Pyramid: Volume

With $V = \frac{1}{3}\cdot{}G\cdot{}h$ and $G = a^{2}$ for the square base, i.e. $V = \frac{1}{3}\cdot{}a^{2}\cdot{}h$. Example: $a = 6 cm$ and $h = 10 cm$ give $G = 36 cm^{2}$ and $V = \frac{1}{3}\cdot{}36\cdot{}10 = 120 cm^{3}$. 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: $a^{2}\cdot{}h = 360 cm^{3}$ would be the enclosing box, the pyramid holds exactly one third of it. Famous application: the Great Pyramid of Giza with $a \approx{} 230 m$ and $h \approx{} 147 m$ reaches about 2.59 million $m^{3}$.

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 $\frac{1}{3}$ 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 h_s 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: h_s = √(h² + (a/2)²). Example: $a = 6 cm$, $h = 10 cm$ gives h_s = √(100 + 9) = √109 ≈ 10.44 cm. Even longer is the lateral edge to the corner: $s = \sqrt{(h^{2} + (\frac{a^{2}}{2}))} \approx{} 10.86 cm$. In exams confusing these three lengths often decides between right and wrong; a labelled sketch protects against it.

Rearrange $V = \frac{1}{3}\cdot{}G\cdot{}h$ for h: $h =\frac{ 3V}{G}$. The factor 3 balances the third in the formula. Example: $V = 400 cm^{3}$ and $G = 100 cm^{2}$ give $h =\frac{ 1200}{100} = 12 cm$. If the edge length a of a square base is given instead of G, first compute $G = a^{2}$. Conversely you find the base area via $G =\frac{ 3V}{h}$, and for a square base the edge $a = \sqrt{G}$. A frequent mistake is computing $\frac{V}{G}$ 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: $V = \frac{1}{3}\cdot{}\pi{}r^{2}\cdot{}h$.