Draw DE and CF perpendicular to AB.

Let AE = x and let BF = y, and DE = CF = h, and EF = CD = 30



Since AB = 75, x + 30 + y = 75
                    x + y = 45


By the Pythagorean theorem

 x² + h² = 39²
 y² + h² = 42²

 x² + h² = 1521
 y² + h² = 1764

Multiply the first equation by -1 and add the two equations:

-x² - h² = -1521
 y² + h² =  1764
----------------
 y² - x² = 243

Factorise:

(y - x)(y + x) = 243, and since x + y = 45, y + x = 45
   (y - x)(45) = 243
     45(y - x) = 243
        y - x = 
        y - x = 5.4

Add these two equations:

 y + x = 45
 y - x =  5.4
-------------
2y     = 50.4
     y = 25.2

Substitute in:

     y² + h² = 1764
(25.2)² + h² = 1764
 635.04 + h² = 1764
          h² = 1128.96
           h = 33.6

Formula for area of trapezium

A = (base1 + base2)(height)/2

A = (75 + 30)(33.6)/2

A = 1764 cm²  

Edwin