Figura w zalaczniku.
[tex]|AB|=\sqrt{(1+4)^2+(-4+4)^2}=\sqrt{5^2}=5\\|BC|=\sqrt{(1-1)^2+(1+4)^2}=\sqrt{5^2}=5\\|CD|=\sqrt{(0-1)^2+(5-1)^2}=\sqrt{(-1)^2+4^1}=\sqrt{1+16}=\sqrt{17}\\|DE|=\sqrt{(-2-0)^2+(1-5)^2}=\sqrt{(-2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt5\\|EF|=\sqrt{(-4+2)^2+(1-1)^2}=\sqrt{(-2)^2}=\sqrt4=2\\|AF|=\sqrt{(-4+4)^2+(1+4)^2}=\sqrt{5^2}=5[/tex]
[tex]Ob=5+5+\sqrt{17}+2\sqrt5+2+5=17+2\sqrt5+\sqrt{17}[/tex]
[tex]\text{Aby obliczyc pole, dzielimy figure na dwie: }\\- \text{Kwadrat o boku 5}\\P_1=5^2=25j^2\\- \text{Trojkat o podstawie EC}\\P_2=\frac{3*4}2=3*2=6j^2\\\\P_f=(25+6)j^2=31j^2[/tex]
