[tex3]\mathsf{x = AM ~e ~y = BC\\
\angle ABC = 90º \\
T. Pit:~\triangle ABC\\
(2x)^2+y^2=(2r)^2 \\
4x^2+y^2=3x^2+(x^2+y^2)=48(I)\\
T,PIt~\triangle BMC: x^2+y^2 = 21(II)\\
\\
De(I) e(II) x = 3\\
y = 2\sqrt{3}\\
\alpha = \angle BAC\\
\triangle AOB (isósceles) ~e~ \triangle OBC(equilátero) \implies \\
\angle ABC = 3\alpha\ \therefore \alpha=30º\\
S_{MBC}=Área~ semicircunferência-S_{segmentoAOB} - S_\triangle{AMC}} [/tex3]
[tex3]\mathsf{Área\ semicircunferência=6\pi\\
A=S_{setor{AOB}} - S_{\triangle{AOB}}\\
S_{setor{AOB}}=4\pi (regra ~de~ três)\\
S_{\triangle{AOB}}\frac{6\cdot2\sqrt{3}\sen{30º}}{2}=3\sqrt{3}
\\ A = 4\pi-3\sqrt{3}
\\ S_{\triangle{AMC}}={3\sen30º\cdot4\sqrt{3}\over2}=3\sqrt{3}
\\S_{MBC} = 6\pi-4\pi+3\sqrt{3}-3\sqrt{3}=\boxed{\color{red}2\pi}}[/tex3]
(Soluçao:null-
viewtopic.php?f=4&t=89404&p=246408&hili ... ea#p246408)