[tex]wzor\ na\ pole\ rownolegloboku :\\\\P=a*h\\\\a- dlugosc\ podstawy\\\\h- wysokosc\ rownolegloboku \\\\dane:\\\\a=\frac{7}{10}\ dm\\\\h=1\frac{2}{3}\ dm\\\\P=\frac{7}{10}*1\frac{2}{3}=\frac{7}{\not{10}^2}*\frac{\not{5}^1}{3}=\frac{7}{6}=1\frac{1}{6}\ dm^2\\\\odp.\ Pole\ tego\ rownolegloboku\ wynosi\ 1\frac{1}{6}\ dm^2.[/tex]
