Consider a pyramid P₁ with base areas A₁ height h₁. Volume V₁=⅓A₁h₁.
Make a frustum by slicing off the pyramidal top P₂ such that its base is parallel to the base of P₁ and its height is h₂. Its volume V₂=⅓A₂h₂.
The volume V of the frustum=V₁-V₂=⅓(A₁h₁-A₂h₂).
The difference in height (distance between bases) h=h₁-h₂, making h₁=h+h₂
P₁ and P₂ are similar figures so h₁/h₂=√(A₁/A₂)=∛(V₁/V₂).
If r=h₁/h₂, h₁=rh₂, so h=rh₂-h₂, and h₂=h/(r-1).
But r=√(A₁/A₂) therefore h₂=h/(√(A₁/A₂)-1)=h√A₂/(√A₁-√A₂).
Substituting for h₂:
1/(√A₁-√A₂) can be rationalised:
V=(h/3)(A₁+A₂+√(A₁A₂)). Note that the base areas are symmetrically referenced here, so it doesn’t matter which one is bigger.
So V=(h/3)(A₁+A₂+√(A₁A₂)) is the general volume of a frustum, where A₁ and A₂ are the base areas of the two triangular bases. h is the height of the frustum=9in.
If A₁ is the smaller base area, then A₁=½(8)(8√3/2)=16√3in².
If x is the larger base length, then A₂=½x(x√3/2)=¼x²√3.
Therefore x=10in, the length of the lower base.