Square problemThe 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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tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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 theĀ perimeter of the square:

PĀ = 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.

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