Question

Ajutor incerc sa o fac de o ora​

Answer 1

$\it b) \\ \\ \ell_3=6\sqrt3 \Rightarrow \begin{cases}\it R=\dfrac{2}{3}\cdot\dfrac{\ell \sqrt3}{2}=\dfrac{6\sqrt3\cdot\sqrt3}{3}=6\\ \\ \it a_3=\dfrac{R}{2}=\dfrac{6}{2}=3\\ \\ \it h_3=R+a_3=6+3=9\\ \\ \it \mathcal{A}_3 =\dfrac{\ell\cdot h}{2}=\dfrac{6\sqrt3\cdot9}{2}=\dfrac{54\sqrt3}{2}=27\sqrt3 \end{cases}$

Answer 2

$\it \mathcal{A}=108\sqrt3\ \ \ \ \ (1)\\ \\ \mathcal{A}=\dfrac{\ell^2\sqrt3}{4}\ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow \dfrac{\ell^2\sqrt3}{4}=108\sqrt3 \Rightarrow \ell^2\sqrt3=4\cdot108\sqrt3|_{:\sqrt3}\Rightarrow \ell^2=4\cdot108 \Rightarrow\\ \\ \\ \Rightarrow \ell^2=4\cdot36\cdot3=2^2\cdot6^2\cdot3 \Rightarrow \ell=2\cdot6\cdot\sqrt3=12\sqrt3\\ \\ R=\dfrac{\ell\sqrt3}{3}=\dfrac{12\sqrt3\cdot\sqrt3}{3}=\dfrac{12\cdot3}{3}=12\\ \\ a_3=\dfrac{R}{2}=\dfrac{12}{2}=6\\ \\ h=R+a=12+6=18$