Question

A wire that is 80 centimeters long is shown below. The wire is cut into two pieces, and each piece is bent and formed into the shape of a square. 80 cm wire Suppose that the side length (in centimeters) of one square is x cm. Express the perimeter of the big square in terms of x. (1 point)

Answer 1

Answer:

The perimeter of the second(big) square = $\mathbf{4 \times \begin {pmatrix} \dfrac{80-x}{4} \end {pmatrix}}$

Step-by-step explanation:

From the information given:

The length of the piece of the wire  = 80 cm

When being cut into two pieces;

the length of the first wire = x

Then the length of the second(big) wire will be = 80 - x

However, when each piece is bent to a square shape

The length of one side of the first square = $\dfrac{x}{4}$

The length of one side of the second(big) square; $\dfrac{80-x}{4}$

Area of a square = l² ( where l is the length of the sides)

For the first square

A₁ = $(\dfrac{x}{4})^2$

A₁ = $\dfrac{x^2}{16}$

The area of the second(big) square is:

A₂ = $(\dfrac{80-x}{4})^2$

A₂ = $(\dfrac{(80-x)^2}{4})$

The perimeter of the first square = $4 \times \dfrac{x}{4}$

The perimeter of the second(big) square = $\mathbf{4 \times \begin {pmatrix} \dfrac{80-x}{4} \end {pmatrix}}$