The area of a rectangle is 27ft squared, and the length of the rectangle is 3ft less than twice the width. Find the dimensions of the rectangle.

Respuesta :

Answer:

width =4.5ft

length=6ft

Step-by-step explanation:

w = x

L = 2x-3

area = L*w

27 = x(2x-3)

[tex]27=2x^{2} -3x[/tex]

[tex]2x^{2} -3x-27=0[/tex]

[tex]2x^{2} +6x-9x-27=0[/tex]

2x(x+3)-9(x+3)=0

(2x-9)(x+3)=0

2x-9=0

2x=9

x=9/2

x=4.5

L=2x-3=2*4.5-3 =9-3=6

The dimensions of the rectangle are; width = 4.5ft and length = 6ft.

What is the area of the rectangle?

The area of the rectangle is the product of the length and width of a given rectangle.

The area of the rectangle = length × Width

The length of the rectangle is 3ft less than twice the width.

Let width = x

Length = 2x - 3

The area of the rectangle = length × Width

27 = x(2x-3)

2x² - 3x = 27

2x² - 3x - 27 = 0

2x(x + 3) - 9(x + 3) = 0

(2x-9)(x+3)=0

The solutions are;

2x-9=0

2x=9

x=9/2

x = 4.5

(x+3) = 0

x = -3

The width cannot be negative.

So,

Length = 2x-3 = 2*4.5-3 = 9 - 3

Length = 6

Hence, The dimensions of the rectangle are; width = 4.5ft and length = 6ft.

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