The length of a rectangle is 6 more than the width. The area is 720 square feet. Find the length and width of the rectangle.

Respuesta :

Given :-

  • The length of a rectangle is 6 more than the width

  • Area of rectangle = 720 feet²

[tex] \\ [/tex]

To find:-

  • Length
  • Width

[tex] \\ [/tex]

Let :-

  • Width of rectangle = x
  • Length of rectangle = 6 + x

[tex] \\ [/tex]

Solution:-

So to find length and width of rectangle, we have to find value of x.

[tex] \\ [/tex]

We know:-

[tex] \bigstar \boxed{ \rm Area~of~rectangle=Length\times{}Width}[/tex]

By using this formula we can find value of x.

[tex] \\ [/tex]

So:-

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area~of~rectangle=Length\times{}Width \\ [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf 720=(x + 6)(x)[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf 720=x(x) +x( 6)[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf 720= {x}^{2} +x( 6)[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf 720= {x}^{2} +6x[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf0 = {x}^{2} +6x - 720[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf {x}^{2} +6x - 720 = 0[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf {x}^{2} +30x - 24x - 720 = 0[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf ({x}^{2} +30x )- 24x - 720 = 0[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf x({x}+30 )- 24x - 720 = 0[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf x({x}+30 ) + (- 24x - 720) = 0 \\ [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf x({x}+30 ) - 24(x+ 30) = 0 \\ [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf (x - 24)({x}+30 )= 0 \\ [/tex]

[tex] \\ [/tex]

We got two solutions:-

first solution

  • x - 24 = 0
  • x = 24

Second solution :-

  • x + 30 = 0
  • x = -30

Length can't be taken negative so 24 is the value of x.

[tex] \\ [/tex]

Value of width:-

  • Width of rectangle = x
  • Width = 24 feet

[tex] \\ [/tex]

Value of Length:-

  • Length of rectangle = x + 6
  • Length = 24 + 6
  • Length = 30 feet

ACCESS MORE
EDU ACCESS
Universidad de Mexico