Răspuns:
a) ΔAFC ~ ΔABE (U.U.U.);
b) 24√3 cm²;
c) 192π cm²
[tex]a) AE \: diametru => ∠A \: B \: E = 90° \\ AF ⊥ BC => ∠A \: F \: C = 90° \\ ∠A \: E \: B = ∠A \: C \: F = 60° \\ (subintind \: acelasi \: arc \: AB) \\ ∠E \: A \: B = ∠C \: A \: F = 30° \\=> ΔA \: F \: C ... \: ΔA \: B \: E \: (U.U.U.) \\ \\ b) ducem \: OM ⊥ AB \: si \: avem \\ AM = MB = \frac{AB}{2} = 12 cm \\ tg(∠O \: A \: M) = tg(30°) = \frac{OM}{AM} \\ = > \: \ \frac{ \sqrt{3} }{3} = \frac{OM}{12} \\ OM = \frac{12 \sqrt{3} }{3} = 4√3 cm \\ aria(ΔA \: O \: B) = \frac{OM×AB}{2} \\ = \frac{4 \sqrt{3} \times 12 }{2} \\ aria(ΔA \: O \: B) = 24 \sqrt{3} cm² \\ \\ c) sin(∠O \: A \: M) = sin(30°) = \frac{OM}{AO} \\ \frac{1}{2} = \frac{4 \sqrt{3} }{AO} \\ AO = 8 \sqrt{3} cm \\ aria \: cercului = \pi×r² = \pi×AO² \\ = \pi×(8 \sqrt{3} )² = 192\pi cm²[/tex]
