Odpowiedzi

2009-12-07T13:12:48+01:00
A) dl. r podst. stożka
l² = r² + h²
r² = l² - h²
l = 5cm
h = 4cm
r² = 5² - 4²
r² = 25 - 16
r² = 9
r = 3
b) pole przekroju osiowego
Ppo = r*h
Ppo =3*4
Ppo = 12cm²
c) V stożka
V = ⅓Pp*h
Pp = πr²
Pp = π*3²
Pp = 9πcm²
V = ⅓*9π*4
V = 12πcm³
d) Pb
Pb = πrl
Pb = π*3*5
Pb = 15πcm²
e)PP
Pp = πr²
Pp = π*3²
Pp = 9πcm²
f)PC
Pc = Pp + Pb
Pc = 9π + 15π
Pc = 24πcm²