Question

Va rog cu modul de rezolvare!
Dau coroana!!​

Answer 1

$DN=BM;~AB=AD;~ AN=AM\\\implies ΔADM∼ΔABM\\\implies A_{ΔADM}=\tfrac{AD×DN}{2}=\tfrac{\ell×\tfrac{2\ell}{3}}{2}=\tfrac{\tfrac{2\ell²}{3}}{2}=\tfrac{\ell²}{3}\\\ A_{ΔABM}=A_{ΔADM}=\tfrac{\ell²}{3}\\\\\iff A_{AMCN}=A_{Pătrat}-2×A_{ΔADM}\\ A_{AMCN}=\ell²-2×\tfrac{\ell²}{3}\\ A_{AMCN}= \tfrac{3\ell²}{3}-\tfrac{2\ell²}{3}\\ \boxed{A_{AMCN}=\tfrac{\ell²}{3}}$