[tex]\it \mathcal{A}_b=\dfrac{\ell^2\sqrt3}{4}=\dfrac{8^2\sqrt3}{4}=\dfrac{64\sqrt3}{4}=16\sqrt3\ cm^2\\ \\ \\ \mathcal{V}=\mathcal{A}_{b}\cdot h \Rightarrow h=\dfrac{\mathcal{V}}{\mathcal{A}_b}=\dfrac{64\sqrt3}{16\sqrt3}=4\ cm[/tex]
[tex]\it \mathcal{A}_\ell = 3\cdot \ell\cdot h=3\cdot8\cdot4=96\ cm^2\\ \\ \mathcal{A}_t=\mathcal{A}_\ell+2\cdor\mathcal{A}_b=96+2\cdot16\sqrt3=96+32\sqrt3\ cm^2[/tex]
[tex]Fie\ M- mijlocul\ laturii\ AC \Rightarrow B'M-median\breve a\ \^{i}n\ \Delta B'AC-isoscel \Rightarrow B'M\perp AC[/tex]
[tex]\it BM - median\breve a\ \^{i}n\ \Delta ABC-echilateral \Rightarrow BM\perp AC[/tex]
[tex]\it [\widehat{(B'AC),\ (ABC)}]= [\widehat{B'M,\ BM}]=\widehat{B'MB}\\ \\ BM\ -\ \^{i}n\breve a l \c{\it t}ime\ pentru\ \Delta ABC-echilateral \Rightarrow BM=\dfrac{\ell \sqrt3}{2}=\dfrac{8\sqrty3}{2}=4\sqrt3\ cm[/tex]
[tex]\it tg (\widehat{B'MB})=\dfrac{B'B}{BM}=\dfrac{^{\sqrt3)}4}{\ 4\sqrt3}=\dfrac{4\sqrt3}{4\cdot3}=\dfrac{\sqrt3}{3}\\ \\ \\ Dar,\ tg30^o=\dfrac{\sqrt3}{3},\ deci\ \widehat{B'MB} =30^o \Rightarrow [\widehat{(B'AC),\ (ABC)}]=30^o[/tex]
