Trying to figure this out too...
V=h/3 (axa + axb + bxb)
A hollow truncated cone is a geometric shape that is cone-shaped. The formula to calculate the volume is s^2=h^2 + (R-r)^2.
The formula for the volume of a truncated square pyramid with height h, and top edge a cm and bottom edge b cm is V = 1/3*(a2 + ab + b2)*h.
If the area of the base of the tetrahedron is A square units and the vertical height is h units, then the volume is V = 1/3*A*h cubic units. If the tetrahedron is regular, with sides of length of length s units, then V = sqrt(2)/12*s3 cubic units.
The volume(cm3) of a tetrahedron is 1/3 (area of the base)X height
V=h/3 (axa + axb + bxb)
A hollow truncated cone is a geometric shape that is cone-shaped. The formula to calculate the volume is s^2=h^2 + (R-r)^2.
A formula is a string of letters and numbers and other mathematical characters. It has no volume.
no
V = (1/3*Pi*h) * (R12 + R22 + R1*R2) Where R1 and R2 are the radii of the bases, and h is equal to the height of the truncated cone.
It is a set of mathematical operations which have to be carried out, using some measures of an object and possibly mathematical constants, to find the total amount of space which an object occupies.
The formula for the volume of a truncated square pyramid with height h, and top edge a cm and bottom edge b cm is V = 1/3*(a2 + ab + b2)*h.
If the area of the base of the tetrahedron is A square units and the vertical height is h units, then the volume is V = 1/3*A*h cubic units. If the tetrahedron is regular, with sides of length of length s units, then V = sqrt(2)/12*s3 cubic units.
6s2
The volume(cm3) of a tetrahedron is 1/3 (area of the base)X height
The volume of a tetrahedron is one-sixth of the volume of a parallelepiped because a tetrahedron can be thought of as a pyramid with a triangular base. When a tetrahedron is inscribed within a parallelepiped, it occupies one-sixth of the space defined by the parallelepiped's volume. Since a parallelepiped can be divided into six such tetrahedra, this means the volume of the tetrahedron is 1/6 of the parallelepiped. However, if the parallelepiped is defined by its full height and includes the whole base area, the tetrahedron's volume is one-sixteenth of the total volume when considering the full dimensions of the parallelepiped.
The individual who earlier posted an answer provided the mathematical formula for volume -.-