Question

Ajutor ! Coroana ! ex : 6,

Answer 1

    
[tex]\displaystyle \\ 6) \\ \texttt{Se da: }\\ A_{\Delta echilaterat} = 9 \sqrt{3} ~cm^2 \\ \\ \texttt{Se cere: } \\ a) ~l = ? \\ b)~ R = ? \\ c) ~ r = ? \\ \\ \texttt{Rezolvare: } \\ \\ a) \\ A_{\Delta echilaterat} = \frac{l^2 \sqrt{3}}{4} = 9 \sqrt{3} ~cm^2 \\ \\ \Longrightarrow ~~l^2 = \frac{4 \times 9 \sqrt{3}}{\sqrt{3}} =4 \times 9 =36 \\ \\ l = \sqrt{36} = \boxed{6~cm} \\\\\\ b) \\ \\ h = \frac{ l\sqrt{3} }{2} = \frac{ 6\sqrt{3} }{2} = \boxed{3\sqrt{3} ~cm}[/tex]


[tex]\displaystyle \\ \\ R = \frac{2}{3}\times h = \frac{2\times 3\sqrt{3} }{3}= \boxed{2\sqrt{3} ~cm } \\ \\ c) \\ \\ r = a_3 = \frac{1}{3}\times h = \frac{1\times 3\sqrt{3} }{3}= \boxed{\sqrt{3} ~cm } [/tex]