Let A, B, C and D are the vertices of the trapezoid, so the trapezoid is ABCD.


Let AB be the longer base of 30 cm and CD be the other base of 16 cm long.


Let the point O be the intersection of diagonals.


Then the triangles  AOB,  BOC,  COD  and  AOD  are right angled triangles.



Since the trapezoid ABCD has equal lateral sides, the triangle AOB is isosceles right angled triangle.

Since it base AB is 30 cm long,  its sides (the legs)  AO and  BO  are   cm long.



Similarly, the triangle COD is isosceles right angled triangle.

Since it base CD is 16 cm long,  its sides (the legs) CO and  DO  are   cm long.



Now the areas of the triangles are


    for triangle  AOB:    = 225 cm^2;

    for triangle  COD:    =  64 cm^2;

    for triangle  BOC:    = 120 cm^2;

    for triangle  AOD:    = 120 cm^2.


The total area of the trapezoid is the sum of areas of triangles  225 + 64 + 120 + 120 = 529 cm^2.