SOLUTION: A 44 cm piece of wire is cut and 2 squares are formed. The total are of both squares is 65 cm2. What’s the perimeter of each square?
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Question 1192931: A 44 cm piece of wire is cut and 2 squares are formed. The total are of both squares is 65 cm2. What’s the perimeter of each square? Found 2 solutions by greenestamps, Theo:Answer by greenestamps(13200) (Show Source):
To make the two squares, we need four pieces of one length and four pieces of another length. So cut the 44cm wire into 4 parts each of length 44/4=11cm; then divide each of those parts into two pieces.
(1) Solving the problem informally....
The total area of the two squares is 65cm^2, so we need two perfect squares whose sum is 65. There are two possible pairs.
65 = 64+1 = 8^2+1^2; but 8+1 is not 11.
65 = 49+16 = 7^2+4^2. 7+4=11, so the side lengths of the two squares are 4 and 7, which makes the perimeters 16 and 28.
(2) Algebraically....
We know the sum of the lengths of the sides of the two squares is 11cm, so
let x = side length of first square
then 11-x = side length of second square
The sum of the areas of the two squares is 65cm^2:
or
If we choose x=4 for the side length of the first square, then the side length of the second square is 11-4=7; if we choose 7 for the side length of the first square, then the side length of the second square is 11-7=4. Either way, we find the side lengths of the two squares are 4 and 7, making the perimeters 16 and 28.
You can put this solution on YOUR website! let x = the length of one side of one of the squares.
let y = the length of one side of the other square.
the area of both squares is given by the equation x^2 + y^2 = 65
the perimeter of both squares is given by the equation 4 * (x + y) = 44
in the second equation, solve for x + y to get:
x + y = 11
in that equation, solve for y to get:
y = 11 - x
in the first equation, replace y with (11 - x) to get:
x^2 + y^2 = 65 becomes x^2 + (11-x)^2 = 65
simplify to get:
2x^2 + 121 -22 * x = 65
order the terms in descending order of degree to get:
2x^2 - 22 * x + 121 = 65
subtract 65 from both sides of the equation to get:
2x^2 - 22x + 56 = 0
divide both sides of the equation by 2 to get:
x^2 - 11x + 28 = 0
factor to get:
(x - 4) * (x - 7) = 0
solve for x to get:
x = 4 or x = 7
when x = 4, x^2 + y^2 = 65 becomes 16 + y^2 = 65.
solve for y^2 to get y^2 = 65 - 16 = 49
solve for y to get y = 7
when x = 7, x^2 + y^2 = 65 becomes 49 + y^2 = 65
solve for y^2 to get y^2 = 65 - 49 = 16
solve for y to get y = 4
your solution is therefore:
x = 4 and y = 7 or x = 7 and y = 4
to confirm, replace x and y in the original equations to get:
x^2 + y^2 = 65 becomes 4^2 + 7^2 = 65 which becomes 16 + 49 = 65 which becomes 65 = 65, confirming the value of x and y are good in the first equation.
4x + 4y = 44 becomes 4*4 + 4*7 = 44 which becomes 16 + 28 = 44 which becomes 44 = 44, confirming the value of x and y are good in the second equation.
this works with x = 4 and y = 7.
it also works with x = 7 and y = 4.
the solution is that the perimeter of one of the squares is 16 and the perimeter of the other square is 28.
