# rectangle length 4 m more than width; area 45 m^2

#### jjrocks170

##### New member
the length of a rectangle is 4 meters more than the width. the area of the rectangle is 45 meters squared. find the length and width.

#### stapel

##### Super Moderator
Staff member
jjrocks170 said:
First, look up the formula for the area A of a rectangle having length L and width w.

Have you worked with variables at all?

Thank you!

Eliz.

#### Loren

##### Senior Member
>the length of a rectangle is 4 meters more than the width. the area of the rectangle is 45 meters squared. find the length and width.

You say "the area of the rectangle is 45 meters squared." This means the area is 45 m times 45 m = 2025 square meters. Or maybe you meant 45 square meters which can be represented as
$$\displaystyle 45m^2$$.
I hope you come to recognize the difference in the two expressions.
$$\displaystyle 45m^2$$ means 45 square meters.
$$\displaystyle (45m)^2$$ means 45 meters squared.

#### sgtpepper

##### Junior Member
Okay so you start with what you know. Hopefully you know that the formula for the area of a rectangle is:

A = L * W

where A is area, L is length, and W is width. You know A = 45m^2. Also, you know that (W + 4m) is equal to the length, L. So, rewriting your equation with the given values, you get:

45m^2 = (W + 4)*W

Now, all you have to do is solve for W. You can do this two ways with a calculator of some sort that solves for variables (e.g. a graphing calculator), or by using the quadratic formula:

Ax^2 + Bx + C = 0

[-B +or- sqrt(B^2 - (4*A*C))] / (2*A) = x

First you have to get our Area formula into the Ax^2 + Bx + C format (I'm dropping the units, m, for clarity's sake) :

45 = (W + 4)*W
0 = (W + 4)*W - 45
0 = W^2 + 4*W - 45

so now,

A = 1
B = 4
C = -45

plug these values into the quadratic formula

[-(4) +or- sqrt((4)^2 - (4*(1)*(-45)))] / (2*(1)) = W

you get two solutions for W (because of the +or- part of the quadratic formula) :

5m and -9m

http://tinyurl.com/yo6jpk

Obviously, the width cannot be a negative number, right? So, 5m must be the width. Therefore the length (W + 4) is 9m. Plug these values back into the area formula to make sure they work:

A = 5m * 9m = 45m^2

( m*m = m^2 )

It does work! Hope this helps somewhat. When looking at the quadratic formula remember that I switched x with W to fit our problem. Also A and B are just the coefficients of X^2, X, and the constant is C.