What is the area of an isosceles trapezoid, if the lengths of its bases are 16 cm and 30 cm and the diagonals are perpendicular to each other?

Call the point of intersection of the diagonals point X.

Each base is the hypotenuse of an isosceles right triangle whose sides are the diagonals and whose 90° angle is at X. The altitude of that triangle (⊥ distance to the base from X) is half the length of the hypotenuse. Then the height of the trapezoid is half the sum of the base lengths.

The area of the trapezoid is the product of the height and half the sum of the base lengths, hence is the square of half the sum of the base lengths.

... Area = ((16 cm +30 cm)/2)² = (23 cm)² = 529 cm²