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Pb - pole wycinka kołowego = παr₁²/360 = 180π9/360 = 9π/2 cm² =
r₁ wycinka = l = tworzącej stożka
ponieważ Pb (pole powierzchni bocznej stożka) = πrl to 9π/2 = πrl
r - promień podstawy
9π/2 = πrl
9π/2 = 3πr
9π = 6πr
r = 9π/6π = 1,5
Pp - pole podstawy = πr² = π(1,5)² = 2,25π cm²
Pc - pole powierzchni całkowitej = Pb + Pp = 9π/2 + 2,25π = 4,5π + 2,25π =
= 6,75π cm²