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?

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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?

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Answer 1 · 2018-03-09T14:42:19

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

Explanation:

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

Tony B

The inner area is $35 m \times 50 m = 1750 m^{3}$ $35 mxx50 m=1750 m^3$

The outer strip area is
$4 x^{2} + (2 \times 35 x) + (2 \times 50 x) = 4 x^{2} + 170 x$ $4x^2+(2xx35x)+(2xx50x)=4x^2+170x$

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

$4 x^{2} + 170 x + 1750 = (2 \times 1750)$ $4x^2+170x+1750 = (2xx1750)$

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

Subtract 3500 from both sides

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

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

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

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.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} dd \to dd \frac{- 85 \pm \sqrt{85^{2} - 4 (2) (- 875)}}{2 (2)}$ $x=(-b+-sqrt(b^2-4ac))/(2a)color(white)("dd")->color(white)("dd") (-85+-sqrt(85^2-4(2)(-875)))/(2(2))$

$x = \frac{- 85 \pm \sqrt{14225}}{4}$ $x= (-85+-sqrt(14225 ))/4$

$x = \frac{- 85 \pm \sqrt{5^{2} \times 569}}{4} \leftarrow Note that 569 is a prime number$ $x= (-85+-sqrt(5^2xx569 ))/4 larr" Note that 569 is a prime number"$

$x = \frac{- 85 \pm 5 \sqrt{569}}{4} \leftarrow Exact values$ $x=(-85+-5sqrt(569))/4 larr" Exact values"$

The negative final value is not logical so is dismissed.

$x = \frac{- 85 + 5 \sqrt{569}}{4} \leftarrow Exact solution$ $x=(-85+5sqrt(569))/4 larr" Exact solution"$

$x~~ 8.567151....$

$x=(-85+5sqrt(569))/4~~8.567$ to 3 decimal places

Tony B

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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?

1 Answer

Explanation:

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