Question: A rectangular piece of land is 40m long and 25m wide. A path of uniform width and 426 m² area surround the land from outside. Find the width of the path.

Solution:

Let x m represent the uniform width of the path.

Given,
Length of the rectangular land (l) = 40m
Breadth of the rectangular land (b) = 25m

We know,
The uniform width surrounds the rectangle in length and breadth by 2 times. 
So,
The length of the whole rectangular land including the uniform width is (l+2x)m.
Similarly,
The breadth of the whole rectangular land including the uniform width is (b +2x)m.

The area of the outer path (A) is 426 m².
Using formula,
$A = 2x ( l+b +2x)$
$or, 426 = 2x (40+25+2x)$
$or, 426 = 2x (65 +2x)$
$or, 426 = 130x +4x²$
$or, 4x² +130x -426 = 0$ - (i)

Comparing equation (i) with ax²+bx+c = 0, we get,
a = 4, b = 130, c = -426

Using formula of quadratic equation,
$or, x = \frac{-b \pm \sqrt{b² -4ac}}{2a}$
$or, x = \frac{-130 \pm \sqrt{(-130)² - 4×4×(-426)}}{2×4}$
$or, x = \frac{-130 \pm \sqrt{16900 + 6816}}{8}$
$or, x = \frac{-130 \pm \sqrt{23716}}{8}$
$or, x = \frac{-130 \pm 154}{8}$

Taking positive sign,
$or, x = \frac{-130+154}{8}$
$or, x = \frac{24}{8}$
$\therefore x = 3$

Taking negative sign,
$or, x = \frac{-130 -154}{8}$
$or, x = \frac{-284}{8}$
$\therefore x = - \frac{71}{2}$

Since, distance is always positive, x = 3.

The required width of the uniform path is 3m.

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