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❗️❗️❗️❗️PLS ❗️❗️The area of a rectangle is 35. If the length is 2 more than the width, find the dimensions of the rectangle. [Only an algebraic solution!!!!]​

Respuesta :

Given :

  • The area of a rectangle is 35. If the length is 2 more than the width.

To Find :

  • The dimensions of the rectangle.

Solution :

  • Let's assume the width of the rectangle as " x "
  • As, the length is 2 more than the width, So the Length will be "(x + 2)"

We know that,

  • [tex] \bf{ Length \times width = Area }[/tex]

So, Substituting the given values :

[tex]\qquad \sf \: { \dashrightarrow (x + 2) \times x = 35}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {x}^{2} + 2x = 35}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {x}^{2} + 2x - 35 = 0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {x}^{2} + 7x + 5x - 35 = 0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow {x}(x + 7) - 5(x + 7) = 0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow ({x}- 5)(x + 7) = 0}[/tex]

[tex]\qquad \sf \: { \dashrightarrow \: x =5 , x = - 7}[/tex]

Note :

  • The width can't be a negative number, so we can't choose x = -7, So we need to choose x = 5

Hence ,

  • Width = 5
  • Length = 5 + 2 = 7

Therefore ,

  • The dimensions of the rectangle are 5 and 7

Given

  • Area of rectangle = 35
  • Length = x+2
  • Width = x

To find

  • Dimensions of the rectangle = ?

Solution

  • [tex] \sf \purple \dashrightarrow \orange{area = length \times width}[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{35= (x+2)\times x}[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{35 = {x}^{2} + 2x}[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{35 - {x}^{2} - 2x = 0}[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{ - 35 + {x}^{2} + 2x = 0}[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{ {x}^{2} + 2x - 35 = 0 }[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{ {x}^{2} + 7x - 5x - 35 = 0 }[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{ x \times (x + 7) - 5(x + 7) = 0 }[/tex]
  • [tex] \sf \purple \dashrightarrow \orange{ x = - 7, x = 5 }[/tex]

Since, The dimension cannot be in negative we will consider the value of x as 5...

Now,

[tex] \tt \pink \multimap \blue{length = x + 2 = > 5 + 2 = > 7}[/tex]

[tex] \tt \pink \multimap \blue{width = x = > 5}[/tex]

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