Answer
$$360\ m, \ 80\bar{0}0\ m^2$$
Work Step by Step
The perimeter of the given quadrilateral is
$$sum\ of \ sides=2(79.2+101)=2(18)=180.2=360.4\ m\approx 360~m$$
The area is given by
$$length \times width=101\times 79.2=7999.2\ m^2\approx 80\bar{0}0~m^2$$
