<PRE><B>Question 90935</B>

The length of a is 4 centimeters more than the width. If the length is increased by 8 centimeters and the width is decreased by 4 centimeters, the area will remain unchanged. Find the orignal demensions of the rectangle




    Let the width of the rectangle be = x

  Then the length of the rectangle   = x+4  cm 

     Therefore the Area of the rectangle A = x.(x+4) sq cms

      Altered length of the rectangle = x+4+8 = x+12  cms

       Altered width of the rectangle = x-4 cms

      Area of the rectangle  = A = (x-4).(x+12)  sq cms

      The Areas remain unaltered 

              x.(x+4) = (x-4).(x+12)

               x^2+4x = x^2-4x+12x-48 
              x^2-x^2+4x+4x-12x+48 = 0

                    -12x+8x+48 = 0

                    48 = 4x

                    x = 48/4 = 12


        The original width = 12 cms 

        The original length  = 12+4 = 16 cms
    </PRE>