Find the area of an isosceles trapezoid, knowing that the diagonals are mutually

Find the area of an isosceles trapezoid, knowing that the diagonals are mutually perpendicular and their lengths are 16 cm.

The perpendicularity of the diagonals forms two similar right-angled isosceles triangles in an isosceles trapezium, in which the bases of the trapezoid are the hypotenuses.

Let the diagonals be divided by the point of intersection into segments m and (16 – m).

From the Pythagorean theorems, we obtain an expression for the sum of the bases of a trapezoid:

a + b = (2 * x ^ 2) ^ (1/2) + (2 * (16 – x) ^ 2) ^ (1/2) = x * 2 ^ (1/2) + (16 – x ) * 2 ^ (1/2) = 16 * 2 ^ (1/2);

The height in such a trapezoid is equal to half the sum of the bases:

h = 8 * 2 ^ (1/2);

The area is equal to:

S = 1/2 * 16 * 8 * 2 = 128 cm².