Square problemThe area of a square is 45 more square inches than its perimeter. What is the length of each side of the square?
For a square with sides s, the perimeter is equal to 4*s and the area is s^2. In the problem, the area of the square is 45 more than its perimeter.
If the side of the square has length s, s^2 = 4*s + 45
s^2 - 4s - 45 = 0
s^2 - 9s + 5s - 45 = 0
s(s - 9) + 5(s - 9) = 0
(s + 5)(s - 9) = 0
s = -5 and s = 9
Eliminate the negative root as length cannot be negative.
The required length of the side of the square is 9 units.
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.