The length of a rectangle is 4 feet shorter than its width. the area of the rectangle is 42 square feet. find the length and width. round your answer to the nearest tenth of a foot

Respuesta :

Let the length of the rectangle to be a value, x.

Since you are told that the length is 4 feet shorter than width, then, this can only mean that 
width = (x+4)
The area of a rectangle = length × width. 
                                      = x(x+4) = 42
                                      =x² + 4x=42
 From here form the quadratic equation then solve it to get the value of x (length of the rectangle)

x²+4x-42=0
Using the quadratic formula,


x = -4 + or - (√(16-4×1×-42))÷2

At this point I had a problem with inserting an equation in this platform. So, i have done it have attached a file that explains the above statement. Kindly find it in the attachment below.  

x = (-4+or - 13.56)÷2
  = 9.56÷2
  =4.78 feet
The length to the nearest foot = 5 feet.
The width to the nearest foot = 5+4 = 9 feet.

Length is [tex]\boldsymbol{4.4}[/tex] feets and width is equal to [tex]\boldsymbol{8.8}[/tex] feets.

Area of a rectangle

The area of a two-dimensional region, form, or planar lamina in the plane is the quantity that expresses its extent.

A quadrilateral having four right angles is known as a rectangle.

Let [tex]\boldsymbol{w}[/tex] feets denotes width of a rectangle.

Length of a rectangle [tex]=\boldsymbol{w-4}[/tex] feets

Area of a rectangle [tex]=\boldsymbol{42}[/tex] square feet

Length [tex]\times[/tex] Width [tex]=42[/tex]

      [tex]w(w-4)=42[/tex]

[tex]w^2-4w-42=0[/tex]

[tex]w=\frac{4\pm \sqrt{16+168}}{2}[/tex]

   [tex]=2\pm \sqrt{46}[/tex]

As dimension can not be negative, [tex]2-\sqrt{46}[/tex] is rejected.

So,

[tex]w=2+ \sqrt{46}[/tex]

   [tex]=\boldsymbol{8.8}[/tex] feets

Length [tex]=8.8-4[/tex]

            [tex]=\boldsymbol{4.4}[/tex] feets

So, length is [tex]4.4[/tex] feets and width is equal to [tex]8.8[/tex] feets.

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