to jest straszne pomocy dawid: PRZEKĄTNA PRZEKROJU OSIOWEGO WALCA MA DŁ. 40 CM I TWORZY Z JEGO PODSTAWĄ KĄT α TAKI ,ŻE sinα=√32.oblicz pole powierzchni całkowitej i objętości tego walca.

Janek191: p = 40 cm

	√3		1
sin α =		⇒ α = 60o ⇒ cos α =
	2		2

r − promień walca h − wysokość walca Mamy

h
	= sin α
p

czyli

h		√3
	=		⇒ h = 20 √3
40		2

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

2r
	= cos α
p

2r		1
	=		⇒ 2r = 20 ⇒ r = 10
40		2

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Mamy h = 20 √3 cm r = 10 cm zatem P_c = 2 P_p + P_b = 2 π r² + 2π r*h P_c = 2 π *10² + 2 π *10*20 √3 = 200 π + 400 √3 π = 200 π *( 1 + 2 √3) −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− V = P_p *h = π r² * h = π* 10² * 20 √3 = 2000 √3 π −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Odp. P_c = 200 π * ( 1 + 2 √3 ) cm² V = 2000 √3 π cm³ =====================