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If the pool is 26 feet around, then its circumference is 26 feet. If the pool is 26 feet across, then its circumference is 81.7 feet.

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Q: What is the circumference of a 26 ft round pool?

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fifty-six The circumference of an 18-foot diameter circular swimming pool is: 56.55 feet

6 pi ft

956 gallons.

Assuming you are asking for the surface area of a cylinder of height 84 in and circumference (of end) of 26 in: 12 in = 1 ft circumference = 2 x π x radius → radius = circumference ÷ (2 x π) surface_area = 2 x area_of_ends + curved_area = 2 x π x radius2 + height x circumference = 2 x π x (circumference ÷ (2 x π))2 + height x circumference = circumference2 ÷ (2 x π) + height x circumference = (26 ÷ 12 ft)2 ÷ (2 x π) + (84 ÷ 12 ft) x (26 ÷ 12 ft) ≈ 15.91 sq ft

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37 feet and 8.5 inches

40 sq ft

fifty-six The circumference of an 18-foot circular swimming pool is: 56.55 feet

A little over 75 feet. 24 x 3.14159

fifty-six The circumference of an 18-foot diameter circular swimming pool is: 56.55 feet

If the diameter is 25 feet then the circumference is 78.5 feet (approx).

The circumference of the pool is 24 ft so its radius is 24/(2*pi) = 3.8 ft.Then radius of pool + edging = 3.8 + 2 = 5.8 ft.So circumference around edging = 36.6 feet = 36.6/3 = 12.2 yards.The circumference of the pool is 24 ft so its radius is 24/(2*pi) = 3.8 ft.Then radius of pool + edging = 3.8 + 2 = 5.8 ft.So circumference around edging = 36.6 feet = 36.6/3 = 12.2 yards.The circumference of the pool is 24 ft so its radius is 24/(2*pi) = 3.8 ft.Then radius of pool + edging = 3.8 + 2 = 5.8 ft.So circumference around edging = 36.6 feet = 36.6/3 = 12.2 yards.The circumference of the pool is 24 ft so its radius is 24/(2*pi) = 3.8 ft.Then radius of pool + edging = 3.8 + 2 = 5.8 ft.So circumference around edging = 36.6 feet = 36.6/3 = 12.2 yards.

6 pi ft

484/7 = 69 and one seventh ft

120

6 ft X 6ft pool would be 36 sq ft 6 ft round pool would be 28.26 sq ft

If the 15 ft is the distance around the edge, then:Distance_round_edge = 15 ft If the 15 ft is the diameter of the round pool, then:Distance_round_edge = π x diameter = π x 15 ft= 15π ft≈ 47.12 ftIf the 15 ft is the radius of the round pool, then:Distance_round_edge = 2 x π x radius = 2 x π x 15 ft= 30π ft≈ 94.25 ft