[tex]\it a)\ \ AC=\ell \sqrt2=4 \Rightarrow \ell=2\sqrt2\ cm;\ \ \ \mathcal{A}=\ell^2=(2\sqrt2)^2=8\ cm^2\\ \\ \\ b)\ \ \widehat{ACD}=45^o\ \ (unghi\ exterior\ pentru\ \Delta DCE) \Rightarrow \widehat{E}=30^o \\ \\ \\ Teorema\ sinusurilor\ pentru\ \Delta DCE:[/tex]
[tex]\it \dfrac{CE}{sin15^O}=\dfrac{CD}{sin30^o} \Rightarrow \dfrac{CE}{sin15^o}=\dfrac{2\sqrt2}{\dfrac{1}{2}}\Rightarrow \dfrac{CE}{sin15^o}=4\sqrt2 \Rightarrow\\ \\ \\ \Rightarrow CE=4\sqrt2\cdot sin15^o=4\sqrt2\cdot\dfrac{\sqrt6-\sqrt2}{4}=\sqrt{12}-2=2\sqrt3-2 \\ \\ \\ CE < 1,5 \Leftrightarrow 2\sqrt3-2 < 1,5\bigg|_{+2} \Leftrightarrow 2\sqrt3 < 3,5 \Leftrightarrow (2\sqrt3)^2 < 3,5^2 \Leftrightarrow \\ \\ \\ \Leftrightarrow 12 < 12,25\ \ (A)[/tex]
