[tex]krawedz\ podstawy:\ a=4\\krawedz\ boczna:\ b=2\sqrt{10}\\wysokosc\ trojkata:\ h=?\\\\h^2+(\frac{a}{2})^2=b^2\\\\h^2=b^2-(\frac{a}{2})^2\\\\h^2=(2\sqrt{10})^2-2^2\\\\h^2=40-4\\\\h^2=36\\\\h=\sqrt{36}\\\\h=6[/tex]
[tex]pole\ trojkata:\\\\P=\frac{1}{2}a*h\\\\P=\frac{1}{\not{2}^1}*\not{4}^2*6=12\ \ [j^2]\\\\obwod\ trojkata:\\\\O=a+2b\\\\O=4+2*2\sqrt{10}=4+4\sqrt{10}=4(1+\sqrt{10})[/tex]
