The volume of  an equilateral triangular frustrum is 183√3 in^3.  The upper base, 9 in. above the lower base,  has an edge which measures 8 in.  Compute for the edge of the lower base.
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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.

by Top Rated User (823k points)

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