# The area of a square is 45 more square inches than its perimeter. What is the length of each side of the square?

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Let the side of a square be x.

Then its perimeter p = 4x.

It is given that area of the square is more than its perimeter by 45 numerically.

Therefore x ^2 = 4x+45.

Subtract 4x+45 from both sides .

x^2-4x-45 = 0.

x^2-9x+4x-45 = 0.

x(x-9)+5(x+9) = 0.

(x-9)(x+5) = 0.

So x= 9.

So the side of the square = 9 inches.

Tally: Area of the square = 9^2 = 81. Perimeter of the square = 4*9 = 36 inch.

So 81 = 36+45 .

Let x be one side of the square.

We'll write the formula for the area of the square:

A = x^2

We'll write the formula for the perimeter of the square:

P = 4x

Now, we'll write mathematically the condition imposed by enunciation:

x^2 = 4x + 45 (area is equal to the perimeter plus 45)

We'll subtract both sides 4x + 45:

x^2 - 4x - 45 = 4x + 45 - 4x - 45

We'll eliminate like terms:

x^2 - 4x - 45 = 0

We'll pply the quadratic formula:

x1 = [4+sqrt(16 + 180)]/2

x1 = (4+14)/2

x1 = 9

x2 = (4-14)/2

x2 = -5

Since the length of the side of the square cannot be negative, we'll reject the second root x2 = -5.

**The length of the side of the square is x = 9.**