SOLUTION: Solve the given problem. 1. The area of a rectangle is 48 square meters. Find its dimension if the shorter side is four less than the diagonal.

Let x and y be the sides of the rectangle.

Then from the description, we have these two equations


    xy = 48                (1)     (for the area)

and

    x^2 + y^2 = (x+4)^2    (2)      (for the diagonal)


From equation (2), we have

    x^2 + y^2 = x^2 + 8x + 16,

          y^2 =       8x + 16.    (3)


Now we square both sides of the equation (1)

    x^2 * y^2 = 48^2              (4)


and substitute expression (3) into equation (4), replacing y^2 there.  We get then

    x^2 * (8x + 16) = 48^2


Cancel common factor 8 in both side.  You get

    x^2 * (x + 2) = 48*6,

or

    x*2 * (x + 2) = 288.


It is easy to solve MENTALLY in your head by "trying and error" :  x = 6.


Indeed, 6^2 * (6+2) = 36*8 = 288.


So, the dimensions of the rectangle are 6 and  = 8 meters.


You may check it on your own that all conditions of the problem are satisfied.