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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.
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