Question 367117: A PIECE OF WIRE 44 CM LONG IS CUT INTO TWO PARTS AND EACH PART IS BENT TO FORM A SQUARE. AF THE TOTAL ARE OF THE TWO SQUARES IS 65 CM SQUARE, FIND THE PERIMETER OF THE TWO SQUARES
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the piece of wire is 44 cm long.
cut it into 2 parts.
the first part is x cm long
the second part is (44-x) cm long.
the perimeter of the first square will be x.
the perimeter of the second square will be (44-x).
each side of the first square will be x/4.
each side of the second square will be (44-x)/4.
the area of the first square becomes (x/4)^2.
simplify this to get x^2/16
the area of the second square becomes ((44-x)/4)^2
simplify this to get (44-x)^2/16
simplify further to get (1936 - 88x + x^2)/16
reorder the terms in the numerator to get (x^2 - 88x + 1936) / 16.
the sum of the areas of the 2 squares is equal to 65 cm.
this means that:
x^2/16 + (x^2 - 88x + 1936) / 16 = 65
multiply both sides of this equation by 16 to get:
x^2 + (x^2 - 88x + 1936) = 1040
simplify by combining like terms to get:
2x^2 - 88x + 1936 = 1040
divide both sides of this equation by 2 to get:
x^2 - 44x + 968 = 520
subtract 520 from both sides of this equation to get:
x^2 - 44x + 448 = 0
use the quadratic formula to solve this quadratic equation to get:
x = 28 or x = 16
either one of these values of x should satisfy the quadratic equation.
it does.
if x = 28, then the length of one of the pieces of wire is 28 and the length of the other piece of wire would have to be 44 - 28 = 16.
id x = 16, then the length of one of the pieces of wire is 16 and the length of the other piece of wire would have to be 44 - 16 = 28.
this confirms that the length of the pieces of wire are 16 and 28.
this means the the perimeter of the squares formed by the pieces of wire is also equal to 16 and 28 since the perimeter of the square of each piece of wire is the length of each piece of wire.
the sides of each of the squares formed by the pieces of wire are:
16/4 = 4 for one of the squares.
28 / 4 = 7 for the other of the squares.
the sum of the areas of the squares is equal to 4^2 + 7^2 which becomes 16 + 49 which becomes 65.
since the sum of the areas of the two squares is supposed to equal 65, then the lengths and the perimeters of the 2 square is correct.
your answer is that the perimeters of the two square is 16 for one of the squares and 28 for the other square.
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