Question

A piece of wire 32cm Long is cut into two parts. Each part is bent to form a square. Given that the total area of the two squares is 34cm square, find the perimeter of each square

Answer 1

The possible perimeters for the squares are 20 cm and 12 cm, depending on how the wire is cut.

Let's denote the length of one part of the wire as x and the length of the other part as 32−x. Each part is bent to form a square.

The perimeter (P) of a square is given by 4 × side length.

The side length of the first square (
`S_1` ) is x/4, and the side length of the second square (
`S_2` ) is (32−x)/4.

The total area of the two squares is given as
`34cm^2` , so we can write the equation:

`S^2_1 + S^2_2 = 34`

Substitute the expressions for
`S_1 and S_2` :

`((x)/(4))^2 + (((32-x))/(4))^2 =34`

Now, solve for x:

`(x^2)/(16) + ((32-x)^2)/(4) =34`

Multiply both sides by 16 to get rid of the denominators:

`x^2 +(32-x)^2=544`

Expand and simplify:

`x^2+1024-64x+x^2 =544`

Combine like terms:

−64x+480=0

Divide the entire equation by 2 to simplify:

−32x+240=0

Now, factor the quadratic equation:

(x−20)(x−12)=0

So, x=20 or x=12.

If x=20, the lengths of the two parts are 20 cm and

32−20=12 cm.

If x=12, the lengths of the two parts are 12 cm and

32−12=20 cm.

Now, we can find the perimeter of each square by multiplying the side length by 4:

For x=20, the perimeter is
`4 * (20)/(4) = 20 cm`

For x=12, the perimeter is
`4 * (12)/(4) = 12 cm`

So, the possible perimeters for the squares are 20 cm and 12 cm, depending on how the wire is cut.

Answer 2

We can get two equations for the two unknowns by using the fact that the perimeters sum to the original length, and that the square's areas are the sum of the squares of the side lengths.