Odpowiedź:
Wzór na objętość stożka to:
V=1/3 π[tex]r^{2}[/tex] *H
Wysokość czyli h obliczamy z twierdzenia Pitagorasa:
[tex]H^{2}[/tex] +[tex]3^{2}[/tex] =[tex]12^{2}[/tex]
[tex]H^{2}[/tex] =144-9
[tex]H^{2}[/tex]=135
H= [tex]\sqrt{135}[/tex] = [tex]\sqrt{9*15} =3\sqrt{15}[/tex] cm
V= 1/3 [tex]\pi[/tex][tex]3^{2} *3\sqrt{15}[/tex]
v= [tex]9\sqrt{15}[/tex] [tex]cm^{3}[/tex]
Pc= [tex]\pi r^{2} + \pi rl[/tex]
Pc=[tex]\pi 3^{2} + \pi *3*12[/tex]
Pc= 9[tex]\pi + 36\pi =45\pi cm^{2}[/tex]
Szczegółowe wyjaśnienie:
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Odpowiedź:
Wzór na objętość stożka to:
V=1/3 π[tex]r^{2}[/tex] *H
Wysokość czyli h obliczamy z twierdzenia Pitagorasa:
[tex]H^{2}[/tex] +[tex]3^{2}[/tex] =[tex]12^{2}[/tex]
[tex]H^{2}[/tex] =144-9
[tex]H^{2}[/tex]=135
H= [tex]\sqrt{135}[/tex] = [tex]\sqrt{9*15} =3\sqrt{15}[/tex] cm
V= 1/3 [tex]\pi[/tex][tex]3^{2} *3\sqrt{15}[/tex]
v= [tex]9\sqrt{15}[/tex] [tex]cm^{3}[/tex]
Pc= [tex]\pi r^{2} + \pi rl[/tex]
Pc=[tex]\pi 3^{2} + \pi *3*12[/tex]
Pc= 9[tex]\pi + 36\pi =45\pi cm^{2}[/tex]
Szczegółowe wyjaśnienie: