A footpath of uniform width runs all around the outside of a rectangular field $30 m$ long and $24 m$ wide. If the path occupies $360 m^{2}$ , find its width

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Solution

Let the width of the path be $x$ .
$∴$ length of the rectangular field and path $= 2 x + 30$
and breadth $= 2 x + 24$
Area of rectangular field $= (30 \times 24)$
$= 720$
Area of the rectangular field and path $= (2 x + 30) (2 x + 24)$
Now,
$(2 x + 30) (2 x + 24) - 720 = 360$
$\Rightarrow 4 x^{2} + 48 x + 60 x + 720 - 720 = 360$
$\Rightarrow 4 x^{2} + 108 x - 360 = 0$
$\Rightarrow x^{2} + 27 x - 90 = 0$
$\Rightarrow x^{2} + 30 x - 3 x - 90 = 0$
$\Rightarrow x (x + 30) - 3 (x + 30) = 0$
$\Rightarrow (x + 30) (x - 3) = 0$
$\Rightarrow x = - 30$ and $x = 3$
Negative value is not valid,
$∴$ Width is $3 c m$

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A footpath of uniform width runs all around the outside of a rectangular field 30 m long and 24 m wide. If the path occupies 360m2, find its width

A footpath of uniform width runs all around the outside of a rectangular field $30 m$ long and $24 m$ wide. If the path occupies $360 m^{2}$ , find its width