Find the area of an isosceles trapezoid whose diagonals are perpendicular and the sum of the bases is 20 cm.

Draw the height HK through the point of intersection of the diagonals.

Since the trapezoid is isosceles, its diagonals at point O are divided into equal segments.

ОВ = ОС, ОА = ОD, then triangles BОС and AOD are rectangular and isosceles.

OH and OK heights, medians and bisectors of triangles BOC and AOD, then in right-angled triangles BON and AOK the acute angles are 45, and then the triangles are right-angled and isosceles. OH = BH = BC / 2, OK = AK = AD / 2.

Then HK = BC / 2 + AD / 2 = (BC + AD) / 2.

By condition, (ВС + АD) = 20 cm, then КН = 20/2 = 10 cm.

Savsd = (ВС + АD) * КH / 2 = 20 * 10/2 = 100 cm2.

Answer: The area of ​​the trapezoid is 100 cm2.