Explicație pas cu pas:
a) r = 6 => OA ≡ OB = 6
m(∢AOB) = m(arcAB) = 120°
[tex]Aria_{sectorAOB} = \frac{\pi {r}^{2} \cdot m(\angle AOB)}{360 \degree } = \\ = \frac{\pi \cdot {6}^{2} \cdot 120 \degree}{360 \degree} = \frac{36\pi}{3} = \bf 12\pi[/tex]
b) AB² = OA² + OB² - 2×OA×OB×cos(∢AOB) = 6² + 6² - 2×6²×cos120° = 2×6²×[1 - (-sin30°)] = 72×(1 + ½) = 108
[tex]\implies \bf AB = 6 \sqrt{3} [/tex]
c) m(∢ACB) = ½×m(arcAB) = 60°
CA = CB => m(arcCA) = m(arcCB)
360° - m(arcAB) = 360° - 120° = 240°
=> m(arcCA) = m(arcCB) = 120°
m(∢ABC) = ½×m(arcAC) = 60°
m(∢BAC) = ½×m(arcBC) = 60°
=> ΔABC este echilateral
[tex]Aria_{\triangle ABC} = \frac{ {AB}^{2} \cdot \sqrt{3} }{4} = \frac{108 \cdot \sqrt{3} }{4} = \bf 27 \sqrt{3} \\ [/tex]
