The middle line of an isosceles trapezoid = 5 cm, and the diagonals are mutually perpendicular. find: area of the trapezoid.

Let’s draw the height PН through the point of intersection of the diagonals. Since in an isosceles trapezoid its diagonal, at the point of intersection, are divided at the bases into equal segments, then ОА = ОD, ОВ = OC. Then the triangles AOD and BOС are rectangular and isosceles. The heights of OH and OP are also the bisectors of the angle O, then the angle AOH = HAO = 45. Then OH = AH = AD / 2.

Similarly, in the BOC triangle, PO = BO = BC / 2.

Then PH = PO + HO = AD / 2 + BC / 2 = (AD + BC) / 2.

The height of the trapezoid is equal to the midline of the trapezoid, PH = KM = 5 cm, then:

Savsd = KM * PH = 5 * 5 = 25 cm2.

Answer: The area of the trapezoid is 25 cm2.