Estude! Com Prof. Caju

A area da região hachurada na figura acima é igual a:

A) [tex3]\frac{7\pi }{8}[/tex3]
B) [tex3]\frac{7\pi }{6}[/tex3]
C) [tex3]\frac{6\pi }{7}[/tex3]
D) [tex3]\frac{5\pi }{8}[/tex3]
E) [tex3]\frac{5\pi }{16}[/tex3]

JohnnyEN,

Calcule o ângulo entre as retas r e s

Fórmula do ângulo agudo entre 2 retas:

[tex3]tg\alpha =\left |\frac{m_s-m_r}{1+m_r.m_s} \right |=\\

\frac{1-(-\frac{1}{\sqrt{3}})}{1+1(-\frac{1}{\sqrt{3}})}=\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\\
\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\sqrt{3}+1(\sqrt{3}+1)}{2}=\frac{3+2\sqrt{3}+1}{2}=2+\sqrt{3}\\
(*tg75^o= tg(45^o + 30^o)\\
\therefore tg\alpha = 75^o \rightarrow \alpha = \frac{5\pi }{12}\\\
\\O\ ângulo\ entre\ as\ retas = \pi -\frac{5\pi }{12}=\frac{7\pi }{12}\\
S = S_{setor1}-S_{setor2}(*S_{setor}=\frac{\alpha r^2}{2})=\frac{7\pi }{12}.\frac{2^2}{2}-\frac{7\pi}{12}.\frac{1^2}{2}=\frac{7\pi }{6}-\frac{7\pi }{24}\rightarrow \\
S = \frac{21\pi }{24}=\boxed{\color{red}\frac{7\pi}{8}}
[/tex3]