\(x-the\ length\ \ \ \wedge\ \ \ x>0\\y-the\ width\ \ \ \Rightarrow\ \ \ \ y= \frac{4}{5} x\\ \\A=x\cdot y=x\cdot \frac{4}{5} x= \frac{4}{5} x^2\ \ \ \wedge\ \ \ A=320\ m^2\\ \\ \frac{4}{5} x^2=320\ /\cdot \frac{5}{4} \\ \\x^2= \frac{320\cdot5}{4} \ \ \ \Rightarrow\ \ \ x^2=400\ \ \ \Rightarrow\ \ \ x=20\ \ \ \Rightarrow\ \ \ y=\frac{4}{5}\cdot20=16\\ \\Ans.\ The\ length=20\ m\ \ \ and\ \ \ the\ width=16\ m\)
