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Pp - pole podstawy stożka
Pb - pole powierzchni bocznej stożka
O - obwód przekroju osiowego stożka
Pp = πr²
Pb = πrl
O = 2r + 2l = 30 => r + l = 15
πrl = 4πr² => 4r²-rl = 0 => r(4r-l) = 0
r = 0 ∨ l = 4r
sprzeczność r = 3 ∧ l = 12
V = 1/3 πr²H
H² = l² - r² = 144 - 9 = 135
H = √135 = 3√15
V = 1/3π * 9 * 3√15 = 9π√15
odp: V = 9π√15