the length of a rectangular court is 11 ft longer than twice the width. the perimeter of the court is 82 ft. determine the length and width of the court

Respuesta :

Answer:

  • Length of court = 31 ft
  • Width of court = 10 ft

Step-by-step explanation:

Given,

  • Length of Rectangular court is 11 ft longer than twice it's Width. Perimeter of the rectangular court is 82 feet²

Let's Assume:

  • Width of rectangle = x feet
  • Length of rectangle = 2x + 11 feet

We know that ,

  • Perimeter of rectangle = 2(Length + Width)

[tex]~[/tex]

[tex]\underline{ \pmb{ \frak{Substituting \: the \: required \: values: }}} \\ \\ [/tex]

[tex] \: \: \: \: \dashrightarrow \: \: \: \sf 82 = 2(x + 2x + 11) \\ [/tex]

[tex] \: \: \: \: \dashrightarrow \: \: \: \sf 82 = 2(3x + 11) \\ [/tex]

[tex] \: \: \: \: \dashrightarrow \: \: \: \sf 82 = 6x +22 \\ [/tex]

[tex] \: \: \: \: \dashrightarrow \: \: \: \sf 82 - 22 = 6x \\ [/tex]

[tex] \: \: \: \: \dashrightarrow \: \: \: \sf 60 = 6x \\ [/tex]

[tex] \: \: \: \: \dashrightarrow \: \: \: \sf \dfrac{60}{6} = x \\ [/tex]

[tex]\: \: \: \: {\purple{\dashrightarrow \: \: \: { \boxed{ \pmb{ \sf {10 = x}}}}}}[/tex]

Hence,

  • Width of court = x = 10 feet

  • Length of court= 2x + 11 = 2(10) + 11 = 31 feet

Answer :

  • Length of rectangular court = 31 ft

  • Breadth of rectangular court = 10 ft

Step-by-step explanation:

Given : -

  • Perimeter of rectangular court is 82 ft.

  • Length = 11 ft longer than twice the width

To Assume : -

  • Let width be x ft

  • Let length be ( 2x + 11 ) ft ( Because in question it is given that length 11 more than twice it's breadth .

Formula Used : -

[tex] \boxed{ \sf{Perimeter \: of \: rectangle = 2(l+b ) }}[/tex]

Solution : -

We know the given perimeter of rectangle that is 82 ft. So equating it with the perimeter formula :

[tex] \longmapsto \: 2(l + b) = 82[/tex]

Step 1 : Substituting the value of length and breadth :

[tex] \longmapsto \: 2(2x + 11 + x) = 82[/tex]

Step 2 : Transposing 2 to right hand side :

[tex] \longmapsto \: 2x + 11 + x = \cancel\frac{82}{2} [/tex]

Step 3 : By cancelling 82 by 2 we get ;

[tex] \longmapsto \: 2x + 11 + x = 41[/tex]

Step 4 : Adding values of left hand side :

[tex] \longmapsto \: 3x + 11 = 41[/tex]

Step 5 : Transposing 11 to right hand side :

[tex] \longmapsto \: 3x = 41 - 11[/tex]

Step 6 : By subtracting 11 from 41 we get :

[tex] \longmapsto \: 3x = 30[/tex]

Step 7 : Transposing 3 to right hand side :

[tex] \longmapsto \: x = \cancel \frac{30}{3} [/tex]

Step 8 : Cancelling 30 by 3 , We get :

[tex] \longmapsto \: \bold{x = 10}[/tex]

Therefore values,

  • Width = x = 10 ft

  • Length = 2x + 11 = 2(10) + 11 = 31 ft

Verification : -

We can verify , our answer by substituting values of length and width that we have find above . If th left hand side and right hand side are equal then our answer is correct.

  • 2 ( l + b ) = 82

  • 2 ( 31 + 10 ) = 82

  • 2 ( 41 ) = 82

  • 82 = 82

  • L.H.S = R.H.S

  • Hence, Verified .

Therefore , our values for length and width are correct / valid .

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