometry worksheet for finding the area and perimeter of rectangular shapes with various dimensions.

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 = 42 in, Width = 12 in
- Area:
\[
\text{Area} = 42 \, \text{in} \times 12 \, \text{in} = 504 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (42 \, \text{in} + 12 \, \text{in}) = 2 \times 54 \, \text{in} = 108 \, \text{in}
\]

#### Rectangle 2:
- Dimensions: Length = 29 yd, Width = 16 yd
- Area:
\[
\text{Area} = 29 \, \text{yd} \times 16 \, \text{yd} = 464 \, \text{yd}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (29 \, \text{yd} + 16 \, \text{yd}) = 2 \times 45 \, \text{yd} = 90 \, \text{yd}
\]

#### Rectangle 3:
- Dimensions: Length = 47 ft, Width = 26 ft
- Area:
\[
\text{Area} = 47 \, \text{ft} \times 26 \, \text{ft} = 1222 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (47 \, \text{ft} + 26 \, \text{ft}) = 2 \times 73 \, \text{ft} = 146 \, \text{ft}
\]

#### Rectangle 4:
- Dimensions: Length = 40 ft, Width = 22 ft
- Area:
\[
\text{Area} = 40 \, \text{ft} \times 22 \, \text{ft} = 880 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (40 \, \text{ft} + 22 \, \text{ft}) = 2 \times 62 \, \text{ft} = 124 \, \text{ft}
\]

#### Rectangle 5:
- Dimensions: Length = 31 in, Width = 19 in
- Area:
\[
\text{Area} = 31 \, \text{in} \times 19 \, \text{in} = 589 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (31 \, \text{in} + 19 \, \text{in}) = 2 \times 50 \, \text{in} = 100 \, \text{in}
\]

#### Rectangle 6:
- Dimensions: Length = 39 ft, Width = 19 ft
- Area:
\[
\text{Area} = 39 \, \text{ft} \times 19 \, \text{ft} = 741 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (39 \, \text{ft} + 19 \, \text{ft}) = 2 \times 58 \, \text{ft} = 116 \, \text{ft}
\]

#### Rectangle 7:
- Dimensions: Length = 22 yd, Width = 14 yd
- Area:
\[
\text{Area} = 22 \, \text{yd} \times 14 \, \text{yd} = 308 \, \text{yd}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (22 \, \text{yd} + 14 \, \text{yd}) = 2 \times 36 \, \text{yd} = 72 \, \text{yd}
\]

#### Rectangle 8:
- Dimensions: Length = 26 yd, Width = 21 yd
- Area:
\[
\text{Area} = 26 \, \text{yd} \times 21 \, \text{yd} = 546 \, \text{yd}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (26 \, \text{yd} + 21 \, \text{yd}) = 2 \times 47 \, \text{yd} = 94 \, \text{yd}
\]

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Rectangle 1:} & \text{Area} = 504 \, \text{in}^2, \, \text{Perimeter} = 108 \, \text{in} \\
\text{Rectangle 2:} & \text{Area} = 464 \, \text{yd}^2, \, \text{Perimeter} = 90 \, \text{yd} \\
\text{Rectangle 3:} & \text{Area} = 1222 \, \text{ft}^2, \, \text{Perimeter} = 146 \, \text{ft} \\
\text{Rectangle 4:} & \text{Area} = 880 \, \text{ft}^2, \, \text{Perimeter} = 124 \, \text{ft} \\
\text{Rectangle 5:} & \text{Area} = 589 \, \text{in}^2, \, \text{Perimeter} = 100 \, \text{in} \\
\text{Rectangle 6:} & \text{Area} = 741 \, \text{ft}^2, \, \text{Perimeter} = 116 \, \text{ft} \\
\text{Rectangle 7:} & \text{Area} = 308 \, \text{yd}^2, \, \text{Perimeter} = 72 \, \text{yd} \\
\text{Rectangle 8:} & \text{Area} = 546 \, \text{yd}^2, \, \text{Perimeter} = 94 \, \text{yd} \\
\end{array}
}
\]