the footpath of uniform width runs all around inside of a rectangular field 45m long and 36m wide if the area of the path is 234m find width
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ASsuming width of path to be 'x'm
Therefore, exclusive of the footpath:
=> length of field = 45-x-x = (45-2x)m
=> width of field = 36-x-x = (36-2x)m
Total area (exclusive of path)
= (45-2x)(36-2x)
= (1620-90x-72x+4x^2)
= (4x^2-162x+1620) sq.m
Total area (inclusive of path)
= (45)(36)
= 1620 sq.m
Area of footpath = 234m
Therefore,
Area of footpath = Area exclusive of path - Area inclusive of path
=> 234 = 1620-(4x^2-162x+1620)
=> 234 = -(4x^2-162x)
=> -234 = 4x^2-162x (Multiplying equation with -1)
=> 0 = 4x^2-162x+234
=> 0 = 2x^2-81x+117 (Dividing equation by 2)
=> 0 = 2x^2-78x-3x+117
=> 0 = 2x(x-39)-3(x-39)
=> 0 = (2x-3)(x-39)
=> 2x-3 = 0 or x-39 = 0
=> x = 1.5m or x = 39m
Discarding x = 39m, because that's longer than the breadth of the field.
Therefore, width of path = 1.5m
Therefore, exclusive of the footpath:
=> length of field = 45-x-x = (45-2x)m
=> width of field = 36-x-x = (36-2x)m
Total area (exclusive of path)
= (45-2x)(36-2x)
= (1620-90x-72x+4x^2)
= (4x^2-162x+1620) sq.m
Total area (inclusive of path)
= (45)(36)
= 1620 sq.m
Area of footpath = 234m
Therefore,
Area of footpath = Area exclusive of path - Area inclusive of path
=> 234 = 1620-(4x^2-162x+1620)
=> 234 = -(4x^2-162x)
=> -234 = 4x^2-162x (Multiplying equation with -1)
=> 0 = 4x^2-162x+234
=> 0 = 2x^2-81x+117 (Dividing equation by 2)
=> 0 = 2x^2-78x-3x+117
=> 0 = 2x(x-39)-3(x-39)
=> 0 = (2x-3)(x-39)
=> 2x-3 = 0 or x-39 = 0
=> x = 1.5m or x = 39m
Discarding x = 39m, because that's longer than the breadth of the field.
Therefore, width of path = 1.5m
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