a footpath of uniform width runs round the inside of a rectangular field 32 m long and 24 m wide . if the path occupies 208 m^2 , find the width of the footpath
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area of field = 32*24= 768m^2
area of footpath = 208m^2
area of remaining field = 768-208= 560m^2
let width of footpath be x
area of remaining part = l*b
=32-2x*24-2x=560
=768-48x-64x+4x^2=560
=768-112x+4x^2=560
=-112x+4x^2=560-768
=4x(-28+x)=-208
=x(-28+x)=-52
=-28x+x^2=-52
=(-14)^2-28x+x^2=-52+(-14)^2
=196-28x+x^2=-52+196
=(x-14)^2=144
=√(x-14)^2=√144
=x-14=12
=x=12+14
=x=26
Therefore,width of the footpath= 26m
area of footpath = 208m^2
area of remaining field = 768-208= 560m^2
let width of footpath be x
area of remaining part = l*b
=32-2x*24-2x=560
=768-48x-64x+4x^2=560
=768-112x+4x^2=560
=-112x+4x^2=560-768
=4x(-28+x)=-208
=x(-28+x)=-52
=-28x+x^2=-52
=(-14)^2-28x+x^2=-52+(-14)^2
=196-28x+x^2=-52+196
=(x-14)^2=144
=√(x-14)^2=√144
=x-14=12
=x=12+14
=x=26
Therefore,width of the footpath= 26m
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26 cannot be the answer because 24 is the breadth and width cannot be more
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