Question

a rectangle has an area that is numerically twice its perimeter. If the length is twice the width, what are its dimensions

Answer 1

Width = 6 units

Length = 12 units

Explanation:

First thing first :)

The area of a rectangle is given by the formula:

`A=wl`

where

`A` is the area of the rectangle

`w` is the width

`l` is the length

On the other hand, the perimeter of a rectangle is given by:

`P=2(w+l)`

where

`P` is the perimeter

`w` is the width

`l` is the length

We know from our problem that the area of our rectangle is twice its perimeter, so:

`A=2(2)P`

`A=4P`

`wl=4(w+l)`

We also know that its length is twice its width, so
`l=2w`. Let's replace that value in our previous equation and simplify:

`w(2w)=4(w+2w)`

`2w^2=4(3w)`

`2w^2=12w`

Subtract
`12w` from both sides of the equation:

`2w^2-12w=12w-12w`

`2w^2-12w=0`

Factor
`w`:

`w(2w-12)=0`

Equate both factor to zero:

`w=0, 2w-12=0`

A length can't be zero, so discard
`w=0` and solve for the other one:

`2w-12=0`

`2w=12`

`w=(12)/(2)`

`w=6`

Know that we have the width of our rectangle, we can find its length:

`l=2w`

`l=2(6)`

`l=12`

We can conclude that the width of our rectangle is 6 units and its length is 12 units.