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Answered 2015-04-08 16:09:02

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.)

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