SOLUTION: a square and a rectangle have the same are. what is the numerical value of the rectangles perimeter? if the squares sides are x and the rectangle's length is 2x-3 and width is x-3.

Question 610249: a square and a rectangle have the same are. what is the numerical value of the rectangles perimeter? if the squares sides are x and the rectangle's length is 2x-3 and width is x-3.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
x^2 = (2x-3)(x-3)

x^2 = 2x^2 - 6x - 3x + 9

0 = 2x^2 - 6x - 3x + 9 - x^2

0 = x^2 - 9x + 9

x^2 - 9x + 9 = 0

Now use the quadratic formula to solve for x

x = (-b+-sqrt(b^2-4ac))/(2a)

x = (-(-9)+-sqrt((-9)^2-4(1)(9)))/(2(1))

x = (9+-sqrt(81-(36)))/(2)

x = (9+-sqrt(45))/2

x = (9+sqrt(45))/2 or x = (9-sqrt(45))/2

x = (9+3*sqrt(5))/2 or x = (9-3*sqrt(5))/2

x = 7.85410196624969 or x = 1.14589803375031

Since each dimension is positive, this means that the only solution is approximately x = 7.85410196624969

Perimeter of the rectangle
P = 2L + 2W

P = 2(2x-3) + 2(x-3)

P = 2(2*7.85410196624969-3) + 2(7.85410196624969-3)

P = 35.1246117974981

So the perimeter of the rectangle is approximately 35.1246117974981 units
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