[tex]\text{Sprawdzmy wspolczynniki kierunkowe prostych AB i CD}[/tex]
[tex]\left \{ {{2=8a+b} \atop {2=a+b/*(-1)}} \right. \\\\\left \{ {{2=8a+b} \atop {-2=-a-b}} \right. \\2-2=8a-a\\0=7a /:7\\a=0 - \text{funkcja stala}\\\\\left \{ {{9=2a+b /*(-1)} \atop {9=5a+b}} \right. \\\left \{ {{-9=-2a-b} \atop {9=5a+b}} \right. \\-9+9=-2a+5a\\0=3a/:3\\0=a - \text{funkcja stala}[/tex]
[tex]\text{Te proste sa podstawami, poniewaz sa rownolegle}[/tex]
[tex]a=|AB|=\sqrt{(8-1)^2+(2-2)^2}=\sqrt{7^2}=7\\b=|CD|=\sqrt{(5-2)^2+(9-9)^2}=\sqrt{3^2}=3\\\text{Cala figura znajduje sie powyzej osi OX, wiec: } \\h=|y_a-y_c|=|2-9|=|-7|=7\\\\P=\frac{(7+3)*7}{2}=\frac{10*7}2=5*7=35j^2\\\underline{P=35j^2}[/tex]
[tex]b=|AC|=\sqrt{(2-1)^2+(9-2)^2}=\sqrt{1^2+7^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt2\\c=|BD|=\sqrt{(5-8)^2+(9-2)^2}=\sqrt{(-3)^2+7^2}=\sqrt{9+49}=\sqrt{58}[/tex]
[tex]Ob=a+b+c+d\\Ob=3+7+5\sqrt2+\sqrt{58}\\\\\underline{Ob=10+5\sqrt2+\sqrt{58}}[/tex]
