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V= 1/3πr²H
II stożek 2r, 1/2H
V₁= 1/3 π(2r)² ·1/2H=1/3π·4r²·1/2H= 2/3πr²H
V₁/V= ( 2/3πr²H): (1/3πr²H)= 2/3 : 1/3= 2/3·3/1= 2
Objętość stożka wzrośnie 2 razy.