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
Here's what I got.
Explanation:
You know that you're working with a piece of wire that is
#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
Similarly, the length of one side of the second square will be
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 = 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
Similarly, if if the perimeter of the first square is
