Question

Question #06a16

Answer 1

see explanation.

Explanation:

see Fig 1,
given that height `h` of cone `=20` cm and slant height `h_s=25` cm,
let `r_c` be the radius of the base of the cone,
`r_c=sqrt(AC^2-AD^2)=sqrt(25^2-20^2)=15` cm
See Fig 2,
the largest possible hemisphere inside the cone touches the cone at `E`, so `AB` is tangent to the hemisphere at `E`,
`=> DeltaAED and DeltaADB` are similar,
`=> (ED)/(AD)=(DB)/(AB)`
`=>` radius of hemisphere `r_h=ED=(20xx15)/25=12` cm

Volume of the cone `V_c=1/3*pi*r_c^2*h`
`=1/3*pi*15^2*20=1500pi " cm"^3`
Volume of hemisphere `V_h=2/3*pi*r_h^3`
`=2/3*pi*12^3=1152pi " cm"^3`
Volume of the remaining portion `=V_c-V_h=1500pi-1152pi=348pi " cm"^3`