Question 89945


    The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (length and the width) of the rectangle

    Let the width of the rectangle be = w = x  cm
        the length of the rectangle be = l = (x+1) cm
        diagonal of the rectangle = d = 4 cm
                 therefore diagonal^2 = l^2+w^2   (use pythagoras theorem)

                                  4^2 = (x+1)^2+x^2

                                   16 = x^2+2x+1+x^2
                             2x^2+2x+1-16 = 0
                              2x^2+2x-15 = 0 

             compare with the std form  , a=2 ,b = 2  &c = -15
                    x = -2+-sq rt 2^2-4*2*-15/2*2
                      = -2+-sq rt 4+120/4
                      = -2+-sq rt 124/4

                      = 2(-1+-sq rt 31)/4
                      (-1+-sq rt 31)/2

          width of the rectangle = w = (-1+-sq rt 31)/2  cm
          length of th rectangle = l = 1+(-1+-sq rt 31)/2  cm