Estude! Com Prof. Caju

Na figura, se AM = 5 e MC = 4.
Calcular R.
A) 3,75m
B) 3,76m
C) 4,75m
D) 2,75m
E) 1,75m

[tex3]\triangle OAD_{(isósceles)}\\.
\therefore \angle CDM = 180^o−(90^o+\angle ODA)=90^o−\angle OAD= \angle DAO \implies \triangle DCA \sim \triangle MCD\\
\frac{AC}{CD}=\frac{CD}{CM}⇒CD^2=AC⋅CM=36⇒CD=6\\
\triangle OAC: (R+CD)^2=R^2+AC^2⇒R=\frac{AC^2−CD^2}{2⋅CD}=\frac{81−36}{12}=\boxed{\frac{15}{4}}[/tex3]
(Solução:LuisFuentes)