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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!!!!]​

User Llamarama
by
8.0k points

2 Answers

10 votes

Given

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

To find

  • Dimensions of the rectangle = ?

Solution


  • \sf \purple \dashrightarrow \orange{area = length * width}

  • \sf \purple \dashrightarrow \orange{35= (x+2)* x}

  • \sf \purple \dashrightarrow \orange{35 = {x}^(2) + 2x}

  • \sf \purple \dashrightarrow \orange{35 - {x}^(2) - 2x = 0}

  • \sf \purple \dashrightarrow \orange{ - 35 + {x}^(2) + 2x = 0}

  • \sf \purple \dashrightarrow \orange{ {x}^(2) + 2x - 35 = 0 }

  • \sf \purple \dashrightarrow \orange{ {x}^(2) + 7x - 5x - 35 = 0 }

  • \sf \purple \dashrightarrow \orange{ x * (x + 7) - 5(x + 7) = 0 }

  • \sf \purple \dashrightarrow \orange{ x = - 7, x = 5 }

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

Now,


\tt \pink \multimap \blue{length = x + 2 = > 5 + 2 = > 7}


\tt \pink \multimap \blue{width = x = > 5}

User Leopoldo
by
8.6k points
4 votes

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,


  • \bf{ Length * width = Area }

So, Substituting the given values :


\qquad \sf \: { \dashrightarrow (x + 2) * x = 35}


\qquad \sf \: { \dashrightarrow {x}^(2) + 2x = 35}


\qquad \sf \: { \dashrightarrow {x}^(2) + 2x - 35 = 0}


\qquad \sf \: { \dashrightarrow {x}^(2) + 7x + 5x - 35 = 0}


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


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


\qquad \sf \: { \dashrightarrow \: x =5 , x = - 7}

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
User Georgi Gerganov
by
8.0k points

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