# The length of a rectangle is 5 m more than its width. If the area of the rectangle is 15 m2, what are the dimensions of the rectangle, to the nearest tenth of a metre?

Dec 16, 2016

$\text{length " = 7.1 m" }$ rounded to 1 decimal place
$\text{width" color(white)(..) = 2.1m" }$ rounded to 1 decimal place

#### Explanation:

$\textcolor{b l u e}{\text{Developing the equation}}$

Let length be $L$
Let width be $w$

Let area be $a$

Then $a = L \times w$ ............................Equation(1)

But in the question it states:
"The length of a rectangle is 5m more than its width"$\to L = w + 5$

So by substituting for $L$ in equation(1) we have:

$a = L \times w \text{ "->" } a = \left(w + 5\right) \times w$

Written as: $a = w \left(w + 5\right)$

We are told that $a = 15 {m}^{2}$

$\implies 15 = w \left(w + 5\right) \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left({1}_{a}\right)$
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$\textcolor{b l u e}{\text{Solving for value of width}}$

Multiply out the bracket

$15 = {w}^{2} + 5 w$

Subtract 15 from both sides

${w}^{2} + 5 w - 15 = 0$

Not that $3 \times 5 = 15$ However, $3 \pm 5 \ne 5$

So using the standardised formula:
$y = a {x}^{2} + b x + c \text{ where } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

a=1; b=5; c=-15

$\implies x = \frac{- 5 \pm \sqrt{{\left(5\right)}^{2} - 4 \left(1\right) \left(- 15\right)}}{2 \left(1\right)}$

$\implies x = - \frac{5}{2} \pm \frac{\sqrt{85}}{2}$

A negative value is not logical so we use

$x = - \frac{5}{2} + \frac{\sqrt{85}}{2} \text{ "=" } 2.109772 . .$

$\textcolor{g r e e n}{\text{The question instructs that are to use the nearest 10th}}$

$\text{width "= x=2.1" }$ rounded to 1 decimal place
$\textcolor{red}{\text{ } \uparrow}$
$\textcolor{red}{\text{ This comment is very important}}$
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$\textcolor{b l u e}{\text{Solving for value of length}}$

$a = L \times w \text{ } \to 15 = L \times 2.109772 . .$

$\implies L = \frac{15}{2.109772} . . = 7.1 .9772 . .$

length$= 7.1$ rounded to 1 decimal place