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

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.

To construct a trapezium ABCD with given measurements and conditions, follow these steps:

1. Draw a straight line segment AB of length 8 cm. This will be the longer parallel side of the trapezium.

2. At point A, construct an angle with a measure of 70 degrees. To do this, use a protractor and draw an arc from point A to intersect the line AB.

3. Mark the intersection point as point D. Thus, you have AD as the shorter non-parallel side of the trapezium.

4. Draw a line segment CD of length 5 cm starting from point D. This will be the shorter parallel side of the trapezium.

5. Mark the point of intersection of line CD and line AB as point C.

Now, you have constructed a trapezium ABCD with AB = 8cm, CD = 5cm, angle DAC = 70 degrees, angle ABC = 50 degrees, and AB is parallel to CD.