Calculate the volume of water in a cylindrical dam that is 80% full, given outer diameter, wall thickness, height, and percentage full.

The outer diameter is 6 m, and the wall thickness is 0.35 m on each side.

$$\text{Inner diameter} = \text{Outer diameter} - 2 \times \text{Wall thickness}$$ $$\text{Inner diameter} = 6 \text{ m} - 2 \times 0.35 \text{ m}$$ $$\text{Inner diameter} = 6 \text{ m} - 0.70 \text{ m}$$ $$\text{Inner diameter} = 5.3 \text{ m}$$

$$\text{Inner radius} (r) = \frac{\text{Inner diameter}}{2}$$ $$r = \frac{5.3 \text{ m}}{2}$$ $$r = 2.65 \text{ m}$$

The dam is a cylinder with height $h = 1.7 \text{ m}$.

$$\text{Volume of cylinder} = \pi r^2 h$$ $$\text{Volume}_{full} = \pi \times (2.65 \text{ m})^2 \times 1.7 \text{ m}$$ $$\text{Volume}_{full} = \pi \times 7.0225 \text{ m}^2 \times 1.7 \text{ m}$$ $$\text{Volume}_{full} \approx 3.14159 \times 7.0225 \times 1.7 \text{ m}^3$$ $$\text{Volume}_{full} \approx 37.500 \text{ m}^3$$

$$\text{Volume}_{water} = 0.80 \times \text{Volume}_{full}$$ $$\text{Volume}_{water} = 0.80 \times 37.500 \text{ m}^3$$ $$\text{Volume}_{water} = 30.000 \text{ m}^3$$

Given that $1 \text{ m}^3 = 1 \text{ kl}$.

$$\text{Volume}_{water} = 30.000 \text{ kl}$$

The volume of water in the dam is $\boxed{\text{30.000 kl}}$.

Given height $h = 8 \text{ cm}$.
Given area $A = 1800 \text{ mm}^2$.
Since $1 \text{ cm} = 10 \text{ mm}$, then $1 \text{ cm}^2 = (10 \text{ mm})^2 = 100 \text{ mm}^2$.

$$A = \frac{1800 \text{ mm}^2}{100 \text{ mm}^2/\text{cm}^2}$$ $$A = 18 \text{ cm}^2$$

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$18 \text{ cm}^2 = \frac{1}{2} \times \text{base} \times 8 \text{ cm}$$

$$18 \text{ cm}^2 = 4 \text{ cm} \times \text{base}$$ $$\text{base} = \frac{18 \text{ cm}^2}{4 \text{ cm}}$$ $$\text{base} = 4.5 \text{ cm}$$

The base length of the triangle is $\boxed{\text{4.5 cm}}$.