[tex]Zadanie\ 10\\\\a=\mid DE\mid=5\\h=\mid AD\mid=12\\\\P _\bigtriangleup= \frac{1}{2}\cdot a\cdot h\\\\P_{AED}=\frac{1}{2}\cdot 5\cdot {12} = 6\cdot5=30\\\\Pole\ trojkata\ AED\ jest\ rowne\ 30-odpowiedz\ A\\\\c=\mid AE\mid=?\\\\c^{2}=a^{2}+h^{2}\\\\ c^{2}=5^{2}+12^{2}\\\\ c^{2}=25+144\\\\ c^{2}=169\\\\ c=\sqrt{169}=13\\\\Ob._{ABCE}=\mid AB\mid+\mid BC\mid+\mid CE\mid+ \mid AE\mid \\\\ \mid CE\mid=20-5=15\\\\ Ob._{ABCE}=20+12+15+13=60\\\\Obwod\ czworokata\ ABCE\ jest\ rowny\ 60-odpowiedz\ D[/tex]