Pyramid Volume

Any pyramid fills exactly one-third of the prism that shares its base and height. Not roughly a third — exactly, and for any base shape at all. You can see the square case with your hands: three identical pyramids, each with the base of a cube and its apex at one of the cube's top corners, fit together to fill that cube with nothing left over. The general case follows from Cavalieri's principle. Slice both solids at any height and compare the cross-sections; if they match at every level, the volumes match. At a fraction $x$ of the way down from the apex, a pyramid's cross-section is a scaled copy of the base with linear scale $x$, so its area is $x^2 B$. Add those up from apex to base — which is to say integrate $x^2$ from 0 to 1 — and you get $\tfrac{1}{3}$. The one-third is the integral of a square, which is also why cones share it.

Because only the cross-sections matter, leaning the apex sideways changes nothing. An oblique pyramid holds exactly what the upright one of the same base and height holds, the same way a leaning stack of paper holds the same paper.

A worked instance: the tapered hopper under a feed bin, square at 1.2 m on a side and drawing down to a point 900 mm below, has $B = 1.44$ m² and $V = \tfrac{1}{3}(1.44)(0.9) = 0.43$ m³, about 430 litres. And the monumental one — the Great Pyramid at Giza, 230 m on a side and originally 146.6 m tall, encloses $\tfrac{1}{3}(52{,}900)(146.6) \approx 2.6$ million m³ of stone.

Two errors, and the first is specific to this page. $B$ is the base area, not the base edge. Typing 230 for the Great Pyramid instead of 52,900 understates the answer by a factor of 230, and because the field accepts any number it will not object. Compute the base area first, in whatever shape the base actually is. The second error is the height: $h$ is measured straight up from the base plane to the apex, never along the slope of a face. At Giza the slant height up the middle of a face is 186 m against a true height of 146.6 m, so using it would inflate the volume by 27%. The slant is the number you can put a tape on, which is exactly why it keeps getting entered. If the slant height $l$ and the base half-width $a$ are what you have, recover the true height with $h = \sqrt{l^2 - a^2}$ before coming back here.