Question

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

Answer 1

Let's find the width of the path.

Let the width of the footpath = $x$

The total area of rectangle ABCD = area of footpath + area of EFGH

$\begin{array}{rcl}50\times 38& =& 492+\left[\left(50-2x\right)\times (38-2x)\right]\\ 1900& =& 492+\left[1900-100x-176x+4x^2\right]\\ 4x^2-176x+492& =& 0\\ x^2-44x+123& =& 0\\ (x-3)(x-41)& =& 0\\ x& =& 3orx=41\end{array}$

$x=41m$ is not acceptable as the width cannot be greater than the breadth..

Therefore, the width of the path is $3m$.