In an isosceles trapezoid, the angle between the diagonals is 90 °, the height of the trapezoid is 8 cm. Find the area.

Let’s draw the CM height through the point O, the point of intersection of the diagonals.

The BOC triangle is rectangular and isosceles since the diagonals of the isosceles trapezoid intersect at right angles, OB = OC.

We divide the height OK in half, the angle OBK in the triangle OBK is 45, then the triangle OBK is rectangular and equilateral, OK = BK = BC / 2.

Similarly, OM = AM = AD / 2.

Then KM = (OK + OM) = (BC + AD) / 2, which is the middle line of the trapezoid.

Then the area of the trapezoid is equal to: Savsd = KM ^ 2 = 8 ^ 2 = 64 cm2.