Question

The length of a rectangle is more than twice its width, and the area of the rectangle is 20. How do you find the dimension?

Answer 1

Length is 10
Width is 2

Explanation:

Let length be `L`
Let width be `W`
Let area be `A`

Given that `L>2W`

Let `L=2W+x`

`A=LxxW` .......................(1)

But `L=2W+x` so by substituting for `L` in equation (1)

`A=(2W+x)xxW`

`A=2W^2+xW`...............(2)

But area is given as `A=20`

Substitute for `A` in equation (2)

`20=2W^2+xW`

`=>2W^2+xW-20=0`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
as `x` is a variable we can find it by default

Set `color(brown)(" "2W^2+xW-20" ")color(blue)( ->" "(2W -4)(W+5))`

Multiply out the brackets to determine the value of `x`

`color(brown)(2W^2+xW-20)color(blue)( -> 2W^2+10W-4W-20)`

`color(brown)(2W^2+xW-20)color(blue)( -> 2W^2+6W-20)`

`color(green)("Thus "x=6" and "W=2" and "cancel(-5)`

However, `W=-5` is not logical so discard it.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(green)("Test "L=2W+x=2(2)+6=10)`

`color(red)(=> area ->LxxW ->10xx2 =20" as given")`