triunghiul echilateral l=9cm R=? apotema indice 3 =? A=?​

[tex]\it R=\dfrac{\ell\sqrt3}{3}=\dfrac{9\sqrt3}{3}=3\sqrt3\ cm\\ \\ \\ a_3=\dfrac{R}{2}=\dfrac{3\sqrt3}{2}\ cm\\ \\ \\ \mathcal{A}=\dfrac{\ell^2\sqrt3}{4}=\dfrac{9^2\sqrt3}{4}=\dfrac{81\sqrt3}{4}\ cm^2[/tex]

Răspuns:

Raza cercului circumscris

[tex]R = \dfrac{2}{3} \cdot h = \dfrac{2}{3} \cdot \dfrac{\ell\sqrt{3} }{2} = \dfrac{9\sqrt{3} }{3} = 3\sqrt{3} \ cm[/tex]

Raza cercului înscris:

[tex]r = \dfrac{1}{3} \cdot h = \dfrac{1}{3} \cdot \dfrac{\ell\sqrt{3} }{2} = \dfrac{9\sqrt{3} }{6} = \dfrac{3\sqrt{3} }{2} \ cm[/tex]

Aria triunghiului:

[tex]A = \dfrac{\ell^2\sqrt{3} }{4} = \dfrac{9^2\sqrt{3} }{4} = \dfrac{81\sqrt{3} }{4} \ cm[/tex]