<Dl> <Dd> V = 1 12 n s 3 cot ⁡ (π n) 1 − 1 4 sin 2 ⁡ π n . (\ displaystyle V = (\ frac (1) (12)) ns ^ (3) \ cot \ left ((\ frac (\ pi) (n)) \ right) (\ sqrt (1 - (\ frac (1) (4 \ sin ^ (2) (\ tfrac (\ pi) (n)))))).) </Dd> </Dl> <Dd> V = 1 12 n s 3 cot ⁡ (π n) 1 − 1 4 sin 2 ⁡ π n . (\ displaystyle V = (\ frac (1) (12)) ns ^ (3) \ cot \ left ((\ frac (\ pi) (n)) \ right) (\ sqrt (1 - (\ frac (1) (4 \ sin ^ (2) (\ tfrac (\ pi) (n)))))).) </Dd> <P> This formula only applies for n = 2, 3, 4 and 5; and it also covers the case n = 6, for which the volume equals zero (i.e., the pyramid height is zero). </P> <P> The surface area of a pyramid is A = B + P L 2 (\ displaystyle A = B+ (\ tfrac (PL) (2))), where B is the base area, P is the base perimeter, and the slant height L = h 2 + r 2 (\ displaystyle L = (\ sqrt (h ^ (2) + r ^ (2)))), where h is the pyramid altitude and r is the inradius of the base . </P>

A pyramid has its vertex at the center