Given h = 20 m,
Side of base = a = 5 m
` :. " Area of base"=(sqrt (3))(4)xx a^(2) xx 6=(6sqrt(3))/(4) xx 5^(2)= (3 sqrt(3))/(2) xx 25 m^(2).`
Volume `=(1)/(3)` Ah, where A = Area of the base and h = Height
`=(1)/(3)xx(3 sqrt(3))/(2) (25) xx 20=sqrt(3) xx 250 = 250 sqrt(3) m^(3)`.
To find slant height, refer to the figure shown In the figure,
O is the vertex of the pyramid and G is the centre of the hexagonal base. H is the mid-point of AB.
OG is the axis of the pyramid.
OH is the slant height of the pyramid.
`triangle OGH` is a right angled triangle.
` :. OH^(2)=GH^(2) +OG^(2)`
`GH = " Altitude of " triangle AGB=(sqrt(3))/(2) a =(sqrt(3))/(2) xx 5=(5 sqrt(3))/(2) m`
` :. OH^(2)=((5 sqrt(3))/(2))^(2) + (20)^(2) = (25xx3)/(4)+400`
`=(75+1600)/(4)=(1675)/(4)`
`implies OH=(sqrt(1675))/(2) m`
` :. " Slant height"=(sqrt(1675))/(2) m`.
