1) objetosc walca jest rowna 72 pi cm3. oblicz srednice jego podstawy gdy wysokosc walca jest rowna 8 oraz pole pow całkowitej tego walca
2) przekroj osiowy walca jest prostokatem o polu rownym 48 i przekatnej majacej dł 10. oblicz objetosc tego walca
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z.1
V = 72 pi cm^3
h = 8 cm
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V = Pp*h = pi*r^2 * h
czyli po podstawieniu
pi *r^2 * 8 = 72 pi / : 8 pi
r^2 = 9
r = 3
====
d = 2*r = 2 * 3 = 6
=================
Pc = 2 Pp + Pb = 2 pi*r^2 + 2 pi*r *h
Pc = 2 pi*3^2 + 2 pi *3*8 = 18 pi + 48 pi = 66 pi
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Odp. Średnica ma długość 6 cm, a pole Pc = 66 pi cm^2
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z.2
P = 48
c = 10
Mmay
P = 2r*h = 48 => 2 r = 48 / h
oraz
(2r)^2 + h^2 = c^2
( 48/ h)^2 + h^2 = 10^2
2304/ h^2 + h^2 = 100 / * h^2
2 304 + h^4 = 100 h^2
h^4 - 100 h^2 + 2 304 = 0
t = h^2
t^2 - 100 t + 2 304 = 0
delta = ( -100)^2 - 4*1* 2304 = 10 000 - 9 216 = 784
p( delty) = 28
t = ( 100 - 28)/2 = 36 lub t = ( 100 + 28)/2 = 64
czyli
h^2 = 36
h = 6
====
h^2 = 64
h = 8
=====
zatem
r = 24 / 6 = 4 lub r = 24/8 = 3
===========================
1)
Dla
r = 3 i h = 8
V = pi*3^2 * 8 = 72 pi
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2) Dla
r = 4 i h = 6
V = pi *4^2 *6 = 96 pi
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