Pyramid: Volume

With V = ⅓·G·h and G = a² for the square base, i.e. V = ⅓·a²·h. Example: a = 6 cm and h = 10 cm give G = 36 cm² and V = ⅓·36·10 = 120 cm³. Important: h is the body height, meeting the centre of the base perpendicularly from the apex, not the slanted face height of the triangular faces. The result carries a cubic unit. As a size check compare with the cuboid: a²·h = 360 cm³ would be the enclosing box, the pyramid holds exactly one third of it. Famous application: the Great Pyramid of Giza with a ≈ 230 m and h ≈ 147 m reaches about 2.59 million m³.

The core of the argument: a triangular prism can be cut completely into three pyramids that pairwise share equal base and equal height and therefore have equal volume. So each one holds a third of the prism. Since every base can be decomposed into triangles, the factor 1/3 carries over to arbitrary pyramids, and the cone is the limiting case with a circular base. By Cavalieri's principle only the cross-section areas at each height matter, which is why oblique pyramids satisfy the formula too. This one-third logic is the same as for the cone; whoever understands one of the two formulas gets the other for free.

The body height h runs perpendicularly from the apex to the centre of the base and belongs in the volume formula. The face height hₛ is the height of a lateral triangle: it runs along the outside from the apex to the midpoint of a base edge and is needed for the lateral surface. In a square pyramid Pythagoras links the two: hₛ = √(h² + (a/2)²). Example: a = 6 cm, h = 10 cm gives hₛ = √(100 + 9) = √109 ≈ 10.44 cm. Even longer is the lateral edge to the corner: s = √(h² + (a²/2)) ≈ 10.86 cm. In exams confusing these three lengths often decides between right and wrong; a labelled sketch protects against it.

Rearrange V = ⅓·G·h for h: h = 3V/G. The factor 3 balances the third in the formula. Example: V = 400 cm³ and G = 100 cm² give h = 1200/100 = 12 cm. If the edge length a of a square base is given instead of G, first compute G = a². Conversely you find the base area via G = 3V/h, and for a square base the edge a = √G. A frequent mistake is computing V/G without the factor 3; the result is then three times too small. A check by substituting into the original formula takes ten seconds and catches exactly this mistake.

Yes, in full generality. The base may be a triangle, rectangle, hexagon or any other polygon; G is simply its area. The apex does not have to sit above the centre either: by Cavalieri's principle only the size of the cross-section at each height matters, and for every pyramid it shrinks quadratically from G to 0. Two pyramids with equal base and equal height therefore have the same volume, no matter how oblique they are; h is always the perpendicular distance of the apex from the base plane. Even the cone follows this logic as a pyramid with a circular base: V = ⅓·πr²·h.