Let h be the height of the pyramid, measured from the center of the base to the apex of the pyramid
volume of a square based pyramid = a2(h/3) = 82(h/3) = 64h/3
where a is the length of a side of the base
Let s be the slant height of the pyramid, then
s = \(\sqrt {{h^2} + {{\left( {a/2} \right)}^2}} \) = \(\sqrt {{h^2} + {{\left( {8/2} \right)}^2}} \) = \(\sqrt {{h^2} + {{\left( 4 \right)}^2}} \) = \(\sqrt {{h^2} + 16} \)
Now the pyramid is cut in two parts by a plane parallel to its base such that height of upper and lower parts are in ratio 1 : 3
Slant height will also be divided in the ratio 1 : 3
So, let the slant height of the upper pyramid be s1, height will be h/4 and base be a1
s1 = \(\frac{{\sqrt {{h^2} + 16} }}{4}\)
So, base of the upper pyramid will be,
\({\left( {\frac{{\sqrt {{h^2} + ;;16} }}{4}} \right)^2}\) = (h/4)2 + (a1/2)2
On solving,
a1 = 2
Then, Volume of upper pyramid = \(\frac{{4{\rm{;x;}}\left( {{\rm{h}}/4} \right)}}{3}\) = h/3
Volume of frustum = volume of actual pyramid - volume of upper pyramid formed by cutting the actual pyramid
volume of frustum = 64h/3 - h/3 = 63h/3
So, the required ratio will be = 63 : 1