Find volume of frustrum.

Question

For frustrum h=1490/121,
r1=63/11 and r2=21/11

Answer 1

Assume that the frustum is conical, because radii are implied.

Volume of large cone (enclosing the frustum) V₁=⅓(πr₁²)h₁; volume of small (top) cone V₂=⅓(πr₂²)h₂, where h=h₁-h₂, i.e., the difference of the heights of the two cones forming the frustum.

r₁=63/11, r₂=21/11, h=1490/121 units (given). [r₁=3r₂.]

Volume of frustum V=V₁-V₂=⅓(πr₁²)h₁-⅓(πr₂²)h₂=(π/3)(r₁²h₁-r₂²h₂).

h₂/r₂=h₁/r₁ because the two cones forming the frustum are similar figures.

h₂=h₁-h, so (h₁-h)/r₂=h₁/r₁,

r₁(h₁-h)=h₁r₂,

r₁h₁-r₁h=h₁r₂,

h₁(r₁-r₂)=r₁h, h₁=r₁h/(r₁-r₂).

From the given info: h₁=(63/11)(1490/121)/(42/11),

h₁=(3/2)(1490/121)=2235/121.

h₂=2235/121-1490/121=745/121. [h₁=3h₂.]

V=(π/3)(r₁²h₁-r₂²h₂)=745(π/3)(21/11)²[27/121)-1/121],

V=26(745π/3)(21/11)²/11²=610.98 cu units approx.