1. ## Pre-algebra

One more and then I'm ready for tomorrow's test!!

The length of a rectangle is one unit more than its width. If the area is 30 sq units, find the dimensions of the rectangle.

2. Originally Posted by splits1
One more and then I'm ready for tomorrow's test!!

The length of a rectangle is one unit more than its width. If the area is 30 sq units, find the dimensions of the rectangle.

You should know how to calculate the area of a rectangle, yes? It's length * width.

let x = width of rectangle
x + 1 = length

length * width = 30

You can plug in the factors, do the multiplication, and solve for x

3. Originally Posted by splits1
One more and then I'm ready for tomorrow's test!!

The length of a rectangle is one unit more than its width. If the area is 30 sq units, find the dimensions of the rectangle.
I'll set you up, and then I'll leave it for you to finish.

Let $w$ and $l$ represent the width and length of the triangle respectively. We know that the length of rectangle is one more unit than its width:

$l=1+w$ -- (1)

and we're told the area of the rectangle is 30 sq units:

$lw=30$ -- (2)

Now substitute (1) into (2) to get $(1+w)w=30\implies w^2+w-30=0$

Can you continue and solve the quadratic equation?

(you'll see that one of the solutions to the quadratic equation won't work)