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l=6
r=½l=½*6=3
Pc=π*3²+π*3*6
Pc=9π+18π
Pc=27π
l=6 cm
r=½l
r=3 cm
Pc=π3²+π3×6
Pc=9π+18π
Pc=27π cm²
Odp.Pole powierzchni całkowitej stożka wynosi 27πcm².