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): You can put this solution on YOUR website!


Conceptually....

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.


Answer by Theo(13342)   (Show Source): 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.

here's what the solution looks like graphically.











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