Calculate the area and perimeter of various triangles with given side lengths.

To solve the problem, we need to calculate the perimeter and area of each triangle. Let's go through each triangle step by step.

---

1. First Triangle

- Sides: 8 cm, 8 cm, 10 cm
- Perimeter: Sum of all sides
\[
\text{Perimeter} = 8 + 8 + 10 = 26 \, \text{cm}
\]
- Area: Use Heron's formula since it is not a right triangle.
- Semi-perimeter \( s \):
\[
s = \frac{8 + 8 + 10}{2} = 13 \, \text{cm}
\]
- Area using Heron's formula:
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( a = 8 \), \( b = 8 \), \( c = 10 \):
\[
\text{Area} = \sqrt{13(13-8)(13-8)(13-10)} = \sqrt{13 \cdot 5 \cdot 5 \cdot 3} = \sqrt{975} \approx 31.23 \, \text{cm}^2
\]

---

2. Second Triangle

- Sides: 4 cm, 7 cm, 9 cm
- Perimeter: Sum of all sides
\[
\text{Perimeter} = 4 + 7 + 9 = 20 \, \text{cm}
\]
- Area: Use Heron's formula.
- Semi-perimeter \( s \):
\[
s = \frac{4 + 7 + 9}{2} = 10 \, \text{cm}
\]
- Area using Heron's formula:
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( a = 4 \), \( b = 7 \), \( c = 9 \):
\[
\text{Area} = \sqrt{10(10-4)(10-7)(10-9)} = \sqrt{10 \cdot 6 \cdot 3 \cdot 1} = \sqrt{180} \approx 13.42 \, \text{cm}^2
\]

---

3. Third Triangle

- Sides: 9 cm, 9 cm, 5 cm
- Perimeter: Sum of all sides
\[
\text{Perimeter} = 9 + 9 + 5 = 23 \, \text{cm}
\]
- Area: Use Heron's formula.
- Semi-perimeter \( s \):
\[
s = \frac{9 + 9 + 5}{2} = 11.5 \, \text{cm}
\]
- Area using Heron's formula:
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( a = 9 \), \( b = 9 \), \( c = 5 \):
\[
\text{Area} = \sqrt{11.5(11.5-9)(11.5-9)(11.5-5)} = \sqrt{11.5 \cdot 2.5 \cdot 2.5 \cdot 6.5} \approx 21.62 \, \text{cm}^2
\]

---

4. Fourth Triangle

- Sides: 10 cm, 11 cm, 12 cm
- Perimeter: Sum of all sides
\[
\text{Perimeter} = 10 + 11 + 12 = 33 \, \text{cm}
\]
- Area: Use Heron's formula.
- Semi-perimeter \( s \):
\[
s = \frac{10 + 11 + 12}{2} = 16.5 \, \text{cm}
\]
- Area using Heron's formula:
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( a = 10 \), \( b = 11 \), \( c = 12 \):
\[
\text{Area} = \sqrt{16.5(16.5-10)(16.5-11)(16.5-12)} = \sqrt{16.5 \cdot 6.5 \cdot 5.5 \cdot 4.5} \approx 51.92 \, \text{cm}^2
\]

---

5. Fifth Triangle

- Sides: 11 cm, 11 cm, 11 cm (Equilateral triangle)
- Perimeter: Sum of all sides
\[
\text{Perimeter} = 11 + 11 + 11 = 33 \, \text{cm}
\]
- Area: Use the formula for the area of an equilateral triangle:
\[
\text{Area} = \frac{\sqrt{3}}{4} \times \text{side}^2
\]
where side = 11 cm:
\[
\text{Area} = \frac{\sqrt{3}}{4} \times 11^2 = \frac{\sqrt{3}}{4} \times 121 \approx 52.39 \, \text{cm}^2
\]

---

6. Sixth Triangle

- Sides: 7 m, 7 m, 6 m (Isosceles triangle)
- Perimeter: Sum of all sides
\[
\text{Perimeter} = 7 + 7 + 6 = 20 \, \text{m}
\]
- Area: Use Heron's formula.
- Semi-perimeter \( s \):
\[
s = \frac{7 + 7 + 6}{2} = 10 \, \text{m}
\]
- Area using Heron's formula:
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( a = 7 \), \( b = 7 \), \( c = 6 \):
\[
\text{Area} = \sqrt{10(10-7)(10-7)(10-6)} = \sqrt{10 \cdot 3 \cdot 3 \cdot 4} = \sqrt{360} \approx 18.97 \, \text{m}^2
\]

---

Final Answers

1. Perimeter: 26 cm, Area: 31.23 cm²
2. Perimeter: 20 cm, Area: 13.42 cm²
3. Perimeter: 23 cm, Area: 21.62 cm²
4. Perimeter: 33 cm, Area: 51.92 cm²
5. Perimeter: 33 cm, Area: 52.39 cm²
6. Perimeter: 20 m, Area: 18.97 m²

\[
\boxed{
\begin{array}{ll}
\text{Triangle 1:} & \text{Perimeter} = 26 \, \text{cm}, \, \text{Area} = 31.23 \, \text{cm}^2 \\
\text{Triangle 2:} & \text{Perimeter} = 20 \, \text{cm}, \, \text{Area} = 13.42 \, \text{cm}^2 \\
\text{Triangle 3:} & \text{Perimeter} = 23 \, \text{cm}, \, \text{Area} = 21.62 \, \text{cm}^2 \\
\text{Triangle 4:} & \text{Perimeter} = 33 \, \text{cm}, \, \text{Area} = 51.92 \, \text{cm}^2 \\
\text{Triangle 5:} & \text{Perimeter} = 33 \, \text{cm}, \, \text{Area} = 52.39 \, \text{cm}^2 \\
\text{Triangle 6:} & \text{Perimeter} = 20 \, \text{m}, \, \text{Area} = 18.97 \, \text{m}^2 \\
\end{array}
}
\]