A piece of wire 44 cm long is cut into two parts and each part is bent to form a square. If the total area of the two squares is 65 sq cm, how do you find the perimeter of the two squares?

1 Answer
Aug 3, 2016

Answer:

Here's what I got.

Explanation:

You know that you're working with a piece of wire that is #"44 cm"# long. You then proceed to cut this piece of wire into two pieces. If you take #x# to be the first piece, you will have

#44 - x -># the second piece

Now, these pieces are used to form two squares. Since a square has four equal sides, the length of one side of the first square will be #x/4#.

Similarly, the length of one side of the second square will be #(44-x)/4#.

The area of a square is given by

#color(blue)(|bar(ul(color(white)(a/a)"area" = "side"^2color(white)(a/a)|)))#

In your case, the area of the first square will be

#A_1 = (x/4)^2 = x^2/16#

The are of the second square is

#A_2 = ((44-x)/4)^2 = (44 -x )^2/16#

The problem tells you that the total area of the square

#A_"total" = A_1 + A_2#

is equal to #"65 cm"^2#, which means that you have

#65 = x^2/16 + (44-x)^2/16#

This is equivalent to

#x^2 + (44-x)^2 = 65 * 16#

#x^2 + 44^2 - 88x + x^2 = 1040#

Rearrange to quadratic equation form

#2x^2 - 88x +896 = 0#

This quadratic has two solutions as given by the quadratic formula

#x_(1,2) = (-(-88) +- sqrt( (-88)^2 - 4 * 2 * 896))/(2 * 2)#

#x_(1,2) = (88 +- sqrt(576))/4#

#x_(1,2) = (88 +- 24)/4 implies {(x_1 = (88 + 24)/4 = 28), (x_2 = (88 - 24)/4 = 16) :}#

Here comes the cool part. You know that the sides of the two squares are

#"For the 1"^("st")" square: "28/4" "color(red)("or")" "16/4" "#

#"For the 2"^("nd")" square: "(44-28)/4 = 16/4" " color(red)("or")" "(44-16)/4 = 28/4 " "#

As you know the perimeter of a square is given by the equation

#color(blue)(|bar(ul(color(white)(a/a)"perimeter" = 4 xx "side" color(white)(a/a)|)))#

This means that the perimeters of the two squares are

#"For the 1"^("st") " square: "4 xx 28/4 = "28 cm " color(red)("or") " "4 xx 16/4 = "16 cm"#

#"For the 2"^("nd")" square: " 4 xx 16/4 = "16 cm " color(red)("or") " "4 xx 28/4 = "28 cm"#

This means that if the perimeter of the first square is #"28 cm"#, then the perimeter of the second square is #"16 cm"#.

Similarly, if if the perimeter of the first square is #"16 cm"#, then the perimeter of the second square is #"28 cm"#.