Question

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?

Answer 1

`"length " = 7.1 m" "` rounded to 1 decimal place
`"width" color(white)(..) = 2.1m" "` rounded to 1 decimal place

Explanation:

`color(blue)("Developing the equation")`

Let length be `L`
Let width be `w`

Let area be `a`

Then `a=Lxxw` ............................Equation(1)

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

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

`a=Lxxw" "->" "a=(w+5)xxw`

Written as: `a=w(w+5)`

We are told that `a=15m^2`

`=>15=w(w+5)....................Equation(1_a)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving for value of width")`

Multiply out the bracket

`15=w^2+5w`

Subtract 15 from both sides

`w^2+5w-15=0`

Not that `3xx5=15` However, `3+-5!=5`

So using the standardised formula:
`y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)`

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

`=>x=(-5+-sqrt((5)^2-4(1)(-15)))/(2(1))`

`=>x=-5/2+-sqrt(85)/2`

A negative value is not logical so we use

`x=-5/2+sqrt(85)/2" "=" "2.109772..`

`color(green)("The question instructs that are to use the nearest 10th")`

`"width "= x=2.1" "` rounded to 1 decimal place
`color(red)(" "uarr)`
`color(red)(" This comment is very important")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving for value of length")`

`a=Lxxw" "-> 15=Lxx2.109772..`

`=>L=15/2.109772.. = 7.1.9772..`

length`= 7.1` rounded to 1 decimal place