A cylinder with a radius of 1.25 meters and a height of 6 meters has a volume of 29.45 cubic meters.
V = 169.65 m3
Volume = 753.98224m3
how to find the dept of the pool if it is 15m long and 6m wide
25m x 6m x 18m = 2,700 cubic m
It is: 6m
V = 169.65 m3
The radius IS given, since height of hemisphere = radius of hemisphere!
Volume = 753.98224m3
I presume the tunnel has an arch shaped cross-section with a semicircle on top of a rectangle. In this case the volume of the tunnel is the volume of the cuboid bottom plus the volume of half cylinder which forms the top. The volume of the half cylindrical top is ½πr²h = ½ × π × (6m)² × 25m = 450π m³ The volume of the cuboid is length × width × height. The length is 25m. The width is the diameter of the top half cylinder which is twice the radius at 2 x 6m = 12 m. The height is not clear. I am going to presume it is to the top of the arch, so that the height of the cuboid is the height less the radius of the cylinder, namely 7m - 6m = 1m. Thus the volume of the cuboid bit is 25m x 12m x 1m = 300 m³ Thus the volume of the tunnel as a whole is 300 m³ + 450π m³ ≈ 1713.72 m³ (If the 7m height refers to the height of the vertical walls, then the volume of the cuboid is 25m x 12m x 7m = 2100 m³ and the volume of the tunnel is 2100 m³ + 450π m³ ≈ 3513.72 m³.)
how to find the dept of the pool if it is 15m long and 6m wide
If it is a rectangular tank, the volume is length times width times height, 6m x 6m x 3m, or 108m3.
Volume pyramid = 1/3 x base_area x height = 1/3 x (6m x 6m) x 6m = 72m3.
Volume = 1/3*pi*52*6 = 50*pi cubic m
Volume= 2.4m * 6m * 8m= 115.20 cubic meters
radius = diameter ÷ 2 volume cylinder = area of end × height → volume = π × radius² × height → volume = π × (diameter ÷ 2)² × height →volume = π × (6m ÷ 2)² × 3 m → volume = 27π m³ ≈ 27 × 3.14 m³ = 84.78 m³ ≈ 84.8 m³ to 1 dp.
286.3
25m x 6m x 18m = 2,700 cubic m