The area of a square is 45 more than the perimeter. How do you find the length of the side?

Jan 15, 2016

Length of one side is 9 units.
Rather than doing a straight factorising approach I have used the formula to demonstrate its use.

Explanation:

As it is a square the length of all the sides is the same.
Let the length of 1 side be L
Let the area be A

Then $A = {L}^{2}$............................(1)

Perimeter is $4 L$........................(2)

The question states: "The area of a square is 45 more than.."

$\implies A = 4 L + 45$.................................(3)

Substitute equation (3) into equation (1) giving:

$A = 4 L + 45 = {L}^{2.} \ldots \ldots \ldots \ldots \ldots \ldots . \left({1}_{a}\right)$

So now we are able to write just 1 equation with 1 unknown, which is solvable.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$4 L + 45 = {L}^{2}$

Subtract ${L}^{2}$ from both sides giving a quadratic.

$- {L}^{2} + 4 L + 45 = 0$
The conditions that satisfy this equation equalling zero gives us the potential size of L

Using $a x + b x + c = 0$ where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = - 1$
$b = 4$
$c = 45$

$x = \frac{- 4 \pm \sqrt{{\left(4\right)}^{2} - 4 \left(- 1\right) \left(45\right)}}{2 \left(- 1\right)}$

$x = \frac{- 4 \pm 14}{- 2}$

$x = \frac{- 18}{- 2} = + 9$

$x = \frac{+ 10}{- 2} = - 5$

Of these two $x = - 5$ is not a logical length of side so

$x = L = 9$

${\text{Check "-> A= 9^2= 81 "units}}^{2}$

$4 L = 36 \to 81 - 36 = 45$

So area does indeed equal sum of sides + 45