How do I solve this using polynomials and how would I solve this without polynomials and how do I know what operations to use?

Question

The area of a rectangle is twice the area of a square. If the rectangle is 6 inches long, and the width of the rectangle is the same length of the side of the square. Find the dimensions of both rectangle and square?

Would the square automathically be side of 4.

Answer 1

Let "s" be the length of the side of the square. Then the area of the square is s²

Since "s" is the length of the side of the square, and it is also the side of the rectangle, then the area of the rectangle is (s)(6) or 6s.

Since the area of the rectangle is twice the area of the square then 6s = 2s²

6s = 2s²

2s² - 6s = 0

2s(s - 3) = 0

s = 0 and s = 3

The s = 0 is meaningless in this case. So s = 3.

The square is then 3 inches x 3 inches and the rectangle is 6 inches x 3 inches

You could also draw a picture and logically determine the side of the square must be half of 6:

If the area of the rectangle is twice that of the square then you should be able to fit two squares in the rectangle:

So the side must be 6 because it takes two to equal the length 6.

Answer 2

Start by assigning variables:

S = side of square and the short side of the rectangle

We are given:

area of rectangle = S*6

area of square = S²

and we are told that the rectangle area is 2 times the area of the square, so

S*6 = 2*S²

divide both sides by S:

6 = 2S

S = 3

CHECK: Area of square = 3² = 9

Area of rectangle = 3*6 = 18