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

3√3 a³ cm³

⇒ Area of base = 6 × \(\frac{\sqrt {3}}{4}\) × (2a)²

                   = 6 \(\sqrt {3}\) a² cm²

⇒ Height = \(\sqrt{(slant\;edge)^2 - (side)^2 }\)

               = \(\sqrt {\frac{25a^2}{4} - 4a^2 }\) 

               = \(\frac{\sqrt {9a^2}}{\sqrt {4}}\) = \(\frac{3}{2}\)a

⇒ Volume of pyramid = \(\frac{1}{3}\) × area of base × height

                                 = \(\frac{1}{3}\) × 6\(\sqrt{3}\)a² × \(\frac{3}{2}\)a

                                 = 3\(\sqrt{3}\) a³ cm³