Question 915319
Outer length of the rectangular field = 38 m
Outer breadth of the rectangular field = 32 m

Outer Area:{{{A[o] = 38m* 32m = 1216 m^2}}}
Let the width of the path be '{{{w}}}' meters

Inner length of the rectangular field:{{{L[I] = (38- 2w)m}}}
Inner breadth of the rectangular field:{{{W[I] = (32-2w)m}}}

Inner area {{{A[I]= (38 -2w)(32-2w) =1216-76w-64w+4w^2 =4w^2-140w + 1216}}}

Area of the path:{{{A[p] = A[o] -A[I] = 1216-(4w^2 - 140w + 1216)= 600}}}


{{{1216-(4w^2-140w + 1216)= 600}}}

{{{1216-(4w^2 -140w + 1216)- 600=0}}}

{{{616 -4w^2 + 140w -1216=0}}}

 {{{-4w^2 + 140w -600=0}}} 

{{{4w^2 -140w +600=0}}}

{{{4(w^2 -35w +150)=0}}}

{{{4(w^2-30w-5w+150)=0}}}

{{{4((w^2-30w)-(5w-150))=0}}}

{{{4(w(w-30)-5(w-30))=0}}}

{{{4(w-5)(w-30)=0}}}

solution


if {{{w-5=0}}}=>{{{w=5}}}

if {{{w-30=0}}}=>{{{w=30}}}

Since the outer breadth of the field is {{{32m}}}, breadth of the rectangular field is {{{5m}}}.