A footpath of uniform width runs all around the inside of rectangular field $50 m$ long and $38 m$ wide.If the area of the path is $492 m^{2}$ ,find its width ?

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Solution

Let the width of footpath $= x$
total area of rectangle $A B C D$ $=$ area of footpath $+$ area of $E F G H$
$\Rightarrow 50 \times 38 = 492 + [(50 - 2 x) \times (38 - 2 x)] \Rightarrow 1900 = 492 + [1900 - 100 x - 176 x + 4 x^{2}] \Rightarrow 4 x^{2} - 176 x + 492 = 0 \Rightarrow x^{2} - 44 x + 123 = 0 \Rightarrow (x - 3) (x - 41) = 0 ∴ x = 3 m$

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A footpath of uniform width runs all around the inside of rectangular field 50m long and 38m wide.If the area of the path is 492m2,find its width ?

Let the width of footpath=xtotal area of rectangle ABCD = area of footpath + area of EFGH⇒50×38=492+[(50−2x)×(38−2x)]⇒1900=492+[1900−100x−176x+4x2]⇒4x2−176x+492=0⇒x2−44x+123=0⇒(x−3)(x−41)=0∴x=3m

A footpath of uniform width runs all around the inside of rectangular field $50 m$ long and $38 m$ wide.If the area of the path is $492 m^{2}$ ,find its width ?