Question

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

Answer 1

• Length of rectangular court = 31 ft

• Breadth of rectangular court = 10 ft

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 : -

`\boxed{ \sf{Perimeter \: of \: rectangle = 2(l+b ) }}`

Solution : -

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

`\longmapsto \: 2(l + b) = 82`

Step 1 : Substituting the value of length and breadth :

`\longmapsto \: 2(2x + 11 + x) = 82`

Step 2 : Transposing 2 to right hand side :

`\longmapsto \: 2x + 11 + x = \cancel(82)/(2)`

Step 3 : By cancelling 82 by 2 we get ;

`\longmapsto \: 2x + 11 + x = 41`

Step 4 : Adding values of left hand side :

`\longmapsto \: 3x + 11 = 41`

Step 5 : Transposing 11 to right hand side :

`\longmapsto \: 3x = 41 - 11`

Step 6 : By subtracting 11 from 41 we get :

`\longmapsto \: 3x = 30`

Step 7 : Transposing 3 to right hand side :

`\longmapsto \: x = \cancel (30)/(3)`

Step 8 : Cancelling 30 by 3 , We get :

`\longmapsto \: \bold{x = 10}`

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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Answer 2

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

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)

`~`

`\underline{ \pmb{ \frak{Substituting \: the \: required \: values: }}} \\ \\`

`\: \: \: \: \dashrightarrow \: \: \: \sf 82 = 2(x + 2x + 11) \\`

`\: \: \: \: \dashrightarrow \: \: \: \sf 82 = 2(3x + 11) \\`

`\: \: \: \: \dashrightarrow \: \: \: \sf 82 = 6x +22 \\`

`\: \: \: \: \dashrightarrow \: \: \: \sf 82 - 22 = 6x \\`

`\: \: \: \: \dashrightarrow \: \: \: \sf 60 = 6x \\`

`\: \: \: \: \dashrightarrow \: \: \: \sf (60)/(6) = x \\`

`\: \: \: \: {\purple{\dashrightarrow \: \: \: { \boxed{ \pmb{ \sf {10 = x}}}}}}`

Hence,

• Width of court = x = 10 feet

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