Determine the area of an isosceles trapezoid in which the diagonals are mutually perpendicular and the height is h

Since, then the condition, the diagonals AC t BD are perpendicular, and the trapezoid is isosceles, then the triangles AOD and BOC are rectangular and isosceles, OA = OD, OB = OC.

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

Then, in a right-angled, isosceles triangle AOD, the height of OH is also its median, since it is drawn from the vertex of the right angle, then it is equal to half the length of the hypotenuse AD, OH = AD / 2.

Similarly, in the BOC triangle, OH = BC / 2.

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

Then Savsd = KH * KH = h ^ 2 cm2.

Answer: The area of ​​the trapezoid is h ^ 2 cm2.