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).
V=(h₂/3)(A₁(h₁/h₂)-A₂)=(h₂/3)(A₁√(A₁/A₂)-A₂).
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:
V=(h/3)(√A₂/(√A₁-√A₂))(A₁√(A₁/A₂)-A₂),
V=(h/3)(A₁√A₁-A₂√A₂)/(√A₁-√A₂).
1/(√A₁-√A₂) can be rationalised:
(√A₁+√A₂)/(A₁-A₂):
V=(h/3)(A₁√A₁-A₂√A₂)(√A₁+√A₂)/(A₁-A₂),
V=(h/3)(A₁²+A₁√(A₁A₂)-A₂√(A₁A₂)-A₂²)/(A₁-A₂),
V=(h/3)(A₁²-A₂²+√(A₁A₂)(A₁-A₂))/(A₁-A₂),
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.
V=183√3in³.
So 3(16√3+A₂+√(16A₂√3))=183√3,
√3(16√3+A₂+√(16A₂√3))=183,
48+A₂√3+√(48A₂√3))=183.
If x is the larger base length, then A₂=½x(x√3/2)=¼x²√3.
Therefore, 48+¾x²+√(36x²)=183,
48+¾x²+6x=183,
¾x²+6x-135=0,
¼x²+2x-45=0,
x²+8x-180=0=(x+18)(x-10).
Therefore x=10in, the length of the lower base.
CHECK
A₂=25√3.
V=(9/3)(16√3+25√3+√((16√3)(25√3))),
V=3(41√3+√1200)=3(41√3+20√3)=183√3.