Arectangle has an area that is numerically twice its perimeter. if the length is twice the width, what are its dimensions

Width = 6 units

Length = 12 units  

Step-by-step explanation:

First thing first :)

The area of a rectangle is given by the formula:

where

is the area of the rectangle

is the width

is the length

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

where

is the perimeter

is the width

is the length

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

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

Subtract from both sides of the equation:

Factor :

Equate both factor to zero:

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

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

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

could u provide us some answer choices or hmm idk maybe a table for us to anser with ? ?