Calculate area and perimeter of irregular shapes with given dimensions.

To solve the problem of calculating the area and perimeter of the given irregular shapes, we will break each shape into simpler geometric figures (rectangles) and then compute the total area and perimeter step by step.

---

Shape 1:

#### Dimensions:
- Larger rectangle: \(6 \, \text{cm} \times 4 \, \text{cm}\)
- Smaller rectangle: \(2 \, \text{cm} \times 4 \, \text{cm}\)

#### Area:
The area of the larger rectangle is:
\[
6 \times 4 = 24 \, \text{cm}^2
\]
The area of the smaller rectangle is:
\[
2 \times 4 = 8 \, \text{cm}^2
\]
Since the smaller rectangle is subtracted from the larger one, the total area is:
\[
24 - 8 = 16 \, \text{cm}^2
\]

#### Perimeter:
The perimeter of the irregular shape can be calculated by summing the outer edges. The shape has the following outer dimensions:
- Top: \(6 \, \text{cm}\)
- Bottom: \(6 \, \text{cm}\)
- Left side: \(6 \, \text{cm}\)
- Right side: \(4 \, \text{cm}\) (top part) + \(2 \, \text{cm}\) (bottom part) = \(6 \, \text{cm}\)
- Inner vertical segment: \(2 \, \text{cm}\)

Thus, the perimeter is:
\[
6 + 6 + 6 + 6 + 2 = 26 \, \text{cm}
\]

#### Final Answer for Shape 1:
\[
\boxed{16 \, \text{cm}^2, 26 \, \text{cm}}
\]

---

Shape 2:

#### Dimensions:
- Larger rectangle: \(6 \, \text{m} \times 4 \, \text{m}\)
- Smaller rectangle: \(2 \, \text{m} \times 4 \, \text{m}\)

#### Area:
The area of the larger rectangle is:
\[
6 \times 4 = 24 \, \text{m}^2
\]
The area of the smaller rectangle is:
\[
2 \times 4 = 8 \, \text{m}^2
\]
Since the smaller rectangle is subtracted from the larger one, the total area is:
\[
24 - 8 = 16 \, \text{m}^2
\]

#### Perimeter:
The perimeter of the irregular shape can be calculated by summing the outer edges. The shape has the following outer dimensions:
- Top: \(6 \, \text{m}\)
- Bottom: \(6 \, \text{m}\)
- Left side: \(6 \, \text{m}\)
- Right side: \(4 \, \text{m}\) (top part) + \(2 \, \text{m}\) (bottom part) = \(6 \, \text{m}\)
- Inner vertical segment: \(2 \, \text{m}\)

Thus, the perimeter is:
\[
6 + 6 + 6 + 6 + 2 = 26 \, \text{m}
\]

#### Final Answer for Shape 2:
\[
\boxed{16 \, \text{m}^2, 26 \, \text{m}}
\]

---

Shape 3:

#### Dimensions:
- Larger rectangle: \(6 \, \text{km} \times 3 \, \text{km}\)
- Smaller rectangle: \(2 \, \text{km} \times 3 \, \text{km}\)

#### Area:
The area of the larger rectangle is:
\[
6 \times 3 = 18 \, \text{km}^2
\]
The area of the smaller rectangle is:
\[
2 \times 3 = 6 \, \text{km}^2
\]
Since the smaller rectangle is subtracted from the larger one, the total area is:
\[
18 - 6 = 12 \, \text{km}^2
\]

#### Perimeter:
The perimeter of the irregular shape can be calculated by summing the outer edges. The shape has the following outer dimensions:
- Top: \(6 \, \text{km}\)
- Bottom: \(6 \, \text{km}\)
- Left side: \(3 \, \text{km}\)
- Right side: \(3 \, \text{km}\)
- Inner horizontal segment: \(2 \, \text{km}\)

Thus, the perimeter is:
\[
6 + 6 + 3 + 3 + 2 = 20 \, \text{km}
\]

#### Final Answer for Shape 3:
\[
\boxed{12 \, \text{km}^2, 20 \, \text{km}}
\]

---

Shape 4:

#### Dimensions:
- Larger rectangle: \(6 \, \text{cm} \times 5 \, \text{cm}\)
- Smaller rectangle: \(2 \, \text{cm} \times 1 \, \text{cm}\)

#### Area:
The area of the larger rectangle is:
\[
6 \times 5 = 30 \, \text{cm}^2
\]
The area of the smaller rectangle is:
\[
2 \times 1 = 2 \, \text{cm}^2
\]
Since the smaller rectangle is subtracted from the larger one, the total area is:
\[
30 - 2 = 28 \, \text{cm}^2
\]

#### Perimeter:
The perimeter of the irregular shape can be calculated by summing the outer edges. The shape has the following outer dimensions:
- Top: \(6 \, \text{cm}\)
- Bottom: \(6 \, \text{cm}\)
- Left side: \(5 \, \text{cm}\)
- Right side: \(5 \, \text{cm}\)
- Inner horizontal segment: \(2 \, \text{cm}\)

Thus, the perimeter is:
\[
6 + 6 + 5 + 5 + 2 = 24 \, \text{cm}
\]

#### Final Answer for Shape 4:
\[
\boxed{28 \, \text{cm}^2, 24 \, \text{cm}}
\]

---

Shape 5:

#### Dimensions:
- Larger rectangle: \(8 \, \text{m} \times 6 \, \text{m}\)
- Smaller rectangle: \(6 \, \text{m} \times 2 \, \text{m}\)

#### Area:
The area of the larger rectangle is:
\[
8 \times 6 = 48 \, \text{m}^2
\]
The area of the smaller rectangle is:
\[
6 \times 2 = 12 \, \text{m}^2
\]
Since the smaller rectangle is subtracted from the larger one, the total area is:
\[
48 - 12 = 36 \, \text{m}^2
\]

#### Perimeter:
The perimeter of the irregular shape can be calculated by summing the outer edges. The shape has the following outer dimensions:
- Top: \(8 \, \text{m}\)
- Bottom: \(8 \, \text{m}\)
- Left side: \(6 \, \text{m}\)
- Right side: \(6 \, \text{m}\)
- Inner horizontal segment: \(6 \, \text{m}\)

Thus, the perimeter is:
\[
8 + 8 + 6 + 6 + 6 = 34 \, \text{m}
\]

#### Final Answer for Shape 5:
\[
\boxed{36 \, \text{m}^2, 34 \, \text{m}}
\]

---

Shape 6:

#### Dimensions:
- Larger rectangle: \(20 \, \text{km} \times 15 \, \text{km}\)
- Smaller rectangle: \(5 \, \text{km} \times 20 \, \text{km}\)
- Square: \(10 \, \text{km} \times 10 \, \text{km}\)

#### Area:
The area of the larger rectangle is:
\[
20 \times 15 = 300 \, \text{km}^2
\]
The area of the smaller rectangle is:
\[
5 \times 20 = 100 \, \text{km}^2
\]
The area of the square is:
\[
10 \times 10 = 100 \, \text{km}^2
\]
Since the smaller rectangle and the square are subtracted from the larger one, the total area is:
\[
300 - 100 - 100 = 100 \, \text{km}^2
\]

#### Perimeter:
The perimeter of the irregular shape can be calculated by summing the outer edges. The shape has the following outer dimensions:
- Top: \(20 \, \text{km}\)
- Bottom: \(20 \, \text{km}\)
- Left side: \(15 \, \text{km}\)
- Right side: \(15 \, \text{km}\)
- Inner horizontal segments: \(5 \, \text{km}\) (left) + \(5 \, \text{km}\) (right) = \(10 \, \text{km}\)
- Inner vertical segments: \(10 \, \text{km}\) (top) + \(10 \, \text{km}\) (bottom) = \(20 \, \text{km}\)

Thus, the perimeter is:
\[
20 + 20 + 15 + 15 + 10 + 20 = 100 \, \text{km}
\]

#### Final Answer for Shape 6:
\[
\boxed{100 \, \text{km}^2, 100 \, \text{km}}
\]

---

Final Answers:

1. \(\boxed{16 \, \text{cm}^2, 26 \, \text{cm}}\)
2. \(\boxed{16 \, \text{m}^2, 26 \, \text{m}}\)
3. \(\boxed{12 \, \text{km}^2, 20 \, \text{km}}\)
4. \(\boxed{28 \, \text{cm}^2, 24 \, \text{cm}}\)
5. \(\boxed{36 \, \text{m}^2, 34 \, \text{m}}\)
6. \(\boxed{100 \, \text{km}^2, 100 \, \text{km}}\)