How many gallons of water does a 27 foot round pool that is 52 inches deep hold?

Answer 1

For a pool 52 inches deep, we will assume you don't want it filled to the brim. Let's fill it to 4 inches below the top. Then the water depth will be 48 inches, or 4 feet. (If this is a math assignment problem rather than a practical question, and you want to calculate as though you'd fill the pool to the brim, just rework the problem with a height of 4.333 feet instead of 4 feet.) But there is still the question of whether the pool is 27 feet around or 27 feet across. So we'll do it both ways. If your "27 foot round pool" is 27 feet around (circumference): The circumference of a circle is pi x d. So the diameter of this pool is 27' / pi = 27'/3.1416 = 8.6'. The radius of this circle is half the diameter, or 4.3'. The area is pi x r2 = 3.1416 x (4.297')2 = 58 square feet. The volume of water in the pool is the area times the height: 58 square feet x 4 feet = 232 cubic feet. Since 1 cubic foot holds 7.481 gallons, this pool will hold 232 x 7.481 or 1,736 gallons of water (to the nearest gallon). (At 8.34 pounds/gallon, this much water weighs 14,478 pounds.) If your "27 foot round pool" is 27 feet across (in diameter): Then the radius of the round pool is half that, or 13.5'. The area is pi x r2 = 3.1416 x (13.5')2 = 572.56 square feet. The volume of water in the pool is the area times the height: 572.56 square feet x 4 feet = 2290 cubic feet. Since 1 cubic foot holds 7.481 gallons, this pool will hold 2,290 x 7.481 or 17,133 gallons of water (to the nearest gallon). (At 8.34 pounds/gallon, this much water weighs 142,889 pounds.)