the remains of an ancient ball court include a rectangular playing alley with a perimeter of about 48 M. the length of the alley is 5 times the width. find the length and the width of the playing alley​

Answer:Length is 20 m and width is 4 m.Step-by-step explanation:Given:Perimeter of the rectangular alley, [tex]P=48\textrm{ m}[/tex]Length is 5 times the width.Let width be [tex]x[/tex].So, as per question,Length,[tex]l = 5x[/tex]Now, perimeter of rectangle is given as:[tex]P=2(l+b)[/tex]Plug in 48 for [tex]P[/tex], [tex]5x[/tex] for [tex]l[/tex] and [tex]x[/tex] for b.[tex]48=2(5x+x)\\48=2(6x)\\48=12x\\x=\frac{48}{12}=4[/tex]Therefore, width is 4 m.Length is [tex]5x=5\times 4=20[/tex] m.