Question #06a16

1 Answer
CW
Oct 29, 2017

see explanation.

Explanation:

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see Fig 1,
given that height hh of cone =20=20 cm and slant height h_s=25hs=25 cm,
let r_crc be the radius of the base of the cone,
r_c=sqrt(AC^2-AD^2)=sqrt(25^2-20^2)=15rc=√AC2−AD2=√252−202=15 cm
See Fig 2,
the largest possible hemisphere inside the cone touches the cone at EE, so ABAB is tangent to the hemisphere at EE,
=> 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

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