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

Answer:

Width = 6 units

Length = 12 units  

Step-by-step 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:

$2w-12=0$

$2w=12$

$w=\frac{12}{2}$

$w=6$

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.