The length of a rectangle is twice it's width plus one. If the area is 21cm², find the perimeter of the rectangle ?

Let the width of the rectangle be x cm.

Therefore, the length of the rectangle is 2x + 1 cm.

The area of the rectangle is given by:
Area = length x width
21 = (2x + 1) x

Expanding the equation:
21 = 2x^2 + x

Rearranging the equation:
2x^2 + x - 21 = 0

Solving the quadratic equation, we get:
x = 3 or x = -7/2 (width cannot be negative)

Therefore, the width of the rectangle is 3 cm and the length is 2(3) + 1 = 7 cm.

The perimeter of the rectangle is given by:
Perimeter = 2(length + width)
Perimeter = 2(7 + 3)
Perimeter = 2(10)
Perimeter = 20 cm

Therefore, the perimeter of the rectangle is 20 cm.

Let's assume the width of the rectangle is "w" cm.
According to the given information:
Length = 2w + 1 cm
Area = Length x Width
Substituting the values:
21 = (2w + 1) x w
21 = 2w² + w
2w² + w - 21 = 0
Solving the quadratic equation:
(2w - 7)(w + 3) = 0
Therefore, w = 7/2 cm or w = -3 cm (not valid as the width cannot be negative)
So, the width (w) is 7/2 cm.
Length = 2w + 1 = 2(7/2) + 1 = 8 cm
The perimeter of a rectangle is given by:
Perimeter = 2(Length + Width)
Perimeter = 2(8 + 7/2)
Perimeter = 2(23/2)
Perimeter = 23 cm
Therefore, the perimeter of the rectangle is 23 cm.