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

- Dimensions: 8 cm, 8 cm, 10 cm
- Type: Isosceles triangle

#### Perimeter
The perimeter is the sum of all sides:
\[
\text{Perimeter} = 8 + 8 + 10 = 26 \, \text{cm}
\]

#### Area
For an isosceles triangle, we can use the formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is 10 cm, and we need to find the height. Using the Pythagorean theorem in one of the right triangles formed by dropping a perpendicular from the apex to the base:
\[
\text{Height}^2 + \left(\frac{\text{base}}{2}\right)^2 = \text{side}^2
\]
\[
\text{Height}^2 + \left(\frac{10}{2}\right)^2 = 8^2
\]
\[
\text{Height}^2 + 5^2 = 64
\]
\[
\text{Height}^2 + 25 = 64
\]
\[
\text{Height}^2 = 39
\]
\[
\text{Height} = \sqrt{39}
\]
Now, calculate the area:
\[
\text{Area} = \frac{1}{2} \times 10 \times \sqrt{39} = 5\sqrt{39} \, \text{cm}^2
\]

---

2. Second Triangle

- Dimensions: 4 cm, 7 cm, 9 cm
- Type: Scalene triangle

#### Perimeter
The perimeter is the sum of all sides:
\[
\text{Perimeter} = 4 + 7 + 9 = 20 \, \text{cm}
\]

#### Area
We use Heron's formula for the area of a triangle:
\[
s = \frac{a + b + c}{2} \quad \text{(semi-perimeter)}
\]
\[
s = \frac{4 + 7 + 9}{2} = 10
\]
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
\text{Area} = \sqrt{10(10-4)(10-7)(10-9)}
\]
\[
\text{Area} = \sqrt{10 \times 6 \times 3 \times 1} = \sqrt{180} = 6\sqrt{5} \, \text{cm}^2
\]

---

3. Third Triangle

- Dimensions: 9 cm, 9 cm, 5 cm
- Type: Isosceles triangle

#### Perimeter
The perimeter is the sum of all sides:
\[
\text{Perimeter} = 9 + 9 + 5 = 23 \, \text{cm}
\]

#### Area
Using the formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is 5 cm, and we need to find the height. Using the Pythagorean theorem in one of the right triangles formed by dropping a perpendicular from the apex to the base:
\[
\text{Height}^2 + \left(\frac{\text{base}}{2}\right)^2 = \text{side}^2
\]
\[
\text{Height}^2 + \left(\frac{5}{2}\right)^2 = 9^2
\]
\[
\text{Height}^2 + 2.5^2 = 81
\]
\[
\text{Height}^2 + 6.25 = 81
\]
\[
\text{Height}^2 = 74.75
\]
\[
\text{Height} = \sqrt{74.75}
\]
Now, calculate the area:
\[
\text{Area} = \frac{1}{2} \times 5 \times \sqrt{74.75} = \frac{5\sqrt{74.75}}{2} \, \text{cm}^2
\]

---

4. Fourth Triangle

- Dimensions: 10 cm, 11 cm, 12 cm
- Type: Scalene triangle

#### Perimeter
The perimeter is the sum of all sides:
\[
\text{Perimeter} = 10 + 11 + 12 = 33 \, \text{cm}
\]

#### Area
We use Heron's formula for the area of a triangle:
\[
s = \frac{a + b + c}{2} \quad \text{(semi-perimeter)}
\]
\[
s = \frac{10 + 11 + 12}{2} = 16.5
\]
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
\text{Area} = \sqrt{16.5(16.5-10)(16.5-11)(16.5-12)}
\]
\[
\text{Area} = \sqrt{16.5 \times 6.5 \times 5.5 \times 4.5}
\]
\[
\text{Area} = \sqrt{2541.5625} \approx 50.41 \, \text{cm}^2
\]

---

5. Fifth Triangle

- Dimensions: 11 cm, 11 cm, 11 cm (equilateral triangle)
- Height: 9 cm

#### Perimeter
The perimeter is the sum of all sides:
\[
\text{Perimeter} = 11 + 11 + 11 = 33 \, \text{cm}
\]

#### Area
For an equilateral triangle, the area can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
\[
\text{Area} = \frac{1}{2} \times 11 \times 9 = 49.5 \, \text{cm}^2
\]

---

6. Sixth Triangle

- Dimensions: 7 m, 7 m, 6 m (isosceles triangle)
- Height: 7 m

#### Perimeter
The perimeter is the sum of all sides:
\[
\text{Perimeter} = 7 + 7 + 6 = 20 \, \text{m}
\]

#### Area
Using the formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
\[
\text{Area} = \frac{1}{2} \times 6 \times 7 = 21 \, \text{m}^2
\]

---

Final Answers

1. First Triangle:
- Perimeter: \( 26 \, \text{cm} \)
- Area: \( 5\sqrt{39} \, \text{cm}^2 \)

2. Second Triangle:
- Perimeter: \( 20 \, \text{cm} \)
- Area: \( 6\sqrt{5} \, \text{cm}^2 \)

3. Third Triangle:
- Perimeter: \( 23 \, \text{cm} \)
- Area: \( \frac{5\sqrt{74.75}}{2} \, \text{cm}^2 \)

4. Fourth Triangle:
- Perimeter: \( 33 \, \text{cm} \)
- Area: \( \approx 50.41 \, \text{cm}^2 \)

5. Fifth Triangle:
- Perimeter: \( 33 \, \text{cm} \)
- Area: \( 49.5 \, \text{cm}^2 \)

6. Sixth Triangle:
- Perimeter: \( 20 \, \text{m} \)
- Area: \( 21 \, \text{m}^2 \)

\boxed{
\begin{array}{ll}
\text{First Triangle:} & \text{Perimeter} = 26 \, \text{cm}, \, \text{Area} = 5\sqrt{39} \, \text{cm}^2 \\
\text{Second Triangle:} & \text{Perimeter} = 20 \, \text{cm}, \, \text{Area} = 6\sqrt{5} \, \text{cm}^2 \\
\text{Third Triangle:} & \text{Perimeter} = 23 \, \text{cm}, \, \text{Area} = \frac{5\sqrt{74.75}}{2} \, \text{cm}^2 \\
\text{Fourth Triangle:} & \text{Perimeter} = 33 \, \text{cm}, \, \text{Area} \approx 50.41 \, \text{cm}^2 \\
\text{Fifth Triangle:} & \text{Perimeter} = 33 \, \text{cm}, \, \text{Area} = 49.5 \, \text{cm}^2 \\
\text{Sixth Triangle:} & \text{Perimeter} = 20 \, \text{m}, \, \text{Area} = 21 \, \text{m}^2 \\
\end{array}
}