Question

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

Answer 1

Hint: In this question we are going to find the width of the footpath with the concept of the rectangle for that we will consider a variable for the width and then according to the question the area of the path will be the area of the rectangle inside the footpath subtracted from the total area of the rectangular field. We will relate the given area to the foundation area. Which will lead us to a quadratic equation, thereby solving the equation we get the required width.

Formula used:
Area of rectangle = \[length \times breadth\]
 Let the quadratic equation be
\[a{x^2} + bx + c = 0\] Then value of x will be given by,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step solution:

First, we will make the diagram of the rectangular field and footpath for better understanding,
Let the rectangle by ABCD where AB is the breadth of the rectangle and
And AD will be the length of the rectangle,
Let the width of the footpath be x m
Now we have to find the length and breadth of footpathinside rectangle.
Length of smaller rectangle will be= length of rectangle ABCD -2 times the
Width of the footpath (as footpath surrounds the smaller rectangle)
As length of rectangle ABCD = 50 m
And width of footpath is assume as x m
So, length of smaller rectangle will be = \[(50 - 2x)\]m
Similarly,
Breadth of smaller rectangle will be= Breadth of rectangle ABCD -2 times the
Width of the footpath (as footpath surrounds the smaller rectangle)
As Breadth of rectangle ABCD = 38 m
And width of footpath is assume as x m
So, Breadth of smaller rectangle will be = \[(38 - 2x)\]m
Now according to question,
Area of footpath = Area of rectangle ABCD – area of smaller rectangle ---------------- [1]
Area of rectangle = \[{\rm{length \times breadth}}\]
Putting the values of length and breadth in above equation that is 50m and 38 m we get,
Area of Rectangle ABCD = \[(50 \times 38){m^2}\]--------------- [2]
Area of smaller rectangle = \[(50 - 2x)(38 - 2x){m^2}\]------------------- [3]
Putting the values of [2] and [3] equation in [1] equation we get,
And the area of footpath is \[492{m^2}\]
We get,
\[492 = (50 \times 38) - (50 - 2x)(38 - 2x)\]
Solving above equation we get opening the brackets by multiplying we get,
\[492 = (50 \times 38 - (50 \times 38 - 100x - 76x + 4{x^2}))\]
Opening the bracket will make sign changes we get,
\[492 = (50 \times 38 - 50 \times 38 + 100x + 76x - 4{x^2})\]
Now subtracting similar terms and writing likes terms together we get,
\[4{x^2} - 100x - 76x + 492 = 0\]
Now, adding like terms we get,
\[4{x^2} - 176x + 492 = 0\]
Divide the above term by 4 we get,
\[{x^2} - 44x + 123 = 0\]
Now using the formula for solving quadratic equation, stated above
Here, a=1 and b=-44 and c =123
Putting the above values in formula,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[x = \dfrac{{44 \pm \sqrt {{{44}^2} - 4(1)(123)} }}{{2(1)}}\]
Now squaring 44 and subtracting the other term from it,
\[x = \dfrac{{44 \pm \sqrt {1936 - 492} }}{{2(1)}}\]
Subtracting 492 from 1936 we get,
\[x = \dfrac{{44 \pm \sqrt {1444} }}{{2(1)}}\]
Calculating the square root of 1444 we get,
\[x = \dfrac{{44 \pm 38}}{{2(1)}}\]
So there are two values one is \[x = 3\] \[x = \dfrac{{44 + 38}}{{2(1)}}\] and other is \[x = \dfrac{{44 - 38}}{{2(1)}}\]
Calculating both we get \[x = \dfrac{{82}}{2}\]or \[x = \dfrac{6}{2}\]
Dividing both by 2 we get,
\[x = 41\]m and \[x = 3\]m
Since, the first value is 41 which is not possible so the value of the width of footpath is 3m

$\therefore$ The value of the width of footpath is 3m.

Note:
Here while finding the length and breadth of the inner triangle we subtract the length and breadth of the outer triangle with 2 times the width of the footpath because the footpath is found in all sides of the rectangular field. That is the foot path is present in two sides of length and two sides of breadth hence the subtraction of width is done twice from the length and breadth.