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r= 4cm
h = 3cm
l - tworząca stożka
V = ¹/₃* π * r² * H = ¹/₃* π * 4*4*3 = 16πcm³
Pc = πr(r+l)
h²+r²=l²
3² + 4² = l²
l² = 9+16
l² = 25
l = √25 = 5
Pc = πr(r+l) = 4π(4+5) = 4π*9 = 36πcm²
Odp. objetosc wynosi 16πcm³, a pole calkowite 36πcm²
Zad. 2
r=3cm
h=5cm
V = ¹/₃* π * r² * H = ¹/₃* π * 3*3*5 = 15πcm³
1ml = 1cm³
15cm³ = 15ml
Odp. zmiesci sie 15 ml.