Question

Area is 320 square feet width is 4/5 length

Answer 1

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