# A playground, which measures 50m by 35m, is to be doubled in area by adding a strip of uniform width around the outside of the existing area. What is the width of the new strip around the playground?

Mar 9, 2018

$x = \frac{- 85 + 5 \sqrt{569}}{4} \approx 8.567$ metres to 3 decimal places

#### Explanation:

With problems of this type it becomes much clearer if you do a quick sketch.

The inner area is $35 m \times 50 m = 1750 {m}^{3}$

The outer strip area is
$4 {x}^{2} + \left(2 \times 35 x\right) + \left(2 \times 50 x\right) = 4 {x}^{2} + 170 x$

Combining the outer strip and the inner area doubles the inner area. So our model becomes:

$4 {x}^{2} + 170 x + 1750 = \left(2 \times 1750\right)$

$4 {x}^{2} + 170 x + 1750 = 3500$

Subtract 3500 from both sides

$4 {x}^{2} + 170 x - 1750 = 0$

Simplifying the numbers a bit. 4 will not divide exactly into 1750. However, all the numbers are even so we can at least halve them

$2 {x}^{2} + 85 x - 875 = 0$

If you can not spot the whole number factors quickly don't waste time in an exam continuing to try and determine them. Use the formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \textcolor{w h i t e}{\text{dd")->color(white)("dd}} \frac{- 85 \pm \sqrt{{85}^{2} - 4 \left(2\right) \left(- 875\right)}}{2 \left(2\right)}$

$x = \frac{- 85 \pm \sqrt{14225}}{4}$

$x = \frac{- 85 \pm \sqrt{{5}^{2} \times 569}}{4} \leftarrow \text{ Note that 569 is a prime number}$

$x = \frac{- 85 \pm 5 \sqrt{569}}{4} \leftarrow \text{ Exact values}$

The negative final value is not logical so is dismissed.

$x = \frac{- 85 + 5 \sqrt{569}}{4} \leftarrow \text{ Exact solution}$

$x \approx 8.567151 \ldots .$

$x = \frac{- 85 + 5 \sqrt{569}}{4} \approx 8.567$ to 3 decimal places