Finding Area and Perimeter of Irregular shapes - Math Worksheets - Free Printable

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

Formula Recap:

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

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

Problem Solutions:

#### 1)
- Dimensions: Length = 9 cm, Width = 5 cm
- Area:
\[
\text{Area} = 9 \, \text{cm} \times 5 \, \text{cm} = 45 \, \text{cm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (9 \, \text{cm} + 5 \, \text{cm}) = 2 \times 14 \, \text{cm} = 28 \, \text{cm}
\]

#### 2)
- Dimensions: Side = 6 in (since it's a square)
- Area:
\[
\text{Area} = 6 \, \text{in} \times 6 \, \text{in} = 36 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 4 \times 6 \, \text{in} = 24 \, \text{in}
\]

#### 3)
- Dimensions: Length = 7 ft, Width = 4 ft
- Area:
\[
\text{Area} = 7 \, \text{ft} \times 4 \, \text{ft} = 28 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (7 \, \text{ft} + 4 \, \text{ft}) = 2 \times 11 \, \text{ft} = 22 \, \text{ft}
\]

#### 4)
- Dimensions: Length = 15 ft, Width = 6 ft
- Area:
\[
\text{Area} = 15 \, \text{ft} \times 6 \, \text{ft} = 90 \, \text{ft}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (15 \, \text{ft} + 6 \, \text{ft}) = 2 \times 21 \, \text{ft} = 42 \, \text{ft}
\]

#### 5)
- Dimensions: Length = 20 in, Width = 10 in
- Area:
\[
\text{Area} = 20 \, \text{in} \times 10 \, \text{in} = 200 \, \text{in}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (20 \, \text{in} + 10 \, \text{in}) = 2 \times 30 \, \text{in} = 60 \, \text{in}
\]

#### 6)
- Dimensions: Length = 60 mm, Width = 8 mm
- Area:
\[
\text{Area} = 60 \, \text{mm} \times 8 \, \text{mm} = 480 \, \text{mm}^2
\]
- Perimeter:
\[
\text{Perimeter} = 2 \times (60 \, \text{mm} + 8 \, \text{mm}) = 2 \times 68 \, \text{mm} = 136 \, \text{mm}
\]

Final Answers:

1. Area = \( 45 \, \text{cm}^2 \), Perimeter = \( 28 \, \text{cm} \)
2. Area = \( 36 \, \text{in}^2 \), Perimeter = \( 24 \, \text{in} \)
3. Area = \( 28 \, \text{ft}^2 \), Perimeter = \( 22 \, \text{ft} \)
4. Area = \( 90 \, \text{ft}^2 \), Perimeter = \( 42 \, \text{ft} \)
5. Area = \( 200 \, \text{in}^2 \), Perimeter = \( 60 \, \text{in} \)
6. Area = \( 480 \, \text{mm}^2 \), Perimeter = \( 136 \, \text{mm} \)

\boxed{
\begin{array}{ll}
1) & \text{Area} = 45 \, \text{cm}^2, \text{Perimeter} = 28 \, \text{cm} \\
2) & \text{Area} = 36 \, \text{in}^2, \text{Perimeter} = 24 \, \text{in} \\
3) & \text{Area} = 28 \, \text{ft}^2, \text{Perimeter} = 22 \, \text{ft} \\
4) & \text{Area} = 90 \, \text{ft}^2, \text{Perimeter} = 42 \, \text{ft} \\
5) & \text{Area} = 200 \, \text{in}^2, \text{Perimeter} = 60 \, \text{in} \\
6) & \text{Area} = 480 \, \text{mm}^2, \text{Perimeter} = 136 \, \text{mm} \\
\end{array}
}