[tex]\large\text{$A) ~A ~\adcute{a}rea ~da ~base ~do ~cone ~ \Rightarrow ~ Ab = 225 ~\pi ~dm^2$}\\\\\large\text{$B) ~ A ~\adute{a}rea ~lateral ~do ~cone ~ \Rightarrow ~ Al = 600 ~\pi ~dm^2$}\\\\\large\text{$C) ~ O ~volume ~do ~cone ~ \Rightarrow ~ V = 1500 ~\pi ~dm^3$}[/tex]
[tex]\large\text{$S\acute{o}lidos ~Geom\acute{e}tricos $}[/tex]
Encontrar o raio da base do cone.
usar o Teorema de Pitágoras
[tex]g^2 = h^2 + r^2\\\\25^2 = 20^2 + r^2\\\\625 = 400 + r^2\\\\r^2 = 625 - 400\\\\r^2 = 225\\\\r = \sqrt{225} \\\\r = 15 ~dm[/tex]
A) a área da base do cone:
[tex]Ab = r^2 ~. ~\pi\\\\Ab = 15^2 ~\pi\\\\Ab = 225 ~\pi ~dm^2[/tex]
B) A área lateral do cone:
[tex]Al = 2\pi ~.~r ~.~ h\\\\Al = 2\pi ~.~15 ~.~ 20\\\\Al = 2\pi ~.~300\\\\Al = 600 ~\pi ~dm^2[/tex]
C) O volume do ~cone
[tex]V = \dfrac{1}{3} ~\pi~. ~r^2~. ~h\\\\\\V = \dfrac{1}{3} ~\pi~. ~15^2~. ~20\\\\\\V = \dfrac{1}{3} ~\pi~.~225 ~. ~20\\\\\\V = \dfrac{1}{3} ~\pi~.~4500\\\\\\V = \dfrac{1}{3} ~\pi~.~4500\\\\\\V = \dfrac{4500}{3} ~\pi\\\\\\V = 1500 ~\pi ~dm^3[/tex]
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