Let us draw the height from the top of the trapezoid ABCD to the base of AD.
Then in the formed right-angled triangle СНD the angle НСD = BCD – BCН = 135 – 90 = 45.
Then the angle НDС = НСD = 45, and the triangle СНD is isosceles, СН = DH.
Since the trapezoid is rectangular, ВA is parallel to CH, and therefore ABCH is a rectangle, then AB = CH = 6 cm, BC = AH = 6 cm.
CH = DH = 6 cm.
Then AD = AH + DH = 6 + 6 = 12 cm.
Then the area of the trapezoid is:
Savsd = (ВС + АD) * СН / 2 = (6 + 12) * 6/2 = 54 cm2.
Answer: Savsd = 54 cm2.
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