Question

04) Um cone circular reto está inscrito em uma esfera. As geratrizes do cone medem cm 10 e o ângulo , formado entre uma geratriz qualquer e a altura do cone, é tal que cosseno do ângulo e 0,9 .



a) Calcule o volume do cone.
b) Calcule o volume da esfera.

Answer 1

$\mathbf{Calculando\ a\ altura\ do\ cone\ (h):}\\\\ \mathrm{\cos{\alpha}=\dfrac{cat.adj.}{hip.}=\dfrac{h}{g}\ \to\ 0,9=\dfrac{h}{10}\ \to\ \boxed{\mathrm{h=9\ cm}}}\\\\\\ \mathbf{Calculando\ o\ raio\ da\ base\ do\ cone\ (r):}\\\\ \mathrm{g^2=h^2+r^2\ \to\ r^2=g^2-h^2\ \to\ r^2=10^2-9^2}\\ \mathrm{r^2=100-81\ \to\ r^2=19\ \to\ \boxed{\mathrm{r=\sqrt{19}\ cm}}}$

$\mathbf{Calculando\ o\ raio\ da\ esfera\ (R):}\\\\ \mathrm{R^2=r^2+(h-R)^2\ \to\ R^2=r^2+h^2-2hR+R^2}\\ \mathrm{2hR=r^2+h^2\ \to\ R=\dfrac{r^2+h^2}{2h}\ \to\ R=\dfrac{g^2}{2h}}\\ \mathrm{R=\dfrac{10^2}{2.9}=\dfrac{100}{18}=\dfrac{50}{9}\ \to\ \boxed{\mathrm{R=\dfrac{50}{9}\ cm}}}\\\\\\ \mathbf{Calculando\ o\ volume\ do\ cone\ (V'):}\\\\ \mathrm{V'=\dfrac{\pi r^2h}{3}=\dfrac{\pi.(\sqrt{19})^2.9}{3}=\pi.19.3\ \to\ \boxed{\mathrm{V'=57\pi\ cm^3}}}$

$\mathbf{Calculando\ o\ volume\ da\ esfera\ (V):}\\\\ \mathrm{V=\dfrac{4\pi R^3}{3}=\dfrac{4\pi(\frac{5.10}{9})^3}{3}=\dfrac{4\pi.125.1000}{729}.\dfrac{1}{3}=\dfrac{5.10^5.\pi}{2187}=}\\\\ \mathrm{=\dfrac{500000\pi}{2187}\ \to\ \boxed{\mathrm{V\approx228,624\pi\ cm^3}}}$