[tex]\it Latura\ triunghiului \ este\ \ \ell = \dfrac{12}{3}=4\ cm\\ \\ \\ \mathcal{A}=\dfrac{\ell^2\sqrt3}{4}=\dfrac{4^2\sqrt3}{4}=\dfrac{4\cdot4\sqrt3^{(4}}{4}=4\sqrt3\ cm^2\\ \\ R=\dfrac{2}{3}\cdot h=\dfrac{2}{3}\cdot\dfrac{\ell\sqrt3}{2}=\dfrac{\ell\sqrt3}{3}=\dfrac{4\sqrt3}{3}[/tex]
[tex]\it r=a_3=\dfrac{1}{2}\cdotR=\dfrac{1}{2}\cdot\dfrac{4\sqrt3}{3}=\dfrac{2\sqrt3}{3}[/tex]
