Explicație pas cu pas:
vezi rezolvarea in foto anexa
[tex]\it \mathcal{A}_{ABC}=\dfrac{1}{2}\cdot\mathcal{A}_{ABMC}=\dfrac{1}{2}\cdot200=100\ cm^2\ \ \ \ \ \ (1)\\ \\ \\ \mathcal{A}_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sin(\widehat{BAC}) =\dfrac{1}{2}\cdot10\cdot AC\cdot sin45^o=5\cdot AC\cdot\dfrac{\sqrt2}{2}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2)\ \Rightarrow\ 5\cdot AC\cdot\dfrac{\sqrt2}{2}=100|_{\cdot2} \Rightarrow 5\sqrt2\cdot AC=200|_{:5} \Rightarrow \sqrt2\cdot AC = 40\Rightarrow\\ \\ \Rightarrow AC=\dfrac{^{\sqrt2)}40}{\ \sqrt2}=\dfrac{40\sqrt2}{2}=20\sqrt2[/tex]
