# The area of a rectangle is 44m^2, and the length of the rectangle is 3m less than the width?

The area of a rectangle is 44m^2, and the length of the rectangle is 3m less than the width.

The area of a rectangle is its length times width, and we're given a value for it, which is 44m^2. So we have our area:
A = l * w = 44m^2
It also says the length of the rectangle is 3m shorter than the width, so we would set the length equal to w - 3.
A = (w - 3)(w) = 44m^2
From here on out, we just have to solve for w, and plug in that value for width and length to get our side length values.
w^2 - 3w = 44
w^2 - 3w - 44 = 0
Use the quadratic formula, which is:
$x=frac{-bpmsqrt{b^2-4ac }}{2a}.$

"a" represents the coefficient for w^2, "b" represents the coefficient for w, and "c" represents the constant number.

x = {3 ± √(-3)^2 - 4(1)(-44)}/2(1)

= (3 ± √9 + 176)/2

= (3 ± √185)/2

So we can have two values: (3 + √185)/2 or (3 - √185)/2.

√185 is estimated to be somewhere between 19 and 20, and if we subtract 19 or 20 from 3, we will get a negative number. Values for length and width cannot be negative, because then how would a rectangle exist? So the second answer is crossed off, and we are left with (3 + √185)/2. This is the value for w.

Width = w = (3 + √185)/2

Length = w - 3 = [(3 + √185)/2] - 3

thanked the writer.