There is a pyramid on a base which is a regular hexagon of side 2a cm. If every slant edge of this pyramid is of length $\frac{5a}{2}$ cm, then the volume of this pyramid is

Volume of pyramid $=\frac{1}{3}\times$ area of the base $\times$ height

Area of the base(Regular hexagon) $=\frac{3\sqrt{3}}{2}\times sid{e}^2=\frac{3\sqrt{3}}{2}\times {\left(2a\right)}^2=6\sqrt{3}{a}^{2}c{m}^2$

Slant height $l=\frac{5a}{2}$

Then height $=\sqrt{l^2-sid{e}^2}=\sqrt{{\left(\frac{5a}{2}\right)}^2-{\left(2a\right)}^2}=\frac{3a}{2}$

Volume $=\frac{1}{3}\times 6\sqrt{3}{a}^2\times \frac{3a}{2}=3\sqrt{3}{a}^{3}c{m}^3$