construction of a trapezium ABCD in which AB=8cm, CD=5cm, angle DAC=70 degree, angle ABC=50 Degree and AB is Parallel to CD

Answer 1

The distance x between parallel lines AB and CD is given by

8 +d/cot70 - d/cot50 = 5

8 -0.475129d = 5
d = 3/0.475129 = 2.10469

You could also do this by construction by laying out line AB with length 8, laying out the directions of sides AD and BC from each end of AB, and finding the line parallel to AB with length CD=5, between the two angles lines AD and BC.