Matematyka
renesmee951
2017-06-27 08:49:37
bryły obrotowe oblicz V i P zalacznik
Odpowiedź
janna1234522
2017-06-27 09:36:23

1] h=wysokosc walca R=srednica podstawy r=promień D=przekątna przekroju osiowego sin 60*=h/D √3/2=h/18 h=9√3 cos60*=R/D 1/2=R/18 R=9 r=4,5 Pp=πr²=π*4,5²=20,25π v=20,25π*h=20,25π*9√3=182,25π√3 j.³ Pb=2πrh=2π*4,5*9√3=81π√3 Pc=2*20,25π+81π√3=40,5π(1+2√3) j.² 2] h=wysokosc=10 l=tworzaca r=promień tgα=r/h r/10=3/5 r=6 10²+6²=l² l=√136=2√34 Pp=π*6²=36π v=1/3*36π*h=12π*10=120π j.³ Pb=πrl=π*6*2√34=12π√34 Pc=12π√34+36π=12π(√34+3)j.²

Piszczello22
2017-06-27 09:37:38

Walec: [latex]V=P_pcdot H=picdot r^2 cdot H[/latex] [latex]P_c = 2pi cdot{r}^2 + 2pi cdor r cdot H = 2pi r(r + H)[/latex] [latex]sin 60^{circ} = cfrac{H}{18}[/latex] [latex]cfrac{sqrt{3}}{2}=cfrac{H}{18}[/latex] [latex]2H=18sqrt{3}[/latex] [latex]H=9sqrt{3}[/latex] [latex]cos 60^{circ}=cfrac{2r}{18}[/latex] [latex]cfrac{1}{2}=cfrac{2r}{18}[/latex] [latex]4r=18[/latex] [latex]r=4,5[/latex] [latex]V=pi cdot(4,5)^2 cdot 9sqrt{3} = 182,25pisqrt{3}[/latex] [latex]P_c = 2pi cdot 4,5 cdot (4,5 + 9sqrt{3}) = 9pi(4,5+9sqrt{3})[/latex] Stożek: [latex]V = cfrac{1}{3}picdot r^2 cdot H[/latex] [latex]P_c = pi cdot r^2 + pi cdot r cdot l = pi cdot r(r + l)[/latex] [latex] analpha=cfrac{r}{10}[/latex] [latex]cfrac{3}{5}=cfrac{r}{10}[/latex] [latex]5r=30[/latex] [latex]r=6[/latex] [latex]l^2=10^2+6^2[/latex] [latex]l^2=100+36[/latex] [latex]l=sqrt{136}=2sqrt{34}[/latex] [latex]V=cfrac{1}{3}pi cdot 6^2 cdot 10 = 120pi[/latex] [latex]P_c=pi cdot 6(6 + 2sqrt{34})=36pi+12pisqrt{34}[/latex]

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