truncated pyramid?

The solution to an angularly truncated pyramid is difficult. In general it requires the evaluation of a number of simple integrals with nasty boundary conditions.

Fortunately, the specific problem here is much easier. By "equilateral pyramid" I assume you mean a square based pyramid with the 4 triangular sides being equilateral. The altitude of each face then is S*sqrt(3)/2 where S is the face length. The height h of the pyramid is determined by solving the right triangle consisting of the pyramid height, the triangle height, and 1/2 the base. This gives a pyramid height of:

h = sqrt((S*sqrt(3)/2)^2 - (S/2)^2) = S * sqrt(1/2)

For your case of S = 600, the height is 424.3 m. Since your plane intersects one of the edges at a height of 455 m, the plane never intersects the pyramid so the answer is the total volume of h*S*S/3 = S^3 * sqrt(1/2)/3 = 5.091 x 10^7 m^3.

ok, so the bottom of the pyramid is a, the top is b. c is the base of the bottom rectangle, d is the base of the top. 1/3(a+cd+b)h so say the bottom of the pyramid is 15, the top is 5. and its 10 high. 15^2=225, cd=75, b^2=25 add those and get 325 3,250 when u multiply by height, then divide by 3 to get 1,083 1/3 or 1,083.333... this was for a square pyramid. im pretty sure this is right but im not totally sure.