That's a good question. I have never thought about that. But yes, for anything excluding circular objects, length and width are used.
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Assuming a circular pool, divide the diameter by 2 to get the radius, then use the formula for the area of a circle. The depth is not relevant for this problem.
Divide the diameter by two to get the radius. Then you can use the standard formula for the area of a circle, which I copy here for your convenience: area = pi x radius squared
The Reynolds number is the relationship between inertial forces (numerator) and viscious forces (demoninator). The terms in the numerator are usually the fluid density, velocity, and "characteristic" dimension (which has units of length). Depending on the type of problem you are trying to solve that characteristic dimension may be a length (flat plate problem) or a diameter (flow in a round tube), or hydraulic diameter (non-circular internal flow). Since a non-circular cross section has more surface area than a circular cross section area there needs to be a way to account for this difference. If you use the hydraulic diameter equation to calculate the hy. dia. of a circular cross section, you will get the diameter of the circle.
You just need to find the circumference of the circular base (or top). It's the same thing. (Hint: use C=pi*diameter to find the circumference).
Only if you have two dimensional objects of specific sizes in specific combinations.