1. Oblicz pole powierzchni całkowitej i objętość stożka o promieniu podstawy 0,4 dm i wysokości 6 cm.
0,4dm = 4cm
r = 4cm
H = 6cm
l = √52
Pc = πr²+πrl
r²+H² = l²
6²+4² = l²
36+16 = l²
l² = 52 |√
Pc = π*6² + π*6*√52
Pc = 36π + 6√52
V = ⅓PpH
V = ⅓πr²H
V = ⅓π*6²*4
V = ⅓π*36*4
V = 12π*4
V = 48π
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0,4dm = 4cm
r = 4cm
H = 6cm
l = √52
Pc = πr²+πrl
r²+H² = l²
6²+4² = l²
36+16 = l²
l² = 52 |√
l = √52
Pc = π*6² + π*6*√52
Pc = 36π + 6√52
V = ⅓PpH
V = ⅓πr²H
V = ⅓π*6²*4
V = ⅓π*36*4
V = 12π*4
V = 48π