The perimeter of a rectangle is 48 m. If the width were doubled and the length were increased by 24 m, the perimeter would be 112 m. What is the length of the original rectangle?

Question

anonymous · Answer

let w be width and l be length

2w+2l=48
      2l=48-2w

4w+2(l+24)=112
4w+2l+48=112
4w+48-2w=64
            2w=16
             w=8

2l=48-16
   =32
 l=16

anonymous · Answer

let l=length
    w=width

2(l+w)=48
   l+w=24
        l=-w+24   (1)

2(l+w)=112
2[(l+24)+2w]=112
2l+48+4w=112   (2)

sustitute the value of length

2(-w+24)+48+4w=112
-2w+48+48+4w  =112
             2w+96    =112
                   2w     =112-96
                     2w   =16
                       w= 8
 
                       l=16