[tex]a)\\\\podstawa\ trojkata:\ a=3\sqrt{2}+5\\wysokosc\ trojkata:\ h=\sqrt{7}\\\\pole\ trojkata:\\\\P=\frac{1}{2}*(3\sqrt{2}+5)*\sqrt{7}= \frac{1}{2}(3\sqrt{14}+5\sqrt{7})\ [j^2]\\\\obwod\ trojkata:\\\\L=3\sqrt{2}+5+5+\sqrt{32}=3\sqrt{2}+10+\sqrt{16*2}=3\sqrt{2}+10+4\sqrt{ 2}=7\sqrt{2}+10[/tex]
