✔ 最佳答案
sector OAD forms the base circle b₁ of the smaller cone c₁
sector OBC forms the base circle b₂ of the larger cone c₂
Let r₁,r₂ -- radius of base circle b₁,b₂ resp.
π=2πr₁, 2π=2πr₂ - - - - -→《C=2πr》
∴ r₁=1/2, r₂=1
∴ area of base circle b₁, b₂ = πr₁²,πr₂² resp.
= π/4,π resp.
Let h₁,h₂ -- height of base circle b₁,b₂ resp.
O___
. . . . . ./ | ↑
. . . .s/. .| h₁ |
. . . ./. . .| |
. .A/--½-|X- h₂
. 4/. . . . | |
B/_____| __↓
1 Y
△OAX ～△OBY
s/(4+s) = (½) /1
=> s=4
From △OAX: h₁²+(½)² = s²
h₁²+¼ = 4²
h₁² = 63/4
h₁ = (3/2)√7
From △OBY: h₂ ²+1² = (4+s)²
h₂ ²+1² = 8²
h₂ ² = 63
h₂ = 3√7
volume of frustum= ⅓ b₂h₂ - ⅓ b₁h₁
= ⅓ π(3√7) - ⅓ (π/4)(3/2)√7
= π√7 - (π/8)√7
= (7π/8)√7
= √7π/8
