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

Height of pyramid

$=\sqrt{\left(\frac{5 a}{2}\right)^{2}-(2 a)^{2}}=\sqrt{\frac{25 a^{2}}{4}-4 a^{2}}$

$=\sqrt{\frac{25 a^{2}-16 a^{2}}{4}}=\sqrt{\frac{9 a^{2}}{4}}=\frac{3}{2} a$ units

Area of base

$=\frac{3 \sqrt{3}}{2} \times(2 a)^{2}=6 \sqrt{3} a^{2}$ sq. units

$\therefore$ Volume of pyramid $=\frac{1}{3} \times$ Area of base $\times$ height

$=\frac{1}{3} \times 6 \sqrt{3} a^{2} \times \frac{3}{2} a=3 \sqrt{3} a^{3}$ cu. units