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.)
what one is bigger 25m or 0.25km
x=width x+25=length 4x+50=130 4x=80 x=20 width is 20 length is 45
25 metres / 0.3048 = 82.02ft
Providing it's a right angle triangle use Pythagoras' theorem: 252+402 = 2225 and the square root of this is about 47.16990566 meters The exact answer is 5 times the square root of 89
25m is 82 feet 0.25 inches.
If the diagonal is 25m and the area is 168m2 then the longest edge of the rectangle will be 24m.
25m
Around 25m in length.
one length down or one length back.
25m x 6m x 18m = 2,700 cubic m
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.)
volume = depth x length x width or volume = depth x area Your area is 1m² and your depth is 25cm or 0.25 m so.... Volume = .25m x 1m² = .25 m³
There are 1500/25 = 60 lengths of 25cm
To cover a hall of 25mx25m is 625 sq m As area of hall is calculated by length x breadth which is nothing but same length is there it is nothing but a square
Typically each lane will be 2.5 meters (8.25 feet) wide. So, just multiply that by the number of lanes the pool has.
A cone that has a radius of 20m and a height of 25m has a volume of 10471.98m3