# Determine the side of a square if the area of the square is 45 more than the perimeter.

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### 3 Answers

We cannot compare the perimeter and the area as they have different dimensions. So we use only the magnitudes here.

Let the side of the square be A.

SO the perimeter is 4*A.

The area is A^2.

As the magnitude of the area is 45 more than that of the perimeter.

A^2 = 4A + 45

=> A^2 - 4A - 45 = 0

=> A^2 + 5A - 9A - 45 = 0

=> A ( A + 5) -9( A + 5) = 0

=> (A + 5)(A -9) = 0

A is either -5 or 9.

But as A is the side of the square it cannot be negative. So we consider only 9

**The side of the square is 9.**

Let the side of the square be x .

Then the perimeter of the square = 4x.

Therefore the area of the square = side*side = x^2.

It is given that area of the square = 45 more than the perimeter.

=> x^2= 45+4x.

=> x^2-4x-45 = 0.

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

=+ x-9 = 0, or x+5 = 0.

So x= 9, or x= -5.

So x= 9 is a practical solution.

So the side of the square is 9 units.

Tally: Area of the square x^2 = 9^2= 81.

Perimeter of the square = 4x = 4*9 = 36.

Area - perimeter = 81-36 = 45.

We'll note as x the side of the square.

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

A = x*x = x^2

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

P = x + x + x + x

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 apply 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.**