[tex]podstawa\ trojkata\ rownoramiennego\ :\ a=6\ \ cm\\pole\ trojkata:\ \ \ P=12\ cm^2\\wysokosc\ trojkata:\ h=?\\ramie\ trojkata:\ b=?\\\\P=\frac{1}{2}ah\\\\\frac{1}{2}*6*h=12\\\\3h=12\ \ |:3\\\\h=4\ cm[/tex]
[tex]stosujac\ tw.\ Pitagorasa\ obliczymy\ ramie\ trojkata:\\\\ b^2=(\frac{a}{2})^2+h^2\\\\b^2=3^2+4^2\\\\b^2=9+16\\\\b^2=25\\\\b=\sqrt{25}\\\\b=5\ cm\\\\obwod:\\\\L=a+2b=6+2*5=6+10=16\ cm[/tex]
